Translations of
MATHEMATICAL
MONOGRAPHS Volume 168
Introduction to Complex Analysis Junjiro Noguchi
American Mathematical Society
Selected Titles in This Series 168 Junjiro Noguchi, Introduction to complex analysis, 1998 167 Masaya Yamaguti, Masayoshi Hate, and Jun Kigami, Mathematics of fractals, 1997 166 Kenji Ueno, An introduction to algebraic geometry, 1997
165 V. V. Iahkhanov, B. B. Lure, and D. K. Faddeev, The embedding problem in Galois theory, 1997
164 E. I. Gordon, Nonstandard methods in commutative harmonic analysis. 1997
163 A. Ya. Dorogovtsev, D. S. Siivestrov, A. V. Skorokhod, and M. I. Yadrenko, Probability theory: Collection of problems, 1997
162 M. V. Boldin, G. I. Slmonova, and Yu. N. Ty urin, Sign-based methods in linear statistical models, 1997 161 Michael Blank, Discreteness and continuity in problems of chaotic dynamics. 1997 160 V. G. OsmolovsklT, Linear and nonlinear perturbations of the operator div, 1997 159 S. Ya. Khavinson, Best approximation by linear superpositions (approodmate nomography), 1997
158 Hideki Omorl, Infinite-dimensional Lie groups, 1997 157 V. B. Kolmanovakil and L. E. Shalkhet, Control of systems with aftereffect, 1996 156 V. N. Shevchenko, Qualitative topics in integer linear programming, 1997 155 Yu. Safarov and D. Vassillev, The asymptotic distribution of eigenvalues of partial differential operators, 1997
154 V. V. Prasolov and A. B. Sossinsky, Knots, links, braids and 3-manifolds. An introduction to the new invariants in low-dimensional topology, 1997
153 S. Kh. Aranson, G. R. Belitsky, and E. V. Zhuzhoma, Introduction to the qualitative theory of dynamical systems on surfaces, 1996 152 R. S. Ismagilov, Representations of infinite-dimensional groups, 1996 151 S. Yu. Slavyanov, Asymptotic solutions of the one-dimensional Schri dinger equation, 1996
150 B. Ya. Levin, Lectures on entire functions, 1996 149 Takashi Sakai, Riemannian geometry, 1996 148 Vladimir I. Plterbarg, Asymptotic methods in the theory of Gaussian processes and fields, 1996
147 S. G. Gindikin and L. R. Volevich, Mixed problem for partial differential equations with quasihomogeneous principal part, 1996 146 L. Ya. Adrianova, Introduction to linear systems of differential equations, 1995 145 A. N. Andrianov and V. G. Zhuravlev, Modular forms and Hecke operators, 1995 144 O. V. Tl,oshidn, Nontraditional methods in mathematical hydrodynamics, 1995 143 V. A. Malyshev and R. A. Mintos, Linear infinite-particle operators, 1995 142 N. V. Krylov, Introduction to the theory of diffusion processes, 1995 141 A. A. Davydov, Qualitative theory of control systems, 1994
140 Aizik I. Volpert, Vitaly A. Volpert, and Vladimir A. Volpert, Traveling wave 139
solutions of parabolic systems, 1994 I. V. Skrypnik, Methods for analysis of nonlinear elliptic boundary value problems, 1994
138 Yu. P. Razmyslov, Identities of algebras and their representations, 1994 137 F. I. Karpelevich and A. Ya. Krelnin, Heavy traffic limits for multiphase queues, 1994 136 Masayoshl Mlyanlshi, Algebraic geometry, 1994 135 Masaru Takeuchi, Modern spherical functions, 1994 134 V. V. Prasolov, Problems and theorems in linear algebra, 1994 133 P. I. Naumkin and I. A. Shishmarev, Nonlinear nonlocal equations in the theory of waves, 1994
132 Hajime Urakawa, Calculus of variations and harmonic maps, 1993 131 V. V. Sharko, Functions on manifolds: Algebraic and topological aspects, 1993 130 V. V. Vershinin, Cobordisms and spectral sequences, 1993 (Continued in the back of this publication)
Introduction to Complex Analysis
Translations of
MATHEMATICAL MONOGRAPHS Volume 168
Introduction to Complex Analysis Junjiro Noguchi Translated by Junjiro Noguchi
American Mathematical Society Providence, Rhode Island
Editorial Board Shoshichi Kobayashi (Chair) Masamichi Takesaki
*f i
(Introduction to complex analysis) by Junjiro Noguchi Copyright © 1993 by Shokabo Publishing Company, Ltd. Originally published in Japanese by Shokabo Publishing Company, Ltd., Tokyo, 1993
Translated from the Japanese by Junjiro Noguchi 1991 Mathematics Subject Classification. Primary 30-01.
Library of Congress Cataloging-in-Publication Data Noguchi, Junjir6, 1948[Fltkuso kaiseki gairon. English[ Introduction to complex analysis / Junjiro Noguchi ; translated by Junjiro Noguchi. p. cm. - (Translations of mathematical monographs ; v. 168) Includes bibliographical references (p. - ) and index. ISBN 0-8218-0377-8 (alk. paper) 1. Functions of complex variables. 2. Mathematical analysis. I. Title. II. Series. QA331.7.N6413 1997 97-14392 515'.9-dc21 CIP
Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication (including abstracts) is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Assistant to the Publisher, American Mathematical Society, P. 0. Box 6248. Providence, Rhode Island 02940-6248. Requests can also
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10987654321
020100999897
TO AKIKO
Contents Preface
xi
Chapter 1. Complex Numbers 1.1. Complex Numbers 1.2. Plane Topology 1.3. Sequences and Limits Problems
1 1
3 9 15
Chapter 2. Complex Functions 2.1. Complex Functions 2.2. Sequences of Complex Functions 2.3. Series of Functions 2.4. Power Series 2.5. Exponential Functions and Trigonometric Functions 2.6. Infinite Products 2.7. Riemann Sphere 2.8. Linear Transformations Problems
Chapter 3. Holomorphic Functions 3.1. Complex Derivatives 3.2. Curvilinear Integrals 3.3. Homotopy of Curves 3.4. Cauchy's Integral Theorem 3.5. Cauchy's Integral Formula 3.6. Mean Value Theorem and Harmonic Functions 3.7. Holomorphic Functions on the Riemann Sphere Problems Chapter 4. Residue Theorem 4.1. Laurent Series 4.2. Meromorphic Functions and Residue Theorem 4.3. Argument Principle ix
91 91
94
99
CONTENTS
4.4. Residue Calculus Problems
106 114
Chapter 5. Analytic Continuation 5.1. Analytic Continuation 5.2. Monodromy Theorem 5.3. Universal Covering and Riemann Surface Problems
117
Chapter 6. Holomorphic Mappings 6.1. Linear Transformations 6.2. Poincare Metric 6.3. Contraction Principle 6.4. The Riemann Mapping Theorem 6.5. Boundary Correspondence 6.6. Universal Covering of C \ {0, 1} 6.7. The Little Picard Theorem
141
117 124
130 138
141
144 149 153 157 163
167
6.8. The Big Picard Theorem
170
Problems
174
Chapter 7. Meromorphic Functions 7.1. Approximation Theorem 7.2. Existence Theorems 7.3. Riemann-Stieltjes' Integral 7.4. Meromorphic Functions on C 7.5. Weierstrass' Product 7.6. Elliptic Functions Problems
179 179 185 192 194
202 210 226
Hints and Answers
229
References
241
Index
245
Symbols
249
Correction List
251
Preface
Complex analysis is an active research subject by itself, but, even more, it provides the foundations for broad areas of mathematics, and plays an important role in the applications of mathematics to engineering. This book is intended to describe a classical introductory part of complex analysis for university students in the sciences or engineering, and in particular to serve as a text or a reference book for juniors, seniors, or first-year graduate students in the sciences. The prerequisites are elementary calculus, from the real numbers through differentiation and integration, and linear algebra; set theory and some general topology are not required, but would be helpful. Historically, the contents of this volume had been discovered by the beginning of the twentieth
century, so they are not very new. Nevertheless, the author tried to arrange the presentation and choose terminology to agree with modern work in the field. Many books have been written on this subject, in many languages. Roughly speaking, they can be divided into two types. One tries to give an understanding
of the subject using intuitive arguments. The other puts more emphasis on rigorous proofs, presenting the subject as a fundamental theory of mathematics. This book falls in the latter group. It is well-known that there is some difficulty in dealing with curves, related to Cauchy's integral theorem, a point on which several published books are somewhat unsatisfactory. To deal with it rigorously, we give detailed descriptions of the homotopy of plane curves. This material on curves is something the reader may well have encountered in previous courses, so it is scattered through the text where needed. Readers whose mathematical background already includes this material may simply confirm the theorems and go on. Since residue theorem is important both in pure mathematics and in applications, we give a fairly detailed explanation of how to apply it to numerical calculations; this should be sufficient for those who are studying complex analysis for applications.
After this book, the student should be ready to take up value distribution theory, the theory of R.iemann surfaces, complex analysis in several variables, the theory of complex manifolds, and other subjects. Complex analysis will also xi
PREFACE
xii
provide fundamental methods for the theory of differential equations, algebraic geometry, number theory, and other fields. The author wrote Chapters 6 and 7 with these transitions in mind.
The author will be very pleased if this book helps students to understand the classical theory of complex analysis, to relish its beauty, and to master the rigorous treatment of mathematical demonstrations. This volume is based on lectures for third-year students given by the author at the Tokyo Institute of Technology. The class contained not just mathematics majors, but also physics and applied physics students. The author is very grateful to all of them. Finally, but not least, the author would like to express his deep thanks to Professor Mitsuru Ozawa, who led him to complex analysis, and to Professor Shingo Murakami, who recommended him to write this book. He is also obliged to Mr. Shuji Hosoki, of the Shokabo publishing company, for the proofreading. April, 1993
at Ohokayama Junjiro Noguchi
Added in English translation: The publication of this English translation was made possible by the sug-
gestion and the recommendation of Professor Katsumi Nomizu. The author expresses his sincere gratitude to him. February, 1997
at Ohokayama Junjiro Noguchi
CHAPTER 1
Complex Numbers
In this chapter we assume that the real numbers are known, and we explain the elementary facts of complex numbers.
1.1. Complex Numbers The set R of all real numbers cannot be enlarged as long as its order " <_ " and the Archimedean axiom are kept (completeness of the system of real numbers).
The order implies that x2 > 0 for all x E R. Therefore the following simple algebraic equation x2 + a = 0
cannot be solved in the system of real numbers for any a > 0. To overcome this inconvenience, we drop the order of the number system, and extend R by taking a solution i of equation (1.1.1) normalized with a = 1, so that we obtain a complex number z = x+iy (= x+yi) (x, y E R). We denote by C the set of all complex numbers. This extension makes it possible not only to solve equation (1.1.1), but also to develop a grand and beautiful theory that cultivates the essence of mathematics. The number i is called the imaginary unit and satisfies i2 + 1 = 0.
(1.1.2)
Then -i is also a solution of (1.1.2), but it does not matter which is the imaginary unit once we choose one solution of (1.1.2). Now we define z = 0 if and only if
x=y=0. For two complex numbers z = x + iy, w = u + iv we define z = w if and only if x = u and y = v. The following operations are defined in C:
z±w
(x±u)+i(y±v),
zw
= (xu - yv) + i(xv + yu).
If z 54 0, we set z-1 = x/(x2 + y2) + i(-y)/(x2 + y2). Then
zz-1 = 1. We also write 11z for z-1, and call it the inverse of z. Thus C carries the four operations of addition, subtraction, multiplication, and division. This fact is 1
1. COMPLEX NUMBERS
2
referred to by saying C forms a field. We may identify z = x + iy E C with the point (x, y) E R2, so that we may also consider C as a 2-dimensional vector space over R. This vector space is called the complex plane, or Gaussian plane, and z is called the complex coordinate. In this complex plane, the x (resp., y) coordinate axis is called the real (rasp., imaginary) axis.
FIGURE 1
A point z = x + iy E C of the complex plane is called a rational point (not rational number!) if x E Q and y E Q. In general, for z = x + iy we set
z=x - iy, which is called the conjugate of z. We have (1.1.3)
z+w=-z+w, zw=zw,
It also follows that
x= I(z+z), y= 2
2i
1(z-z),
which are respectively called the real and imaginary part of z, and denoted by Re z and Im z. If Im z = 0, 2 is called a real number, and if Re z = 0, z is called a purely imaginary number: if and only if z= zisreal z is purely imaginary if and only if z = -z. Note that zz = x2 + y2 ? 0. The non-negative square root of zz is called the absolute value of z, and is denoted by zJ:
Iz,=
x+y2j 0.
It follows that (1.1.4)
1 - zI = IzI,
IzwI = IzI IWI,
Iz±w4
IzI=04=t. z=0.
1.2. PLANE TOPOLOGY
3
For z, w E C we set d(z, w) = Iz - w l; this is called the distance between z and w, and satisfies
d(z,w)=0e--* z=w, d(z, w) = d(w, z), d(zl, z2) + d(z2, z3) >
By making use of the trigonometric functions, cos0 and sin 8, we may write a complex number z = x + iy as follows: (1.1.6)
z =x+iy = r(cos 0 + i sin 0),
0ER,
r=IzI>0.
If z # 0, 8 is uniquely determined up to integral multiples of 21r (that is, 0 is uniquely determined "modulo 2a"). We call 8 the argument of z, and denote it by arg z. The pair (r, 8) is called the polar coordinate system, and (1.1.6) the polar coordinate representation of z. II,
---------------
r sin B
z
r
8
0
FIGURE 2
REMARK. Here we dealt with the trigonometric functions cos8 and sin0 by geometric intuition. This is not consistent with our aim of presenting the theory rigorously. We have to define the trigonometric functions without geometric intuition, and show that they are periodic and satisfy the above stated geometric intuition. This will be done in Chapter II, §5.
1.2. Plane Topology For a point a E C of the complex plane and a real number r > 0, we define the disk of radius r with center a by
0(a; r) = {z E C; d(z, a) < r} In particular, when a = 0, we set
A(r) = 0(0; r) and A(1) is called the unit disk
({Iz - al < r}).
1. COMPLEX NUMBERS
4
A subset A C C of C is said to be bounded if there is r > 0 with A C A(r). The set A is said to be open if for any a E A there is
ar>0with A(a;r)CA. An open set containing a point z E C (resp., a subset B C C) is called a neighborhood of z (resp., B). The disk
A(a;r) itself is an open set, since
A(z;r - Iz - al) C A(a;r) for all z E A(a; r). Hence, A(a; r) is called a disk neighborhood of a. We denote the set of all open subsets of C by A. Then we have
FIGURE 3
(1.2.1) i) 0 E A, C E A; ii) for finitely many A1 ,
, An E A, Al n n A E A; iii) for an arbitrary subfamily {AQ }aer C A, UQEr A. E A.
In general, "defining a topology' means that a family of subsets which satisfies
the properties (1.2.1), i)-iii) is given. A subset A C C is called a closed set if the complement A` is open, and a point z E C an accumulation point of A if for
anyr>0 (0(z; r) \ {z}) n A 0 0. A point of A which is not an accumulation point of A is called an isolated point
of A. The set consisting of all points of A and all accumulation points of A is called the closure of A and is denoted by A. The naming comes from the following theorem.
(1.2.2) THEOREM. i) A subset A C C is closed if and only if A = A.
ii) A=A. iii) A is the smallest set among the closed subsets containing A.
PROOF. i) Assume that A is closed. Let a E A be an accumulation point A, then a E A`, so that .(a; r) C A' for some r > 0; that is, of A. If a
A(a; r) n A = 0. This is absurd. Thus A = A.
To show the converse, we take an arbitrary z ¢ A. By assumption, z is not an accumulation point of A, either. Therefore, there is r > 0 such that A(z; r) n A = 0; that is, A(z; r) C A`. This shows that A' is open. ii) It suffices to show that A C A. Take an arbitrary accumulation point z of A. Since for any r > 0
A(z;r)nA96 0, we may take z1 E 0(z; r) n A. Set r1 = r - Iz1 - zj. Since z1 E A, we have
A(z1;r1)nA#0.
1.2. PLANE TOPOLOGY
5
Hence, A(z; r) n A 54 0, and then z E A.
.4
FIGURE 4
iii) Taking an arbitrary closed set E with E D A, we have to show that E D A. This is easy and is left to the reader.
For a subset A C C, a point z E A is called an interior point of A if there is a neighborhood U of z with U C A. Therefore, the set of all interior points of A, denoted by A, is open. An interior point of A` is called an exterior point of a A. The set 8A = A \ A is called the boundary of A, and a point of it a boundary point. In particular, 8A is a closed set. O
We set
C(a;r) = 80(a; r) = {Iz - al = r}, which is called the circle of radius r with center a; in particular, C(0;1) is called the unit circle. Let B be a subset of A. Then B is called an open set in A if there is an open set U C C with B = A n U. Similarly, B is called a closed set in A if there is a closed set E C C with B = A n E. In this way we get the relative topology of A. If B is an open (resp., closed) subset of A, then A \ B is a closed (resp., open) subset of A. If B has no accumulation point in A, B is said to be discrete in A. In general, a family {Ua}aEr of subsets Ua C C is called a covering of A if A C UaEr U, . In particular, if all Ua are open, the covering is called an open covering of A. In this case, {UQ n A}QEr is called an open covering of A (with respect to the relative topology), too. If there is an open covering {U1, U2} of A satisfying (1.2.3)
U1nA540,
U2nA340, UInU2nA=O,
we say that A is not connected. If there is no such covering, then A is said to be connected. A connected open subset of C is called a domain. (1.2.4) THEOREM. The following three conditions are mutually equivalent. i) A is connected.
1. COMPLEX NUMBERS
6
ii) Let {U1, U2} be any open covering of A with respect to the relative topol-
ogy, such that U1 fl U2 = 0. Then either U1 = A (U2 = 0), or U2 = A (U1 = 0).
iii) If any non-empty subset B of A is open and closed in A, then B = A. PROOF. i) -4:* ii) This is just the definition. ii) iii)
. iii) Put U1 = B and U2 = A \ B. . ii) Let the non-empty U; be B.
Let I be a closed interval of R. Then a continuous mapping 0 : 1 3 t
¢(t) _
x(t) + iy(t) E C (i.e., x(t) and y(t) are real-valued continuous functions on I) is called a curve or arc. In the case where 1 is a general interval (open interval, semi-closed interval, etc.), it is called a general curve. The image C = ¢(I) is also called a curve, but, strictly speaking, 0 should be called a curve. We say that "C is given by 0". In fact, the two curves 01
[0, 2a] 3 t - cost + i sin t E C,
dpi : [0,2w] 3 t -+ cos 2t + i sin 2t E C,
have the same image 01([0, 27r]) = 02((0,2r]). But 01 goes around the unit circle once, and 02 twice, so they should be distinguished. The variable t is called the parameter of 0.
FIGURE 5
A curve
: J -+ C is called a parameter change of 0 if there is a strictly mono-
tone continuous function r : J -F I from J onto I satisfying O(s) = 0(r(s)), s E
J. If I = [To, TI] and 0(To) = 0(T1), then 0 is called a closed curve. In this case, we consider also the following ip as a parameter change. Let J = [So, S1] (So < S1) be an arbitrary closed interval, and take -ro E (To, T1) and ao E (So, S1). We put (1.2.5)
tb(s)=0(ro+ s-So (T1-ro)), \\
VI (
s)
= 0 TO +
So:s
Qo,
ao - So s - ao
Si - ao
(ro
- To
)
,
Qo < s < S 1.
1.2. PLANE TOPOLOGY
7
Furthermore, a parameter change of this Tp is also considered as a parameter change of .0. 0(r.)=d(TJ
1
FIGURE 6
J
FIGURE 7
It is clear that parameter changes of curves give rise to an equivalence relation. In our arguments, it is inconvenient to consider a curve as a continuous mapping. Therefore, we call the equivalence class of 0 a curve, and denote it by C, C((b),
or C(O: I -- C). Here 0 stands for the representative. Still, we may call 0 and the image 0(1) a curve, but the meaning will be clear. ExERCISE 1. Let 0 : [To, T1 ] C be a curve, and [So, S1 ] (So < S1) an arbitrary closed interval. Then there is a parameter change : J -' C of 0 with
J=[So,S11. Let C(0 : [To,T1] -+ C) be a curve. Then O(To) is called the initial point, and 0(T1) the terminal point. The initial and terminal points of C are called the ends of C, and we say that C connects ¢(To) and 4)(T1). If 0 is one to one, C is called a simple curve (or a Jordan curve). If C is closed and 0 is one to one on [To, TI), C is called a simple (or Jordan) closed curve. When C is a closed curve, an arbitrary point of C can be the initial (terminal) point by a parameter change. Thus, in this case we consider that the initial (terminal) point of C is O(TO)), determined if a representative ¢ of C is fixed. If 0 is constant (0(t) then ¢ is called a constant curve. We say that 4) : [To, T1 ] - C is continuously differentiable if the real part x(t) of 4)(t) and the imaginary part y(t) of 0(t) are defined in an open interval containing [To, T1 ], are differentiable there, and their derivatives are continuous. We set T(t) _ 4)'(t) = dt (t) + i y (t). Here, in case 0(To) = O(T1), we also assume that do(To)/dt = d¢(T1)/dt. Moreover, 0 is said to be non-singular if dO(t)/dt # 0 for all t. If there is a partition To = to < t 1 < ... < t, = T1 such that the restrictions 01 [tt _ I, t, ] to [ti _ 1, t;], 1 < j < n, are continuously differentiable, ¢ is said to be piecewise continuously differentiable. A curve C is (piecewise) continuously differentiable if there is such a representative 0 of C. A non-singular curve and a piecewise non-singular curve are similarly defined.
1. COMPLEX NUMBERS
8
C) be a curve. On the image of C, 0(t) moves Now, let C(4 : [To,TI] from the initial point toward the terminal point as the parameter t E [To,T1) increases. Assume that C is not a constant curve. By the above Exercise 1 we may assume [To,T1] = [-1,1]. Put 0(t) = 0(-t). Then ip : [-1,1] -. C is not a parameter change of 0. If 0(-1) iO 0(1), then 0(t) moves from 0(-1) toward '(1) as t moves from -1 toward 1, and ?P(t) from 0(1) toward 0(-1) Therefore we may consider that the curve C(¢) already has an orientation in this sense. .
The curve given by >(i is called the curve with inverse orientation and is denoted
by-C.
o{-))=v())
IX 1)
FIGURE 8
Now, suppose that the terminal point of a curve C1 is the same as the initial point of another curve C2. Suppose that C1 is given by ¢I : [0,1] -. C and C2 by 02 : [l, 2} -. C. Then we define a curve Cs(03 :10,21 -. C) so that it is equal to 01 over [0,1] and to 02 over [1, 2]. The curve C3 is called the sum of C1 and C2, and we write C3 = C1 + C2.
For a subset A C C, C is called a curve in A if its image is contained in A. If two arbitrary points of A can be connected by a curve in A, A is said to be arcwise connected For instance, a disk i(a; r) and C itself are arcwise connected.
Ex itcISE 2. A closed interval [To, TI] (C R C C) is connected in the sense of Theorem (1.2.4). (1.2.6) THEOREM. An open subset of C is connected if and only if it is arcwise connected.
PROOF. The "if' part follows from Exercise 2. To show the "only if' part, we assume that A is a connected open set and A # 0. Fix a point zo E A. Let U1 be the totality of all points of A connected to z1 by curves in A, and let U2 be its complement. For any point a E Uj U = 1, 2), we take r > 0 with A(a; r) C A. Since 0(a; r) is arcwise connected, A(a; r) C Up Thus the U, are both open sets. Since A = U1 U U2 and U1 # 0, we have U2 = 0 by definition,
andsoA=UI. O
1.3. SEQUENCES AND LIMITS
9
A curve 0 : [To, T1 ] -+ C is called a piecewise linear curve (Streclcenzug) if < t = T1 of [To,T1] such that on each there is a partition To = to < t1 < [tj -I , tj ]
is written as
0(t) = 0(t3-1) + aj(t - tj-1),
ti-1 5 t 5 tj,
where a, E C. We immediately obtain from the proof of Theorem (1.2.6)
(1.2.7) COROLLARY. An open subset A of C is connected if and only if two arbitrary points of A can be connected by a piecewise linear curve. In general, let A be a subset of C. For two points z1, z2 E A we write z1 - z2
if there is a connected subset A' C A with z1, z2 E A'. We show that this is an equivalence relation. The reflexive law, "z1 - z2 . z2 - z1", is trivial. Assume that z1 - z2 and z2 - z3. Take connected subsets A', A" C A such that x1, z2 E A' and z2, z3 E A". It suffices to show that A' U A" is connected. Let {U1i U2} be an open covering of A' U A" with respect to the relative topology such that U1 fl U2 = 0. Since {U1, U2} is also an open covering of A', A' C U1 or A' C U2; we may assume A' C U1. In the same way, we see that A" C U1 or A" C U2. Since Z2 E A' fl A" C U1, A" C U1. Therefore, A' U A" C U1, and so its connectedness follows. For z E A we set A2 = {z' E A; z' - z}; this is called the connected component of A containing z. If A.., 34 Ax then A., f1 A., = 0. Taking representative elements za from all connected components of A, we have (1.2.8)
A = UA2,. Z.
This is called the decomposition of A into connected components. In particular, if A is open, then so are the connected components of A, so that their represen-
tatives za = xa + iya may be chosen as rational points. Since there are only countably many such points, we may enumerate them, and hence denote them by zi, i = I,- , N(< oo). Put A; = Az,, i = 1,... , N. Then they are disjoint domains and N
(1.2.9)
A=UAi. i=1
1.3. Sequences and Limits From now on, a complex number is simply called a number. { z } no of numbers (or points) is An infinite sequence (zo, z1, ... , z,, , ... called a sequence. A sequence {z} n o is said to converge to a E C or to have limit a if for an arbitrary e > 0 there is a number np E N such that
Iz - al<e
for all n>no.
1. COMPLEX NUMBERS
10
In this case we write
a = lim n--xzn or zn -+ a (n
(1.3.1)
oo).
Since (1.3.2)
max{IRezl,lImzl} < Izl S IRezl+IImzl
for z E C, (1.3.1) is equivalent to
a = lim n-=Zn,
(1.3.3)
0 = lim yn, n-+oc
where zn = xn + iyn and a = a + if3. A convergent sequence {zn }n o is always bounded (that is, for some M, IznI < M for all n). In particular, for a sequence (r,}"°-0 of real numbers we write
llm rn = +00 n-ao if for an arbitrary K > 0 there is a number no E N such that rn > K for n ? no. If lim zn and lim wn exist, and a, b E C, then we have lim(azn + bwn) = a lim z + b lim wn,
lim(znwn) = (lin1zn)(Iinwn), lnn zn = rim Zn ,
I IiMzn I = lim I zn l
EXERCISE 1. Show the above four equalities.
A sequence {zny}"_p formed by apart of (the order is not changed) is called a subsequence of {zn} 0. If {zn} o converges to a, then so does any of its subsequences. (1.3.4) THEOREM. Every bounded sequence has a convergent subsequence.
This is clear by the corresponding Weierstrass' theorem for real numbers, (1.3.2) and (1.3.3). A sequence {z,,}'o is called a Cauchy sequence if for an arbitrary e > 0 there
is an no ECsuch that
Izn-z,nl<e,
n,m>no.
In this case, {Rezn}
are both Cauchy sequences by (1.3.2). and {Imzn} Thus by Cauchy's theorem for sequences of real numbers we have (1.3.5) THEOREM. A sequence {zn}n o converges if and only if it is a Cauchy sequence.
Let A C C be an arbitrary subset. Note that A is closed if and only if any convergent sequence of points of A has a limit in A. We say that A is compact if any sequence of points of A has a convergent subsequence.
1.3. SEQUENCES AND LIMITS
11
(1.3.6) THEOREM. The following are equivalent for A C C: i) A is compact. ii) A is bounded and closed. iii) Let A C UQEr UQ be an arbitrary open covering. Then there are finitely many U , , , ... , UQ, such that I
AC UUQ,. =1
PROOF. The equivalence of i) and ii) follows immediately from Theorem (1.3.4).
Put S = {Re z; z E A}. Then S is a bounded closed subset of R. Let a be the minimum of S. Then we put
A[a,r]={zEA;a:Re z
a. The corresponding statement for bounded closed subsets of R is known as the Heine-Borel theorem. Therefore A[a, a] is covered by finitely many
UQ,,1SjSk: k (1.3.7)
A[a, a] C U UQ, . j=1
FIGURE 9
It follows that there is a r1 > a such that k
A[a, r1] C U UQ, . j=1
(If there is no such rl, we can find z E A[a,a + 1/n] \ U;=1 UQ,, n = 1, 2, .... By "i) a ii)" shown above we may assume that converges to a. Then a E A[a, a], so that some UQ, 9 a. Thus z E UQ, for large n. This is absurd.) Now, let TO be the supremum of r such that A[a, r] is covered by finitely many UQ . Suppose ro < max S. It follows from the same reason as (1.3.7) that A[ro, rp]
1. COMPLEX NUMBERS
12
is covered by finitely many U. , and hence so is A[ro - 6, ro + 6] for some 6 > 0. Since A[o. ro-6] is covered by finitely many Up, so is A[u, ro+6]. This contradicts
the choice of r0. Thus r(j = maxS. Again by the same reasoning we see that A = A[o, max S] is covered by finitely many U0. Consider the open covering {,3(a;1);a E Al of A. By the assumption there are finitely many points a1, ... , a1 E A such that A C c7(a1; 1) U ... U 6(a,;1).
Therefore A is bounded. We next show that A is closed. Let a be an accumulation point of A. Suppose
a ¢ A. Put
Uo={zEC;Iz-a]>2 U = 1zEC;n+2 <]z - al < 11, n=1,2,..
FIGURE 10
Then A C C \ {a} = Un 0 Un, and so
is an open covering of A. Since a is an accumulation point of A, A cannot be covered by finitely many U,,. This is a contradiction, and hence A is closed.
A subset B of A is said to be relatively compact if the closure B of B is compact and B c A, and in this case we write B c A. EXERCISE 2. In Theorem (1.3.6), iii) we may take relatively compact open subsets V, C= U,,, with A C
u=1 v,.
Let {z,, }n o be a sequence. We call the formal sum En o z a series, and each z,, a term of F,', z,,. For N = 0, 1, 2, ... , the sum sN = En=" zn is called the Nth partial sum. If {sN}N_o converges, E' oz is said to be convergent, and the limit limN_., SN is denoted by E'A z,,: N
n=0
N-ac n-o
1.3. SEQUENCES AND LIMITS
13
If Fn 0 zn and En °_o wn are convergent, and a, b E C, then oc x x E(azn +bwn)=a>zn+b )
n-0
n=0
E E.zn n=0
-F =
wn,
nn=O
zn.
n=0
We say that En '=0 zn satisfies the Cauchy condition if the sequence {sN }N_o of the Nth partial sums is a Cauchy sequence. If E' 0 Iznl converges, we say that 1_0 zn converges absolutely. Let A : Z+ - Z+ be an injective and surjective mapping; that is, A is one to one, and A(Z+) = Z+. Then E,,. llc 0 z.\(,,) is called a series of order change of EOC
Ln=O zn
(1.3.8) THEOREM. i) A series Z0 zn converges if and only if it satisfies the Cauchy condition.
If E° 0 zn converges, then limn-,r z = 0. iii) if En 0 zn converges absolutely, it converges. iv) Any series of order change of an absolutely convergent series converges absolutely, and has the same limit as the original one. EXERCISE 3. Prove the above theorem.
Let E .0 zn and E 0 z;, be two series. Put zizn_,.
wn = 3=0
Then the series En own is called the Cauchy product of G.nx0 zn and E, Similarly to the case of series of real numbers, we have
0
z',.
(1.3.9) THEOREM. If En0 zn and F_n o z;, converge absolutely, then so does their Cauchy product E'0 wn, and
x
wn = E zn n=0
n=O
zn
n=0
A formal product n,=, zn for a sequence {zn}n 0 is called an infinite product. o zn is said to be convergent if the following conditions are satisfied:
An infinite product n
(1.3.10) i) There is an no E N such that an 54 0 for all n > no.
ii) The sequence
{nnnu+m
'c zn }IM=0 converges to a non-zero number.
1. COMPLEX NUMBERS
14
In this case, we set
Zn= (:) (
-)
--0
n=r+p
which is called the limit of the infinite product fn 0 zn. It is clear that the limit, if it exists, does not depend on the choice of no. ao
nx
(1.3.11) THEOREM. i) If an infinite product H zn converges, then lim zn = 1. n=0
ii) An infinite product jjn o zn converges if and only if for an arbitrary e > 0 there is an no E N such that na
fl zn-1 <e
(1.3.12)
for aUn2>n1?no.
n=n,
PROOF. i) Take no as in (1.3.10), i), and put a = limrn_,, Hn no zn. Then a54 0 and
_
zm
lln=no Zn m_1 n=nr, zn
_ aa =1
(m-+oo).
ii) We first show the "only if" part. Take no as above. There is an M > 0 such that 1 <
no+m
m?0.
fJ Zn < M, Al = "=no
(1.3.13)
It follows from Theorem (1.3.5) that for any e > 0 there is an nl such that no+-1
no+m2
ri zn
n=no
Therefore
ri Zn
m2?MI >nl.
<E,
n=no
nq+m2
n=no+m,+l
Z'-1 < no+i Hn=O Znl
< Me.
Since e > 0 is arbitrary, the claim is proved. We next show the "if part. We may assume by the condition that (1.3.13) holds (putting e = 1/2, we may take M = 2). For an arbitrary c > 0 there is an no (> n4) such that nz
11 n1
zn-1 <e
foralln2?n1_no.
It follows from this and (1.3.13) that n, Inriil n2
Zn - fl Zn n=no
n=no
I I
Zn Zn
In=no
I
I
I
I In°ni
Me.
PROBLEMS
15
gives rise to a Cauchy sequence, and so it converges by Hence { m ;b zn } M-o Theorem (1.3.5). By (1.3.13) the limit is not zero. 0
We will discuss infinite products again in the next chapter, after defining the exponential and logarithmic functions for complex numbers.
Problems 1. Write f, 'T1,
1 -
3i in the form of x + iy.
2. In general, put x + iy = u + iv. Express u and v as functions of x and y. 3. To a complex number z = x+iy we assign a real matrix 0(z) _ ( y V). Let O(z1)+¢(z2) and 0(z1)0(z2) stand for the addition and product as matrices. Then show the following. 0(zt ± z2) = O(zl) ± 0(z2), 0(z0O(z2) = O(zlz2),
04)- _ 0(z-1)
(z 34 0).
+ i sin 2, n E N, 0 < j < n - 1, on the 4. Take n points P3 = cos unit circle. Let POP, be the distance between Po and P3. Show that
.,-1-
j=1 POP; = n.
5. Let 0 < r < 1, and 0, E R, n = 0, 1, 2, .... Show that
a
E rn (COS On + i sin On) n=0
converges.
6. Define a sequence {zn}n`_o by zn+1 - zn = a(zn - z.-I) with 0 < 1al < 1. Write the limit lim zn in terms of zo and z1.
7. Let E. C C, a E t, be compact subsets such that EQ, n . . . n EQk # 0 for any finitely many EQ ... , EQk. Show that nQEr EQ 0 0. 8. Let B C A be a discrete subset in A (c C). Show that for any compact subset K C A, K n B is a finite set. 9. Show that the ring R(r1 i r2) = {z E C; r1 < Izj < r2} (0 < r1 < r2) is a domain.
10. Let D C C be a domain, and E C D be a discrete subset. Show that D\ E is also a domain.
CHAPTER 2
Complex Functions
In this chapter we use the properties of complex numbers described in the previous chapter to deal with complex functions, per series and analytic functions, and to study their properties. By making use of power series we define elementary functions such as exponential functions, trigonometric functions, etc. Then we define the Riemann sphere and introduce linear transformations. The material presented in this chapter may be said to form the foundation of complex functions theory.
2.1. Complex Functions Let A c C, and let f : A -. C be a mapping. We call f a complex function defined on A. We set
f (z) = f (z),
Ref=f+f 2
z E A,
'
Imf=f2i-f We call Ref (reap., Imf) the real (reap., imaginary) part of f . The function f is said to be bounded if there is a number M > 0 such that If (z) I < M for all z E A. Let zo E C be an accumulation point of A. Then we say that f has limit a E C at zo, if for an arbitrary e > 0 there is a b > 0 such
that
If(z) - al < 0,
zE(A(zo;6)\{zo}) f1 A,
and write
a = lim f (z). ExERC11E 1. A function f (z) has limit a at zo if and only if for an arbitrary f (z,,) = a. of A \ {zo} converging to zo,
sequence
17
2. COMPLEX FUNCTIONS
18
In particular, if zo E A and f (zo) = lim=_,, f (z), f is said to be continuous at zo. If f is continuous at all points of A, f is said to be continuous on A. Let g be another function on A. If f and g are continuous at zo (or on A) and a, b E C, then af(z) + bg(z),
f(z)g(z)
are continuous there; moreover, if g(zo) # 0, then f(z)/g(z) is continuous at zo, too.
EXERCISE 2. i) f (z) is continuous at zo if and only if Ref (z) and Imf (z) are continuous at zo. ii) If f (z) is continuous at zo, then so is if (z) 1.
We say that a complex function f : A -. C is uniformly continuous if for an
arbitrary or >0there is ab>0such that
Iz-w1
z.wEA,
(2.1.1) THEOREM. i) A continuous complex function f on a compact set A is bounded, and If(z)I attains the maximum. ii) A continuous complex function f on a compact set A is uniformly continuous.
PROOF. By definition, for arbitrary e > 0 and a E A there is a b(a) > 0 such
that z E i (a,b(a)) n A
If(z) - f(a)I < e.
Since {A(a; 6(a) /2); a E A) is an open covering of A, there are finitely many ax, ... , ai E A by Theorem (1.3.6) such that I
AC
tJ A(aj,bj /2),
b; = b(aj).
Therefore, for an arbitrary z E A there is some j with &(a,,5J/2) 9 z, so that
If(z)l =1f(z) - f(ad) + f(a,)l < e + If(aJ)I <e+ max if(ai)Ic 15j<1
Thus f is bounded. Put M = sup{I If (z)1; z E Al < +oo.
If (zn)I = M. Since A is compact, Take a sequence {zn}'n°_0 such that we may assume by taking a subsequence that {zn}n o converges to zo E A. By the continuity of f If (zo)l = n-x lim If (zn)I = Af. Thus i) follows.
2.2. SEQUENCES OF COMPLEX FUNCTIONS
19
To show ii), we put bo = min1 b.. We arbitrarily take z,w E A with Iz - wI < bo/2. Then 0(a,; bj/2) D z for some j, and so w E A(aa;b,). Hence
If(z) - f(w)I
If (z)
- f(a,)I + If(w) -f (a)) I < 2e.
This shows that f is uniformly continuous on A. 0 EXERCISE 3. Let f (z) = z2 and 0 < f < 2. Determine the largest b such that
forz,wEA(1) I z - wI <6==> If (z) - f (w)I
2.2. Sequences of Complex Functions Let {f n }.°_o be an infinite sequence of complex functions defined on A. We simply call it a sequence of complex functions. We say that {f,,} converges
at z E A if the sequence {fn(z)}n o of numbers is convergent. If {fn(z)}n o converges for all z E A, then { fn}n 0 is said to converge on A, and the function f (z) = limn _,,c fn (z) in z E A is called the limit function of { fn}n o. We write
f = lim n-x fn,
or fn,f (n-'oo).
Moreover, we say that {f} 1o converges uniformly to f if for an arbitrary e > 0 there is a number no such that for all n > no and z E A
Ifn(z) - f(Z)1 < f(2.2.1) THEOREM. i) A sequence {fn }n'=0 of complex functions on A converges uniformly if and only if for an arbitrary e > 0 there is a number no such
that for allm,n?no and zEA Ifm(z) - fn(z)1 < f. ii) If a sequence {fn}' o of continuous complex functions fn on A converges
uniformly, then the limit function f = limn, fn is continuous. PROOF. i) This is easy and is left to the reader. ii) Take an arbitrary e > 0. By assumption there is a number no such that
If (Z) - fb(z)I < f,
z E A.
Take a point a E A, and fix it. Since fro is continuous at a, there is a b > 0 such
that Ifno(z) - fno(a)I < f, z E A(a;b). It follows from the above two equations that for z E A(a; 6)
If(z) - f(a)I < if (z) - f,b(z)I + Ifno(z) - f, (a)I + Ifno(a) - f(a)I < 3f.
Thus f is continuous at a. 0
2. COMPLEX FUNCTIONS
20
A sequence (f,, } n o of complex functions on A is said to be uniformly bounded
if there is an M > 0 such that for all n and z E A
If"(z)I s M. A bounded sequence of complex numbers necessarily has a convergent subsequence (Theorem (1.3.4)). If a sequence of complex functions on A is uniformly bounded, then does it have a convergent subsequence? The answer is "No" in general (see problem 3 at the end of this chapter for an example). Furthermore, here we consider sequences of continuous complex functions, and require the limit functions to be continuous. With this in mind we introduce the following new concept: A family F of complex functions on A is said to be equicontinuous if for an
arbitrary c> 0 there is a 6> 0 such that for all f E F (2.2.2)
Z, W E A, Iz - wI < b
If (z) - f (w) I <,E.
In this case, every f E F is, of course, uniformly continuous. The following theorem is called the Ascoli-Arzelu theorem and is widely used in all fields of analysis.
(2.2.3) THEOREM. If a sequence {f"}n o of complex functions on a compact set A is uniformly bounded and equicontinuous, then { f"}1 o has a uniformly convergent subsequence.
PROOF. If A is a finite set, then the claim is clear, so that A may be assumed to be infinite. We take a subset E = {z,,; v = It 2,... } of A such that E = A. We may take such E by the following procedure: By Theorem (1.3.6) A is bounded, and so there is an M > 0 such that IRezI 5 M, IImzI 5 M for all z E A. Set
F1={zEC;IRezISM, IImzISM}. Dividing each side of the closed square F1 into two equal parts, we have four closed squares F2,1, ... , F2,4. Repeating this process n times, we have 4" 1 squares F", J, 1 5 j <_ 4'- 1, and
4--'
U F".i = F1 D A. j=1
Taking one point z",, E A fl F",_, for non-empty A f1 F",,, we set E" = {z",J }i,
which is a finite set. Then set E = U' E. Then it is clear that E satisfies 1
the required property.
Now, for z1 E E the sequence {f"(z1)}n o is bounded, and so by Theorem (1.3.4) it has a convergent subsequence In the same way, has a convergent subsequence Repeating this process, we obtain a subsequence {f"iki.(z)}°_o of {f"(z)}0 o converging at } V o of { f"}'0 o z1, ... , zk for k = 1, 2, .... Thus we get a subsequence {
2.2. SEQUENCES OF COMPLEX FUNCTIONS
21
which converges on E. We write this subsequence as (g,110 for the sake of simplicity. Given e > 0, we choose a 6 > 0 so that (2.2.2) holds. By Theorem (1.3.6) A is covered by finitely many A(a3;b) with aj E A and 1 < j < 1. Since E 0(aj; 6) n E for every j. There is a number Ito E = A, there is a point. such that for all It, p' > Ito (2.2.4)
19v(z,.(J)) -go,(z,(,))I < C,
1Sj
Now, let a E A be an arbitrary point. Then t(a,; b) 3 a for some j. It follows from (2.2.2) and (2.2.4) that for the above p, p' > Ito 9,,'(a)f :519,,(a) - 9p(z,,(j))I +
9v'(z,.(,))l
+ Igo,(z,.(,)) -gv'(a)1 < 3e.
Therefore we infer from Theorem (2.2.1), i) that {g}N o converges uniformly on A.
Let D be an open subset of C and {f }n o be a sequence of complex functions on D. We say that o converges uniformly on compact subsets if it converges uniformly on every compact subset of D. o of complex functions on D converges uniformly on every compact subset if and only if for every a E D there is an r > 0 such that A(a; r) C D and If,, }n_0 converges uniformly on A(a; r). ii) The limit function of a sequence of continuous complex functions on D which converges uniformly on compact subsets is continuous.
(2.2.5) THEOREM. i) A sequence
PROOF. i) The "only if" part is clear. We show the "if" part. Let K be a compact subset of D. By assumption, there is an rd > 0 for every a E K such that A(a;r0) C D and o converges uniformly on 0(a;rQ). Since {I(a;rQ)}QEK is an open covering of K. there are finitely many al,... a, E K such that K C Vj-1 0(a,;rl) with r, = r0,. Since o converges uniformly on U.,=1 t(aj; r,), it also converges uniformly on K. ii) To show continuity we may restrict the complex functions to a disk neighborhood A(z; r) C= D of a given point z E D. The result then follows immediately from Theorem (2.2.1), ii).
We say that a sequence {f, }n o of complex functions on D is uniformly bounded (reap., equicontinuous) on compact subsets if it is uniformly bounded (resp., equicontinuous) on every fixed compact subset of D. (2.2.6) THEOREM. If a sequence {f,,} n o of complex functions on D is uniformly
bounded and equicontinuous on compact subsets of D, then it contains a subsequence converging uniformly on compact subsets of D. We prepare the following for the proof.
2. COMPLEX FUNCTIONS
22
(2.2.7) LEMMA. There are relatively compact subsets U,,, n = 1, 2, ... , of D satisfying i) U. C=
1, n = 1, 2, ... ; ii) for any compact subset K C- D, there is an n E N with K C U,,.
PROOF. If 8D = 0, then D = C, and we may put U = 0(n), n = 1, 2, ... . Assume that 8D 0. We define the boundary distance by d(z; 8D) = inf { Iz - wI; w E 8D},
(2.2.8)
z E D.
Take a sequence w E 8D, n = 0,1, 2, ... , so that
Iz -
-d(z; 8D) (n -oo).
o is bounded, and so it has a converging subsequence by Theorem (1.3.4). Hence we may assume from the beginning that limwn = wo. Since 8D is closed, wo E 8D. Therefore we have
Then {w },°,°
(2.2.9)
d(z; 8D) = Iz - woI = min{Iz - wI; w E 8D} > 0.
Forz'ED Iz' - woI - Iz - woI
Id(z';OD) - d(z;BD)I < Iz' - zI.
(2.2.10)
In particular, d(z; 8D) is continuous. For r > 0 the set defined by Dr. = {z E D; d(z; 8D) > r} is open.
FIGURE 11
Put
U = D11 f1 0(n),
n = 1, 2,... .
Then U are compact, and D = U' U,,. Thus ii) follows from this and Theo1
rem (1.3.6), iii).
0
2.3. SERIES OF FUNCTIONS
23
PROOF OF THEOREM (2.2.6). Take Un. n = 1, 2,... , as in Lemma (2.2.7). o of {fn}n o which conBy Theorem (2.2.3) there is a subsequence verges uniformly on U1. In the same way, we take a subsequence {fn(2)v}V o of {fn(1>v }v o which converges uniformly on U2. Inductively, we take a subsequence which converges uniformly on U,, (ja = 1,2,... ). o of Then the subsequence { f n y 1 } i } o of {f,,} converges uniformly on every U. Hence, {fn(P+1),'}x o converges uniformly on compact subsets of D. 0
EXERCISE 1. Show that the functions f,, (z) = zn, z E A(1), n = 1, 2, ... , are not equicontinuous. EXERCISE 2. Show that if {fn(z)}°' 1 is uniformly bounded and equicontin-
uous on a subset A C C, then so are gn z =
n
2.3. Series of Functions In what follows, functions mean complex functions. Let {fn}no be a sequence 3o f,, is called a series of functions defined on A C C. Then the formal sum o f functions. For a number N (= 0, 1, - - -) the sum En=o f,, is called an Npartial sum or simply a partial sum. If the sequence {sv(z)}N-o converges at a point z E A, the series o fn of functions is said to converge at z, and we write
n
lira sN(x) _ fn(z). N--x n=0 If {sN}N=o converges at all points of A, we say that F_n ofn converges on A and write the limit function as
x A.
lim sN _ n=0
If two series, E o f,, and F_n o g, of functions on A converge at a point z E A (resp., on A), then for constants a, b E C
x
x
x
(afn(z) + bgn (z)) = a > f,, (z) + b> g. (z) n=0
n=0
n=0
at z (resp., on A). If ER`a Ifn(z)I converges at z (resp., on A), Ln o fn is said to converge absolutely at z (resp., on A). The following is clear. (2.3.1)
Theorem (1.3.8), iii) and iv) hold for a series Fx ofn of functions at z E A or on A.
We say that F,n o fn converges uniformly on A if {sN } N=o converges uniformly on A.
2. COMPLEX FUNCTIONS
24
(2.3.2) THEOREM. A series ER o f of functions on A converges uniformly on A if and only if for an arbitrary e > 0 there is a number no such that fj (Z) <e 1=m
for all n ? m >_ no and z E A. Moreover, if the series converges uniformly and all the functions fn are continuous, then the limit function Eno fn is continuous.
PROOF. Assume that F_R o f, = f converges uniformly. Then for an arbitrary e > 0 there is a number no such that n
IE fj(z) -f(z)
j-0
for all n ? no and z E A. Take any n ? m > no. Then we have n
E f,(=)
- ('f(Z)_f(Z) j=o
fj(z) - f(Z)
j=m
)
j=o
m-I E fj(z) - f(z) < 2e.
fj(z) - f(z)
j0
)=0
This proves the "only if" part.
Take an arbitrary e > 0.
Now, we show the "if" part. Put f = E By assumption there is a number no such that 0
n
Ef3(z) - Efj(z) j=0
j=0
for all n ? m > no and z E A. Letting n
E f) (Z) < e
Pm+I
oo, we have
m
f(z) - Efj(z)
< C.
1=0
Thus Eno f, converges uniformly to f . Continuity is a direct consequence of Theorem (2,2.1), ii).
0
We say that a series Fn o M with M 0 is a majorant of a series F,' 0 fn of functions on A if
Ifn(z)I < M.,
z E A, n = 0, 1, ... .
The following theorem is called the majorant test or Weierstrass' M-test. This test is simple but quite useful.
2.4. POWER SERIES
25
o Mn be a majorant of a series En o fn of functions on A. If E' o Mn converges, then E'0 fn converges absolutely and uniformly
(2.3.3) THEOREM. Let
on A.
The proof is clear by definition and by Theorem (2.3.2). EXERCISE 1. Show the above Theorem (2.3.3).
Let T_n o fn be defined over an open set D C C. If the sequence {sN } of its partial sums converges uniformly on compact subsets of D, then =o fn is said to converge uniformly on compact subsets of D. By Theorem (2.2.5), ii) we have
(2.3.4) THEOREM. Let >n o f" be a series of functions on an open set D C C. If En o fn converges uniformly on compact sets of D. then the limit function is continuous.
2.4. Power Series
Let {an}n o be a sequence and c E C. Then the series >no an (z - c)' of functions is called a power series. By a translation we can reduce to the case c = 0, and so we henceforth mainly assume that c = 0. For a power series o anz" and zo E C we consider the following three conditions:
En
(2.4.1)
i) Eno anzo converges.
ii) limn_,, anzo = 0. iii) limn_,,,,Ianzp I < +oo.
Clearly, i) implies ii), and ii) implies iii).
(2.4.2) LEMMA. If any of (2.4.1) holds, then and uniformly on compact subsets of o(Izol).
oanz' converges absolutely
PROOF. It suffices to assume that zo 96 0 and (2.4.1), iii) holds. Then there is an M > 0 such that Janzo"I <_ M,
n = 0, 1,... .
Take arbitrarily 0 < r < IzoI. We are going to show that ER o Ianz"I converges uniformly on A(r). We have Ianz"I < Ianlr" < Ianzo l l r
< M 1J/
(-r
Since r/Izol < 1, E ' M(r/Izol)" converges. By Theorem (2.3.3) En0 IanznI converges uniformly on 0(r).
Let R (< +oo) be the supremum of those r > 0 such that Eno anz" converges. We call R the radius of convergence of E 0 anz", and {z E C; Iz1 = R} the circle of convergence. If R > 0, we call F,n o an z" a convergent power series. The limit n oanz" defines a continuous function on A(R).
2. COMPLEX FUNCTIONS
26
(2.4.3) THEOREM. The radius of convergence of > 0anzn is given by
R=
1
!im n- 3c
IanI
where 1/ + oc = 0 and 1/0 = +oo. PROOF. Suppose that R = 0. Then it follows from Lemma (2.4.2) that for any r > 0 there are infinitely many n with Ianlrn > 1. Hence
nlm "IanI>_T
Letting r \ 0. we have that tun
nx
IanI = +00.
Suppose that R > 0. Take arbitrarily 0 < r < R. By Lemma (2.4.2) there is a number no such that IanrnI < 1 for n no. Therefore
llm "
IanI < f.
Letting r / R, we have that lim " IanI < 1/R. Next we take r > 0 so that noo im " IanI < 1/r. Then there is a number no such that Ian Irn < 1 for all
n-x
n ? no. By Lemma (2.4.2), r <= R. Letting r / 1/ !im
nix
1
n
hm " IanI
lanl, we get
_< R.
00
Thus R = 1/n lim " IanI o 3C (2.4.4) THEOREM. if limn.-..Ian(/Ian+il (5 +oo) exists, then
R = !im n-»
IanI
. Ian+ll
PROOF. Assume that 0 < limn-ac IanI/Ian+lI < +oo. Take arbitrarily 0 < r < limn-x IanI/Ian+il. Then there is a number no such that r < IanI/Ian+1I for n > no. Therefore ,r,n-no
<
Ian,,+1l
lafQI
,
Ian-lI =
Ian,,+1 I
I
Taking the n-th root of both sides, and letting n
r<
oo, we get
1 =R.
Gm "IanI n-x
We let r / limn_ -, Ian I/Ian+i I, so that (2.4.5)
lim
Ianol
IanI
<
Ian+1I =
1
n--x "IanI I
=R
2.4. POWER SERIES
27
Therefore, if lim la, I /Ian + 1 l = + cc, R = +oo, and so they are equal. Suppose
n-x
lim'anl/jan+ll that 0 S lim lan1/lan+11 < too. Take an arbitrary r >n-x n-31-
Similarly to the above there is a number no such that rn-no > lan0l/janl for n > no. It follows that
r>
1
1
lim " land
=R.
_>
_lim_ " lanl
n-x
Letting r \ lim,_-,, janl/Ian+1I, we get (2.4.6)
_a"1
lim
>
Ian+11
> R.
1
Jim "
aanI
n-x
Thus the desired equality follows from (2.4.5) and (2.4.6). By (2.4.5) and (2.4.6) we have (2.4.7) COROLLARY. If Jim
n-x
Ian+ll
(< +oc) exists, then so does Jim
n--x
lan l
"
-Janl (<
+00), and an+11 lim n-x n-x lanl = lim
E
.
For example, the radius of convergence of En ozn is R = limn-x and
x (2.4.8)
F, zn = 1
1
z,
z E c1(1).
n=0
This is verified by letting N - oo in the sum E zn = (1 - zN+1)/(1 - z). Taking the Cauchy product of (2.4.8) with itself and using Ek-0 zazn-k = (n + 1)z", we see that (2.4.9)
E(n + 1)z" = n=0
1
(1 - z)2
The radius of convergence of this power series is 1imn_,x(n + 2)/(n + 1) = 1. Let R be the radius of convergence of a power series E°_0 a"z". Then the series does not converge at any point outside the circle of convergence. On the circle of convergence there are, in general, points where it converges and where it does not converge. For example, the power series (2.4.10)
(-1)"zn+1
1(z) n=0
n+ I
has radius of convergence I and converges at z = 1, but not at z = -1.
2. COMPLEX FUNCTIONS
28
Let f be a function in an open set D C C. The function f is said to be analytic if for any zo E D there are r > 0 with A(zo; r) C D and a power series a an (z - z0)n, converging in 0(zo; r), such that
En
00
an (z - zo)z E A(zo; r).
f (z) =
(2.4.11)
n=0
In this case, f is, of course, continuous in D. We call (2.4.11) the power series expansion or the Taylor series of f about zo. (2.4.12) THEOREM. A function defined by a power series 00
f(z) = E an(z - a), n=0
which converges in 0(a; r) with a E C, is analytic in A(a; r). PROOF. Fix an arbitrary point b E 0(a; r). Then for partial sums we have (2.4.13) N
N
E an(z
- a)' = r, an(z - b + b - a)n
n=0
n=0 N
=
n
E E n=0 m=0
an r\m/l (b
a)n-m(z
(m) (b = n=0 1 {an m=0 N Cm(z-b)',
m =O
where Cm =
EN
n=m an (m)
(b - a)n-'n.
FIGURE 12
a)n-"`
b)m
t (z - b)m 1
2.4. POWER SERIES
29
In the same way we have N IC+nIIz-blm
M=0
NN N m=0 n=m
IanI(n)Ib - aIn-mlz - blm I
x
.I
_ > lanl(Ib - al + lz - bl)n < n=0
E
lanlr'n
< Oc,
n=0
where lb - al + lz - bl < r' < r. The above estimate implies also that for every fixed m the series ,,, an (n)(b - a)n-'n converges absolutely. Thus, setting
cn, = limN, cm and M = En o lan Ir'n, we have for every fixed N N
N
ICmIIz - bum =
:slim > lCm IIz - blur <= M. m=0
M=0
Hence, E Cm (z - b)m converges absolutely on A(b; r - Ibl ), and
Cm(z-b)m,
f(=)=
z E i(b; r - Ibl).
0
m =0
The following theorem is called the identity theorem.
(2.4.14) THEOREM. Let f be an analytic function in D. Let E C D have at least
one accumulation point in D. If the restriction f IE of f to E is identically 0, then so is f in D. PROOF. Let a E D be an accumulation point of E. As in (2.4.11) we expand f:
x
f (z) = > an (z - a)' ,
z E Q(a; r).
n=0
We have that ao = f (a) = 0. Now we want to show that all an = 0. Suppose that it is not the case. Put no = min{n > 1; an # 0}, and f (z) = (z - a)"° {an, + ana...] (z - a) + ... } = (z - a)n°{an + (z - a)9(z)},
n
a,+n(z - a)r 1. Since g(z) is continuous on 0(a;r), there where g(z) _ is a positive r1 < r such that I(z - a)9(z)I < Inn, 1,
z E 0(a: ri ).
It follows that for all z E 0(a; r1) \ (a) (2.4.15)
f (z) 54 0.
Hence E n (0(a; rl) \ (a)) = 0. This contradicts the assumption that a is an accumulation point of E. Now, let D' denote the set of all those points z E D such that there is a disc
A(z; r) C D with f IA(z; r) = 0. Clearly, D' is open and contains a, so that
2- COMPLEX FUNCTIONS
30
D' 0. Let z E D be an accumulation point of V. Since f ID' - 0, z E D' by the fact shown above. Therefore D' f1 D = D', and so D' is open and closed in D. It follows from (1.2.4) that D' = D. We deduce from Theorem (2.4.14) that if f is a non-constant analytic function in D, then the set of solutions of the equation f (z) = w with w E C is discrete in D. The following theorem is a direct consequence of Theorem (2.4.14), but a fundamental property of analytic functions.
(2.4.16) THEOREM. Let f and g be analytic functions in D. If f (z) g(z) - 0 in
D, then f(z) EO inD or g(z) =- 0 in D. EXERCISE 1. What is the radius of convergence of F,n I nlognznq EXERCISE 2. What is the radius of convergence of n^ z"?
EXERCISE 3. Let f (z) = En=0 anz" converge in A(r). Show that f - 0 is equivalent to a" = 0 for all n = 0,1,.... EXERCISE 4. Show Theorem (2.4.16).
2.5. Exponential Functions and Trigonometric Functions The radius R of convergence of the power series z"
n'
(2.5.1)
22
z
zn
n=O
is R = lim"_,_{(n + 1)!/n!} = limn-a(n+ 1) = +oo by Theorem (2.4.4). The analytic function in C defined by (2.5.1) is called the exponential function and denoted by e= or exp z. When z is a real number, the reader should know that
it coincides with the z-th power of the natural logarithm e. Let z, w E C be
two arbitrary points. Then E. " and E o n converge absolutely. By Theorem (1.3.9) the Cauchy product is
x n=0
n
j=0
n '
(n - )
n=0 n j=0
_z."-
(z + w)"
roc n =O
n!
Therefore we have (2.5.2)
ez+U. = e: , e'°.
In particular, e= e-' = e° = 1, so that e' 96 0 for all z E C.
When z = z E R and IxI < 1, the reader should know that the logarithmic function log(1 + x) is given by (2.4.10); that is,
log(1 + x) _ n =O
n
+)1 zn+1
2.5. EXPONENTIAL FUNCTIONS AND TRIGONOMETRIC FUNCTIONS
31
Thus we define the logarithmic function for z E A(1) by (2.5.3)
Z"I
log(l + z) _ n=o
n +)1
The logarithmic function log(1 + z) is analytic in A(l). One may also write (2.5.4)
n
1), (z
logz=n=On+
zEA(1;1).
1
We have e1ogz
=x
(x E A(1;1) f1 R), (Jes - 11 < 1).
loge = x
Since compositions of analytic functions are again analytic (this can be verified directly, but we will prove it in Remark (3.5.9) of the next chapter), the Identity Theorem (2.4.14) implies that (2.5.5)
el°g' = z
(z E A(11 1)),
loge' =z
(Iez-11<1).
For z E C we define trigonometric functions by cos z =
e,z + e-tz 2
e1z - e,
sin z =
They both are analytic in C, and the following hold:
2i
z
2. COMPLEX FUNCTIONS
32
e" = cos z + i sin z
x cost=
(Euler's formula),
(-On
Z2n
(2n)! n=O 1)n
31RZ = n _O
Z 2n+1
(2n +1)!
By these we define other trigonometric functions, tan z = sin z/ cos 2, cot z = cosz/sinz. etc. It follows from (2.5.2) that eu e-'= = 1, so that the following functional equation holds: cost 2 + sin` z = 1.
(2.5.6)
We also infer from (2.5.2) the following addition formulas of trigonometric functions:
cos(z+w) = coszcosw - sinzsinw sin(z + w) = sin z Cos w + Cos z sin w.
For a moment we assume that z is a real number x E R, and prove that coax and sin x are so-called periodic functions. By definition, cos 0 = I > 0. Since _ 4 0' n > 2, (2n - 2)! (4n)! 1
letting x = 2 we have ;
? cost=l - 22 + 2 2! 4! <1-2+3
1
m-2 (4m - 2)!
_
4
24m-2
(4m)!
3.
Therefore cos x has a zero point in 0 < x < 2 by the mean value theorem. We show that this is a unique one. For m >0 and 0 < x <_2 we get 1
(4m + 1)!
_
x2
(4m + 3)!
> 0,
x sin x = = ((4m 1+I )! _ (4mx2+ 3)!) x m=o
1
> 0.
We take 05x
2.5. EXPONENTIAL FUNCTIONS AND TRIGONOMETRIC FUNCTIONS
33
(2.5.7)
Cosx -
y2x)-Cos(x2y+y2x)
cosy=cos(
x2y =2sinx2ysiny2x>0.
Hence cos x is strictly decreasing in the interval [0, 21. This proves that cos x has a unique zero point ao E (0, 2). Set (2.5.8)
r = 2oo.
We call r the ratio of the circumference of a circle to its diameter. It follows from the above arguments that (2.5.9)
sin 2 = 1.
cos 2 = 0,
Again making use of the addition formulas we get (2.5.10)
sin (x + 2) = coax, cos (x + 2)
sin x,
sin (x + r) = -sin:, cos (x + rr) _ - cos x. Thus we obtain the following periodicity: (2.5.11)
sin(x + 2r) = sin x,
cos(x + 2r) = cos x.
(2.5.12) THEOREM. The fundamental period (the positive minimum period) of sin x and cos x is 2r.
PROOF. It follows from (2.5.10) that sin x and coax have the same periods. We deal with sin x. Since sin 0 = 0, sin a = 0 for any period a. By the Identity Theorem (2.4.14) the periods do not accumulate at 0. Therefore there is the
positive minimum period w, so that 0 < w < 2r. If 2r/w is not a natural number,
0<27r-
2rlw<w,
Iw where [21r/w] stands for Gauss' symbol. This shows the existence of a positive period less than w, which contradicts the minimality of w. Set p = 27r/w E N. Since sin x > 0 for 0 < x < r/2(< 2) and sin(rr - x) = sin x, (2.5.13)
sinx>0,
0<x<7r.
Suppose that p > 1. Since sin(x+ r) _ - sin x, p $ 2, so that p ? 3. Thus we see that 0 < w < 2r/3, and so sinw 0 0 by (2.5.13). This is again a contradiction. Hence we get w=2r.
2. COMPLEX FUNCTIONS
34
If 0<x cos y. That is, cos x is strictly decreasing in [0, ir). Therefore the image of the injective continuous mapping
10, r) 30-.cos e+isin0EC just describes the upper half part (Im z >= 0) of the unit circle {z E C; Izf = 1}.
-1
0
FIGURE 14
Therefore the polar coordinate representation (1.1.6) is written as z=re'f'=Iz[e"rgz
Setting w = Iwle` "rg w, we have
zw = I2llu,le:(argz+argw) Henceforth we have
arg(zw) = arg z + arg w (modulo 27r),
and (2.5.14)
log z =log Izl + i argz
forzEA(I;1). EXERCISE 1. Represent ` 1 + f i in polar coordinates. EXERCISE 2. Let z E A(1),z 0 0, and let t be the half line from the origin passing through z. Express in terms of z a point z' E I with Iz'Ilzl = I. ExERCIsE 3. Show that tan(z + r) = tan z for z 34 r/2 + nrr, n E Z.
2.6. Infinite Products The convergence of an infinite product jjn°' 0 z,, is defined by (1.3.10). If this converges, then
lim zn=1.
n-oo Therefore, to investigate the convergence of the infinite product, it is more convenient to replace z by I + zn, and to deal with Hn 0(1 + zn ). If fO 0(1 + zn )
2.6. INFINITE PRODUCTS
35
converges, then lim z = 0, and there is a number no such that Iz,, I < 1/2 for all n n0 and n
fj (I+ZJ)-1 J=no
2
Therefore the values of I + zn (n ? no), f na(1 + zn) are contained in the domain A(1; 1) where the logarithmic function log is defined (cf. (2.5.5)), so that n
n
log(1 + z') = log
fl (1 + z))
)=no
7=no
x
(n
log fl (1 + zJ)
oo).
J=no
On the other hand, if Izn < 1 for n > no for some no and
no
log(1 + zz )
converges, then l+zn =e'Oss(1+=') -
eo= I
(n-+oo)
Hence we have limn_,,, zn = 0 and
(2.6.1)
ft(i+zn)=exP(
log(1 + z)) 1
J=no
J=no
I
exp E log(1 + ))
0 (n -- oo).
J=no
Therefore the infinite product fn o(1 +zn) converges. Thus we have proved the following theorem.
(2.6.2) THEOREM. An infinite product nn a(l + zn) converges if and only if
there is a number no such that Izn I < 1 for n > no and E' n,, log(1 + zn ) converges.
Given this fact, we define the absolute convergence of an infinite product nn o(1 + zn) by the convergence of
log(1+zj)I <+oo =no
2. COMPLEX FUNCTIONS
36
for some number n0. It follows from (2.5.3) that
x Ilog(1+z)i =
(-1)n zn
z
n=O
=Izl.
I
n+1
1
n=1
Since for lzl 5 1/2 2
1
2
n
+
1(-1)nzn
_ 1F( 4n-` i )=-2 1
we deduce that (2.6.3)
2Iz15llog(1+z)l5 2Izl,
zEO(i).
As in Theorem (1.3.8), iv), we consider an infinite product rln 0(1 + za(n)) of order change of an infinite product fl t_0 (I + z,'). (2.6.4) THEOREM. i) An absolutely convergent infinite product converges. ii) An infinite product of order change of an absolutely convergent infinite product again converges absolutely, and the limit does not change.
iii) An infinite product nn o(1 + z1) converges absolutely if and only if Eo 0 zn converges absolutely. PROOF. i) and ii) follow easily from (2.6.1) and Theorem (1.3.8), iii), iv). iii) is clear by the definition and (2.6.3).
Let { f n } n o be a series of functions defined over a subset A C C. Then the
formal product rl' 0 fn is called an infinite product of functions. If rIn 0 fn (z) converges at z E A, then ri"0 f, is said to converge at the point z. If the product converges at all points of A, it is said to converge on A, and the limit function is also denoted by rln o fn. We say that rj;°_0 fn converges uniformly on A if the following conditions are satisfied: (2.6.5) i) There is a number no such that fn(z) # 0 for all n >_ n4) and z E A. ii)
x
{rln°_ o fn J=0 converges uniformly on A to a non-vanishing function m=0
g such that there is a constant b > 0 with lg) > b.
If every f,, is continuous in A and rin o fn converges uniformly on A, then the limit function is continuous there. (2.6.6) THEOREM. An infinite product rlno fn of functions on A converges uniformly on A if and only if for an arbitrary f > 0 there is a number no such that fi(x) - 1 < gy=m
2.7. RIENIANN SPHERE
37
for alln>m>_n0 and z E A. PROOF. Noting that fn -. 1 (n -+ oo) uniformly, we make use of the same arguments as in Theorem (1.3.11), ii). 0
For an infinite product Ijn 0 fn of functions on an open set, we define its uniform convergence on compact subsets as in the cases of sequences and series
of functions. In this case, if every fn is continuous, then the limit function is continuous. By the definition and (2.6.3) we have (2.6.7) THEOREM. Let 11'0(1 + fn) be an infinite product of functions on A. Then fn 0(1 + fn) converges absolutely if and only if the series F° _0 converges. Moreover, if >'o IffI converges uniformly, then so does TIn o(I + fn). EXERCISE 1. Prove the above theorem.
EXERCISE 2. Show that the infinite product n (1 -
converges uni-
formly on compact subsets of C.
2.7. Riemann Sphere A bounded closed set of the complex plane C is compact, but C itself is not compact. On the other hand, it might be convenient to make analytic arguments dealing with sequences or sequences of functions if we can choose a convergent subsequence of them. Let us also consider the following simple function: (2.7.1)
fo(z) = z
By the definition of complex numbers, fo is not defined at z = 0, but is continuous
outside {0}, that is, in C' = C \ (0). The mapping (2.7.2)
fo:C' - C'
is injective and surjective, and the inverse function fo 1 = fo is continuous, too. EXERCISE 1. Show that fo is an analytic function in C.
As z -+ 0, fo(z) does not have a limit in C. It might be useful to have a compact space that contains C and such that lim,,.o fo(z) is defined. The following concept of the Riemann sphere was invented with this in mind. Let (Ti , T2, T3) be the standard coordinate system of the real 3-dimensional euclidean space R3. Let S be the unit sphere with center at the origin:
S={P=(Ti,T2,T3)ER3;T1 +T2+T3 =l}. We identify C with the hyperplane {T3 = 0} of R3 through
The point N = (0, 0, 1) E S is called the north pole and S = (0, 0, -1) E S the south pole. Take a point P = (T1 , T2, T3) 0 N. The line passing through
2. COMPLEX FUNCTIONS
38
N and P intersects C at a unique point z = x + iy = (x, y, 0). The mapping P -{ z is called the stereographic projection (from the north pole N). By simple calculations we have (2.7.3)
z = z(P) = 1
T,
z+z
T, = 1 + JzJ2,
+i 1 TT,3,
z-2
T2 = i(1 + Iz12).
T3 =
Iz12-1 Iz12 + 1-
Henceforth C is identified with S\ {N}, and z (resp.. P) is continuous in P (resp.. z). Through this identification, the topology of C is the same as S \ {N} (that is, convergent sequences of C converge as sequences of S \ {N}, the converse holds, too, and the limits are equal to those corresponding points). T3
FIGURE 15
The sphere S is a bounded closed subset of R3. and hence compact. We write N = oo:
S=CUfool. be a The point oo is called the infinity, or the point at infinity. Let sequence of C. If it is bounded, it contains a subsequence converging to a point of C. If it is unbounded, we can take a subsequence {z,,,,},'`(, of 0 such N in S. A priori, we have Iz,,,,I = +oo. This {z,,,,}' 0 converges to that the complex coordinate z in the subset C of S. We want to introduce a complex coordinate in a neighborhood of oc as well. We consider the stereographic projection from S similar to the one above. Take a point P = (T,,T2,T3) E S \ {S}. and let (x', y', 0) be the point at which the line passing through S and P intersects the hyperplane T3 = 0. In the same way as (2.7.3) we get Ti (2.7.4)
r =
T2 ,
1 + T3
y =
T3.
1 +
Take a new complex plane C with complex coordinate z = z + iy, and set
1 =s',
U=-y'.
2.7. R1EMANN SPHERE
39
(This is natural because of the orientation of the stereographic projection.) We assign P the complex number
= j(p)
T1 _ -T2 +T" + 1 + T3
By computation we obtain (2.7.5)
z(P)i(P) = 1.
P E S\ {N.S}.
It follows that i(oo) = 0. The function ff in (2.7.2) is extended over S to a continuous mapping into S by setting f (0) = oc and fo(oc) = 0. We use the complex coordinate z in neighborhoods of oc. A neighborhood {i E C; kil < r} with r > 0 is called a disk neighborhood of oc. The subspace S\{N. S} is identified with C', which carries two complex coordinates z and i, and by (2.7.2) both coordinates give the same topology. The space S, whose subspace C\ {N} (resp., C \ {S}) is assigned the complex coordinate z (resp., i) is called the Riemann sphere, and denoted by C. The statements on the topology of C such as the convergence of sequences. accumulation points, closures, closed sets, open sets, connectedness, domains, continuous functions and mappings. etc.. continue to bold on C. For instance, let E2 be a mapping. Then f is said to E3 C C, j = 1, 2. be subsets, and let f : E1 be continuous if for an arbitrary point P E E1 and an arbitrary sequence o 0 converges to f (P), i.e., lim,, , f (P,) = f (P). In particular, if f is continuous, injective, and surjective. and if the inverse f -1 E2 - E1 is also continuous, then f is called a homeomorphism. For
in E1 converging to P, { f :
example, fo in (2.7.2) is a homeomorphism. Since the composition of analytic functions is analytic (see Remark (3.5.9) in
the next chapter), it follows from (2.7.5) that if a function in an open subset of C \ {0,o0} is analytic in z. it is analytic in i, and the converse holds, too. Therefore the notion of analyticity is well defined on an arbitrary open subset of C. A circle of C stands for a circle, or a line of C (plus the point at infinity oc to be precise). Here a line of C is considered as a circle passing through ao. EXERCISE 2. Let P (resp., P') be the point of S corresponding to z E C (resp.,
z' E C) by the stereographic projection from N. Express the distance d(P, P') between P and P' in R3 in terms of z and z'. Moreover, express the distance d(P, N) in terms of z. EXERCISE 3. Show that under the stereographic projection circles in C correspond to circles in S which are intersections of S and hyperplanes in R3, and vise versa. (This fact is referred to as the correspondence of circle to circle under stereographic projection.)
2. COMPLEX FUNCTIONS
40
2.8. Linear Transformations A mapping f from C onto itself written as (2.8.1)
f(z)
=
az + b
cz+d
is called a linear fractional transformation, a linear transformation, or a PvMobius transformation. Here the coefficients a, b, c, d E C must satisfy
ad-bc96 0. If c = 0, f (oo) = oo, and otherwise
f(oo) = a,
f
(_d)\ = 0,0_
The mapping f has the inverse f-1: f-1(W )
//
_ dw-b
-cur+a
which is a linear transformation. In particular, f : C C is a homeomorphism. The multiplication of a. b, c, and d by a common non-zero number does not change the linear transformation f (z), and hence we may assume that (2.8.2)
ad - be = 1.
The set of all 2 x 2 complex matrices ( d) satisfying (2.8.2) is denoted by SL(2,C). In general, a set G endowed with an operation a b (a, b E G) satisfying the following conditions is called a group:
i) Two arbitrary elements a, b E G determine a third element a b E G, and (a b) c = a (b c) holds for any three elements a, b, c E C. ii) There is an element e E G such that a e = e a = a for all a E G. (This e is unique and is called the unit element.) iii) For an arbitrary a E G there is an element b E G such that ab = ba = e. (This b is unique; it is called the inverse element of a and is denoted by EXERCISE 1. Show the uniqueness asserted in ii) and iii).
The set SL(2, C) forms a group under matrix multiplication, and is called the special linear group. By a direct computation we see that the composition of two linear transformations is also a linear transformation. It is easily seen that this composition satisfies the above conditions so that the set Aut(C) of all linear transformations is a group. The unit element of Aut(C) is the identity mapping.
2.8. LINEAR TRANSFORMATIONS
41
(2.8.3) THEOREM. 7he mapping
c
d) E SL(2, C) -+
cz + d
E Aut(C)
is a surjective group homomorphism (i.e., 4;(a p) = 44(a) Ker 0 (= $-1(the unit element of C)) is as follows:
The kernel
'1)}.
PROOF. The first half follows from a direct computation. For the latter half we use the identity az + b
cz+d
=
z
and (2.8.2). Then we have b = c = 0, ad = 1, and a/d = 1. Therefore a = d = I
ora=d=-l.
Identifying a E SL(2, C) and -a E SL(2, C), we have the following group:
SL(2, C)/ H0 1)}. By Theorem (2.8.3) the group Aut(C) is written as
Aut(C)
= SL(2, C)/ Hol 'I))-
The right-hand side of this equality is often denoted by PSL(2, C), and called the projective special linear group. (2.8.4) THEOREM. For two triples, (zl, z2, z3) and (wl, w2, w3), of distinct points
of a, there is a unique linear transformation f such that f (z;) = w i = 1, 2, 3. PROOF. We first show existence. If zl = oo, we set f, (z) = 1/z; otherwise, we
set fl(z) = z - zl. Then fl(z1) = 0. If fl(Z3) = oo, we set f2(z) = z; otherwise
we set f2(z) = z/(z - fi(z3)). Then f2 o fi(zl) = 0, and f2 o f,(z3) = oo. Moreover, we set f3(z) = z/f2 o h(22) and f = f3 o f2 o f1. Then f is a linear transformation satisfying f (zl) = 0, f (z2) = 1, and f (z3) = oo, Similarly, there is a linear transformation g satisfying g(wl) = 0, g(w2) = 1, and g(w3) = 00. Then g-1 o f is the required linear transformation. For uniqueness it suffices to show that if a linear transformation f satisfies f (0) = 0, f (1) = 1, and f (oo) = oo, then f (z) = z. This is easy. If c = 0, the linear transformation (2.8.1) gives rise to
f(z) = dz+
b ;
d
otherwise,
f (z)
a
=c
add be
z + d/c
2. COMPLEX FUNCTIONS
42
Therefore an arbitrary linear transformation is represented by a composition of the following linear transformations of three kinds-
(2.8.5)
i) ii)
f (z) = z + b,
bEC
f(z) = az,
a E C' (non-zero multiplication).
iii)
f (z) =
(translation). (inversion).
Z
A circle of C is written by
(2.8.6)
Iz-z1I =k IZ - Z2I
with distinct Z), z2 E C and k > 0, which is the so-called Appollonius' circle.
EXERCISE 2. Show that if k = 1, equation (2.8.6) presents a line passing through (z1 +z2)/2 and perpendicular to the line passing through z, and z2, and
that if k * 1 it presents a circle of C with center at. (z1 - k2z2)/(1 - k2) and with radius I(z, - z2)k/(1 - k2)I. Substituting a linear transformation z = f "1(w) of (2.8.5). i)-viii) into (2.8.6), we obtain
(2.8.7)
Iw- f(zl)I _k' (>0). Iw-f(z2)I
This defines a circle. Thus circles of C are mapped to circles of C by linear transformations. This property of linear transformations is called the correspondence of circle to circle. Take a circle C(a; r). If two points z1, z2 E C satisfy
(2.8.8)
(z1 - a)(z2 - a) = r2,
then z, is called the reflection of z2 with respect to the circle C(a; r), and vise versa. We also say that z1 and z2 are mutual reflections with respect to C(a; r). We define the center a and or, to be mutual reflections. In particular, a point of C(a; r) is the reflection of itself with respect to C(a; r). The points zl and z2 in (2.8.6) are mutual reflections with respect to the circle defined by (2.8.6). This
2.8. LINEAR TRANSFORMATIONS
43
is verified by Exercise 2 and a direct computation.
FIGURE 17
FIGURE 16
A line L of C was defined to be a circle of C. If two points z1,z2 E C are mirror symmetric with respect to L, then they are said to be mutual reflections with respect to L. The points on L, including oo, are reflections of themselves with respect to L. Therefore one deduces the following reflection principle. (2.8.9) THEOREM. Let f be a linear transformation, C a circle of C, and zl the reflection of z2 with respect to C. Then f(zj) and f(z2) are mutual reflections with respect to f (C).
Note that two points zl, z2 E C \ L are mutual reflections with respect to a line L of C if and only if L is defined by (2.8.6) with k = 1. The same holds for a circle C(a; r). That is, two points 21, z2 E C \ C(a; r) are mutual reflections with respect to a circle C(a; r) if and only if C(a; r) is defined by (2.8.6) with suitable a E C and r > 0. In general, a subset F of a group G is called a subgroup if F itself is a group under the operation of C. The unit element is always shared by all subgroups of G. For example, {± (o 0)j is a subgroup of SL(2, C). If a linear transformation f satisfies f(A(1)) = A(1), f is said to preserve s(1). We denote the set of all those f by Aut(A(l)), which is a subgroup of Aut(C). We determine the type of f E Aut(A(1)). For a E 0(1) we look for Oa E Aut(A(1)) such that Oa(C(0;1)) = C(0;1), 00(A(1)) = A(1), and 0.(a) = 0. The reflection of a with respect to C(0;1) is 1/a, and so 0a(1/a) = oo. Hence for a candidate we set
z-a (2.8.10)
oa(z) =
-az + 1'
2. COMPLEX FUNCTIONS
44
FIGURE 18
In fact, oa(a) = 0, and, if Ez) = 1,
z-a
_ Iz-aI
-az+1
la - z)
-- 1.
Furthermore, we see that al2
fzI < 1 b 1 -
=
1z -
Iaz - 112
(1 -
1a12)(1 - 1z12)
> 0.
Paz - 112
It follows that a(0(1)) _ i(1). Note that 0,-' = 0-
.
Let a linear transfor-
mation f preserve A(1). Then g = f o 0 '(o) is a linear transformation, and g(0) = 0. Since the reflection point of 0 with respect to C(0;1) is oo, g(z) = orz with some a E C'. Since g(C(0;1)) = C(0;1), Ial = 1, so that a = eee (0 E R). Therefore ej& _az-a a = f -' (0). A z) = z + l'
(2.8.11) THEOREM. i) A linear transformation f E Aut(A(l)) is written as ff(z (z))= =
e`a
ii) For two arbitrary points
z-a E i(1), there is an f E Aut(A(1)) such
that f (a) =#. PROOF. The first half is already proved. The latter half follows from f =
0'o0.
Property ii) of the above theorem is referred to by saying Aut(A(1)) acts transitively on A(1). Set
H = {zEC;Imz>0}.
2.8. LINEAR TRANSFORMATIONS
45
H is called the upper half plane. Next we find a linear transformation ip E Aut(C)
such that 1(H) = 0(1).
FIGURE 19
We set a condition, V5(i) = 0. Since ?(i(R U {oo}) = C(0;1), it follows from Theorem (2.8.9) that -0(-i) = oo. Hence for a candidate we set z
a
Cz) = z + i
(2.8.12)
By easy computations I7/i(z)I = 1 for z E R, and IO(z)I < 1 if and only if z E H. Let Aut(H) denote the set of all linear transformations preserving H. Then
Aut(H) is a subgroup of Aut(C). For an element f E Aut(H) there is a g E Aut(A(1)) such that f = tP -1 ogoz/i. From this and Theorem (2.8.11) we deduce
that Aut(H) acts transitively on H. Since If (0), f (1), f (oo)} C R U fool, we see that az + bd' ad - be = 1, f (z) - cz + where a, b, c, d E R. Hence f (R U {oo}) = R U {oo} and
Im z > O b Im f (z)
1 faz+b az+b _ Imz = 2i
cz + d
cz + d
Icz + d12
> 0.
It follows that f (H) = H. Let SL(2; R) denote the set of all real matrices of SL(2, C). We hence obtain the following theorem. (2.8.13) THEOREM. i) The mapping
(a b) E SL(2, R) - f (z) =
az + b E Aut(H) cz + d
is a surjective group homomorphism, and Ker D = {± o ° )} . ii) Aut(H) acts transitively on H. We have the following expression:
Aut(H) = SL(2,R)/ { f ( 0
0)1.
2. COMPLEX FUNCTIONS
46
The right-hand side of the above equation is denoted by PSL(2, R). The group SL(2, R) carries subgroups such as
d) a, b, c, d E Z, ad - be = 1
SL(2, Z) _
;
)ESL(2.Z); Cc a
r (n) (o
a)
d)
-0
0) (mod n) }
(n ? 2).
(mod
n) means that a-1, b, c, and d- I are integral multiples of n. The subgroup f (n) or C(n)/ {f (p ° )} is called the principal congruence subgroup, and in particular C(1) = SL(2, Z) or PSL(2, Z) the modular group. These groups play important roles in the theory of elliptic functions and the theory of automorphic forms. Here (ac
i)
EXERCISE 3. Show that I'(n) is a subgroup of SL(2, R).
Problems log(1 + z)
ez - 1
sin z
, and lim 1. Compute the limits lim z , lim :-0 z-.0 2 Z 2. Let f (z) = (z" - 1)/(z - 1). Find the supremum and the infimum of If (z)l on A(1). 3. Let fn (r) = sin nx, n = 1, 2, ... , be a sequence of functions of the real variable x E [0,2ir]. Then {fn(x)}n , is uniformly bounded. Prove that any subsequence of { f"(x)},°,°_1 does not converges on (0, 2ir). (This requires
a knowledge of Lebesgue integration. If the reader finds it too difficult, see the answer in the back of the book.) 4. Find the radii of convergence of the following power series:
x
i) E npz"
00
ii)
(p > 0).
>q"2z"
n=0 iii)
00
iv) 1+Ez"
j(n!)""z".
(tql < 1). n-1
(a+Y)(0+v)
(Gauss' hypergeometric function)
5. (Stolz' domain and Abel's continuity the-
orem)Let (EC(0;1),and let G((;r)denote a subdomain of 0 (1) hedged by two line segments passing through ( and form-
ing an angle 0 < r < 7r/2 with the line passing through 0 and C. We call G((; r) Stolz' domain with vertex (. (a) Show that G((; r) n A((; cos r) C 2
cosy} nA((;cosr). {,E.12-(I1 - Izj <
PROBLEMS
47
On the other hand, for any given K > 1, take r = cos' -21 E (0, n/2). Thr(en, show that
l
(zE0(1);1z(1
Iz)
1
JJJ
(b) Let f (z) = E 0 anzn be a power series converging on 0(1). Prove that if Eno an(' converges, then 00
lim
f(z)=a('. n=0
zEG(S;r)
6. Show that Aut(0(1)) is equicontinuous on compact subsets of A(1) as a family of functions. (The uniform boundedness is clear.) sin z cos z
7. Show that tan z =
is a function on C \ {2n + it/2; n E Z} with
fundamental period tr.
8. Show that if a > 1,
fi (1 - a)n converges uniformly on compact subsets
n=1
of C.
00
9. Show that II (1 - zn) converges absolutely and uniformly on compact subn=1
sets of 0(1).
10. Show that
II 00 (1 + z2n) =
1
1-z n=o 11. The cross ratio of four points zl, .... Z4 of C is defined by /
lZ1,Z2,Z3,Z4) =
(ZI - Z3)(Z2 - Z4) (Z1 - Z4)(z2 - Z3)
Here, if some zJ = oo, the above right hand side is supposed to represent
the limit as z., - oo. Let f E Aut(C) be a linear transformation and ZI, Z2, Z3 E C such that P Z2) = 1, f (z3) = 0, and f (z4) = oo. Show that f (Z) = (Z, Z2, Z3, Z4)
12. Show that for any linear transformation f and four points zr, ... '24 of C (z1, Z2, z3, z4) = (f (zi ), A Z2), f (Z3), A Z4))
(the invariance of the cross ratio).
13. Let f (z) = (az + b)/(cz + d) j9 z, ad - be = 1, be a linear transformation. Prove that if a + d = ±2, then f has one fixed point z (a point z such that f (z) = z); otherwise, f has two fixed points. 14. Let a and 6 be the above fixed points of f . Show that if a 0 Q, w = f (z) is given by
w-a _KQisZ - a z-f3 w-f3
K>0, 9ER.
2. COMPLEX FUNCTIONS
48
The linear transformation f is said to be hyperbolic if e`B = 1, elliptic if K = 1, and loxodromic otherwise. 15. Let or and Q be as above. Suppose that a = t3. Show that w = f (z) is given by 1
1
w-a
z-a
o0),
(a
00).
In this case, f is said to be parabolic. 16. Let f (z) be as in problem 13. Show that if a+d is real, then f is hyperbolic, elliptic, or parabolic according to whether Is + dl > 2, < 2, or = 2; show that if a + d is not real, f is loxodromic. 17. Find a general form for linear transformations preserving 0(R), 0 < R < oo. 18. i) Find a linear transformation which maps 0, 1, oo to i,1 + i, 2 + i. ii) Find a linear transformation which maps -I, i,1 to -2, i, 2.
CHAPTER 3
Holomorphic Functions
In this chapter we first define holomorphic functions, and then prove Cauchy's integral formula, which is the most fundamental and important theorem in complex analysis. One sees the close relationship between the topological structure of a domain and complex analysis on it.
3.1. Complex Derivatives Let D be an open set of C, and let f be a function on D. We say that f is complex differentiable at a point a E D if there is a number a E C such that for an arbitrary > 0 there is ab >0 with f(a + z) - f(z) al
(3.1.1)
h
<e
for all h E A(b) \ {0}. We call a the complex derivative of f at a, and denote it by f'(a) = JL (a). The expression (3.1.1) is equivalent to (3.1.2)
f(a + h) = f(a) + ah + o(h),
a = f'(a),
where o(h) is a term such that limh.o o(h)/h = 0. If f is complex differentiable at a, then f is continuous at a. If f is complex differentiable at every point of D, f is said to be holomorphic. In this case the function f' : z E D f'(z) E C is called the derived function or the derivative of f , and is also denoted by df /dz. If f is holomorphic, so is 1 If outside the set (f = 0}, and
f
( f),
When fl and f2 are holomorphic functions, and a, and a2 are constants, the following linearity is clear:
(aifi + a2f2)' = aifi + a2f2 We have the Leibniz formula:
(fl f2)'=fi f2+fi f2 49
3. HOLOMORPHIC FUNCTIONS
50
EXERCISE 1. Show that for n E Z, (z")' = nz"-I
We set z = x + iy, and consider f (z) as a function f (x, y) in x and y. Set f (x, y) = u(x, y) + iv(x, y),
where u(x, y) (resp., v(x, y)) is the real (resp., imaginary) part of f (x, y). Assume that f is holomorphic. Then by (3.1.2) u(x, y) and v(x, y) are totally differentiable. Letting h --+ 0 with real h and with purely imaginary h, we have
au ax
av .au a5 8x - 8y av
Therefore
su (3.1.3)
5;
=
au = - ax av .
av
Si
8y ,
These are called the Cauchy-Riemann equations. EXERCISE 2. Let u(x, y) and v(x, y) be real valued, totally differentiable functions satisfying the Cauchy-Riemann equations (3.1.3). Show that the complex valued function f (x, y) = u(x, y) + iv(x, y) is holomorphic as a function of
z=x+iy. Partial differential operators 8: = a/az and a= = 0/ft are defined by
a=f-az-2(8x+tay) of aZf-a8-
1
Of _ 1 of
ax
i0y)'
Then (3.1.3) is equivalent to a=f = 0.
(3.1.4)
This is called the a-equation (d-bar-equation). For a holomorphic function f,
f =0Zf.
(3.1.5)
Let +/ : I -. D be a continuously differentiable curve. For t, t' E I with t # t'
f o b(t') - f o t'(t) = , a fi(001) 001) - +'(t) + n
t'-t
f
t'-t
(
(t') - v (t) t,
t
Since limt.-t(o(t') - P(t))/(t` - t) = v(t), (3.1.6)
.f c
00(t) 00-
(3.1.7) THEOREM. If a holomorphic function f on a domain D satisfies f' = 0, then f is constant.
3.1. COMPLEX DERIVATIVES
51
PROOF. If we set f(z) = f(x,y) = u(x,y) + iv(x,y), we have by (3.1.5) or (3.1.3)
8u
8u
8v
8v =
ax=8y=8x8ySince D is connected, u and v are constant.
(3.1.8) THEOREM. Let f be a holomorphic function on a domain D. Then, if Ref or Imf is constant, so is f . PROOF. It is sufficient to deal with the case where Ref is constant. Setting
f = u + iv, we have 8u/8x = 8u/8y - 0. It follows from (3.1.3) that 8v/8x = 8v/8y - 0. Hence Theorem (3.1.7) implies the constancy of f .
Let f be a holomorphic function on a domain D. Let D' be a domain of C containing the image f (C). Let g be a holomorphic function on Y. Fix zo E D arbitrarily, and set wo = f (zo). Then for a small h 9(f (zo + h)) - 9(f (zo)) = g(wo + hf'(zo) + o(h)) - g(wo) = g'(wo)(hf'(zo) + o(h)) + o(hf'(zo) + o(h))
= g'(wo)f'(zo)h + o(h). Therefore
l hm
9(f(zo + h)) - 9(.f(zo)) _ 9V (zo))f (zo) h
Thus we have the following:
(3.1.9) THEOREM. The composite g o f of two holomorphic functions f and g is again holomorphic, and (go f)'(z) = 9 (f(z))f'(z)
We take a convergent power series about a E C:
f (z) = 00 E a, (Z - )n. n=o
Let R > 0 be its radius of convergence. (3.1.10) THEOREM. The above f (z) is a holomorphic function on 0(a; R), and
nan(z - a)'-
f'(z) _ n=1
Here the radius of convergence of the right side is R, too.
3. HOLOMORPHIC FUNCTIONS
52
Since limn
PROOF. Without loss of generality we may assume a = 0.
"n+1=1,
lim " Ian I = Um nx n-oo
Ian I = lim n-oo
`
Ian+1
= lim n-ocV (n + 1)IanI. Thus the radius of convergence of
x
g(z) = >
nanzn-1
n=1
is R. Take an arbitrary point z E t(R). Fix r with IzI < r < R. By Lemma (2.4.2) there is an M > 0 such that (3.1.11)
n = 1,2,... .
Ianrn1 < M,
ForhEi(r-IzI)\{0}we have If (z + hh - f(z)
_
9(z)I
(z + h)n - zn
_
an
h
n=1 n
'C
a E nF, n=2 v=2
_ E Ian n=1
< M x
(n) Zvhn-v-1
n
0o
Ia.I n=2
(IZI + Ihj)n - Zn l
Ihi IZI + IhI
1
ihI(
)
r
IZlvlhln-v-1
v=2
- nlzln-1
n
(v)
I n
n
(r }-r (Lzl)n-'] IZI
rHere we used (3.1.11) in the last estimate. It follows from (2.4.8) and (2.4.9)
that If (z + h) - f(z)
_
h
< M
=
_
1
9(z)I
IzI + IhI
IhI `r - (IzI + IhI) MrIhI
(r - IzI)2(r - (IzI + IhI))
-
IzI
l
r - IzI)
-0
(h
r (r - IzI)2 0)
O
(3.1.12) COROLLARY. An analytic function is holomorphic.
We mention that the converse of the above statement holds, and will be proved in Theorem (3.5.7). It follows from Theorem (3.1.10) that a function f (z) expressed by a convergent power series about a is complex differentiable arbitrarily many times, and the coefficients an are determined by the n-th derivatives f(n) (a) of f(z): (3.1.13)
an = nifi"1(a).
3.1. COMPLEX DERIVATIVES
53
Hence we see that if a function is expanded to a power series, the coefficients of the power series are uniquely determined.
Let f (z) = u(x, y) + iv(x, y) be a holomorphic function of z = x + iy E D. The Jacobian J f (x, y) of f : (x, y) E D -+ (u(x, y), v(x, y)) E R is given by
Jf (x, y) _
(x, y) k (x, y) (x,y)
I7Z(x,y)
The Cauchy-Riemann equations (3.1.3) imply
(z=x+iy).
Jf(x,y)=If'(z)I2
(3.1.14)
Therefore J f (x, y) > 0 everywhere. If f'(z) 4 0, then J f (x, y) > 0, so that f gives rise to an orientation preserving mapping from (x, y)-plane to (u, v)-plane.
That is, to vectors Xj = (aj1,aj2), j = 1,2, at z we assign vectors
Uj =lad,
8u
+aj2
3u
Ot,
M
,ajIT +a, j2ft
7 = 1, 2'
at f (z). If {X1, X2} is a right-hand system (det ( x , ) > 0), then so is (U1, Uz).
X7
X1
f(z)
FIGURE 20
Now let zo E D, and assume f'(zo) 0 0. For the sake of simplicity, we assume that zo = 0 and f (zo) = 0. Let Cj (0) : [0, tj] D), j = 1, 2, be two curves such
that the limit lime....+oOj(t)/t = aj exists. Take 0, = arga,, j = 1,2, so that 0 5 8z - 81 < 27r. Then we call 0 = 8z - 81 the angle pinched by C1 and C2. (3.1.15) THEOREM. Let the notation be as above. Then the angle pinched by f (C1) and f (C2) is equal to that pinched by C1 and C2.
PROOF. Since f is holomorphic, it follows from (3.1.6) that lim
+o
f(0j(t)) = f'(0)aj = If'(0) I t
Iajle'(e;+"9 f'(o))t
Thus the angle pinched by f (C1) and f (C2) is 8z - 81. 0
3. HOLOMORPHIC FUNCTIONS
54
FIGURE 21
The property of f in the above theorem is called conformality. Hence, an injective holomorphic function f with nowhere vanishing f' is called a conformal mapping.
EXERCISE 3. Let u(x, y) and v(x, y) be totally differentiable real functions. Set f (x, y) = u(x, y) + iv(x, y). Show that if J f (z, y) # 0 and f is conformal, then f is holomorphic. EXERCISE 4. Show the following:
dze2=e',
d Iog(1 + z)
+z
zEC. z E O(1).
EXERCISE 5. Let f (z) = f (x, y) be a continuously differentiable function on a domain D C C. Let K be a compact subset of D. Fbr z E K, set
f(z+h) = f(z)+ Lz(z)h+
Lz(z)h+o(h).
Show that o(h)/h -. 0 uniformly in z E K as h -+ 0; i.e., for an arbitrary e > 0,
there is a >0 such that lo(h)/hl
L(C; (d)) = F, I0(ti) - O(tj-1)I j=1
We call each t. a partition point, and define the width I(d)I by
I(d)I = max{t; - tj_1; 1 5 j S I}.
3.2. CURVILINEAR INTEGRALS
55
Let (d') be a partition of [To, T1] such that every partition point of (d) is a partition point of (d'). Then the partition (d) is called a refinement of (d), and L(C; (d)) < L(C; (d')). We define (3.2.1)
L(C) = sup{L(C; (d)); (d) is an arbitrary partition of [To,Ti)} <+00.
Note that L(C) is invariant under the change of parameter of C. EXERCISE 1. Prove this fact.
When L(C) < oo, C is said to have a finite length. For a general curve C(¢ : I - C), we take a bounded closed interval J C I. Let 4IJ be the restriction of ¢ to J, and let CJ J denote the restriction of the curve C to J given by ¢IJ. Define
L(C) = sup(L(CIJ); J C I) < +oo. Assume that 0 : [TO,Ti] - C is a piecewise continuously differentiable curve.
Take apartition (d'):To=t'o
so that 0j[e,-j,e,] are
t1 < ... < t1 = T1 be a continuously differentiable. Let (d) : To = to refinement of (d'). Setting O(t) = 01(t) + i4 (t), we have by the mean value theorem (3.2.2)
(O(ti) - 0(ti-1))' + (02(ti) - -02(ti-1))'
I0(ti) - O(ti-1)1 =
_
(O1(t1
+81(t;
-t,_1)))2+ (02(t,_1 +02(t,
- t;_1)))2 (ti -t;-1)
(0<01<1,0<02<1)
(ti - t;-1) +o(1)(t; - t;_,). Since ml(t) and 0'2(t) are continuous on every closed interval between the par-
tition points of (d'), o(1) -+ 0 uniformly in j as I(d)I -. 0. Thus we have the Riemann integral (3.2.3)
L(C) _ jT
(O,(t))'+ (0'2(t))2 dt.
o
EXERCISE 2. Expressing the half circle C of radius r > 0 as ¢(t) = cos t+i sin t, 0:5t:5
(3.2.4) LEMMA. Let C(¢ [To, T1J - C) be a curve with finite length. i) The real function - t E [To, TiJ -, L(CI [To, t)) E 10, L(C)J
is monotone increasing, and continuous.
3. HOLOMORPHIC FUNCTIONS
56
ii) For S E [0, L(C)[ we take t E and set i(s) = t(t). Then i(s) is independent of the choice oft E jti-1(a), and satisfies I0(s) - 0(s`)I < Is - s'1
(3.2.5)
for s, s' E [0, L(C)1.
PROOF. i) The monotone increasingness is clear. Assume that V) is not continuous at a point to E [To,T1j. Then one of the following must hold: lim L(CI (To, t1) < L(CI [To, to1),
c-to-o
lim L(CI1To,t1) < L(C[[To,to)) 1-t0+0
Since the argument is the same in either case, we assume the first. We have to > To, and generally for t E [To, T1 ] (3.2.6)
L(CI[To,TT[) = L(CI[To,t))+L(CI[t,T11).
Set
Eo = L(CI [To, tot) - _!i o L(CI [To, t1) > 0.
By definition there are infinitely many tj < t f < t3+1 < tj+1 < to, j such that L(CI [tj, tj1) > 2 It follows from (3.2.6) that L(CI [To, to]) = +oo. This is a contradiction. ii) It is clear that 0(s) is defined independently of the choice of t E >G-1(s). The definition of the length of curves immediately implies (3.2.5). 0
An inequality of the type (3.2.5) with right side replaced by KIs - s'I, where K is a positive constant, is called Lipschitz' condition for gyp. In general, 0 is not a parameter change of 0, but it follows that 0(t) = 0(1y(t)).
In particular, if C(¢) is a simple curve, then is a parameter change of 0. The parameter s of 0(s) is called the length parameter, and 0 is called the length parametrization. REMARK. From now on we include the length parametrization of a curve with finite length in parameter changes.
EXERCISE 3. Show that a curve C(o) in C has finite length if and only if there is a change of parameter of C(O) satisfying Lipschitz' condition.
Let D C C be a domain and C(¢ : I --+ D) be a curve with finite length. Let f be a continuous function on D. Let I = [To, T11, and take a partition (d) : To = to < t1 < ... < ti = Ti. For Cj E [t!-1, tj1, 1 < j < 1, we set (3.2.7)
S((d);
Ef(0(ti))(0.(ti) - 0(ti-1)) j=1
3.2. CURVILINEAR INTEGRALS
57
We first take (d) and tj as follows:
T1-To.
``
Since f o 0(t) is uniformly continuous in t E I, for an arbitrary c > 0 there is a number n such that
It - t'I <
Tl
Ifo0(t)-fom(t')I <E.
-TO
Let I(d)I < (TI - To)/2n and . j be as above. If a point tnj lies in the interior of [t,-,, tl,], we add tnj to (d) as a partition point, and divide It,,-,, for the U [tnj,tl,] into two parts. We use the same t;,, E intervals [ty_1, tnj] and [tnj, tl,].
FIGURE 22
< Let tnj_1 = tjo < Thus we have a refinement (d) of (d) and tjkU) = tnj be the partition points of (d) contained in [tnj_1,tnj). Write jµ for 1 < p < k(j). Then f,, associated to rn
k(j)
f o 0(SJµ)(O(tjµ) - O(t3l. - 1)).
S((d); (C7)) j=1 µ=1
It follows that IS((dn); (tnj)) - S((d); %)) I k(j)
n
A e E f o 0(tnj)(O(tnj) - 0(tnj-1)) - E f o 0(40.)(0(t)µ) - 0(t1µ - 1)) ,,AA
,,II
µ=1
J=1
n k(j) _ F, F,(f o 0(tnj) - f
``
1
0(Sjµ))(0(t, ) - 0(t1µ - 1))
j=1 µ=1
By the choice ofjt Itnj - j,,I < (T1 - To)/n. It follows that for all j If oO(tnj) - f o0(51µ)I < E. Therefore we have
IS((dn); (tnj)) - S((d); (!j))I < L(C)E.
3. HOLOMORPHIC FUNCTIONS
58
For arbitrary m 1, m2 > 2n IS((d,,,, ); (t,n,j)) - S((d...,); (tr.,j))I
< IS((dmi ); (4n0)) - S((dn); (tj))I
+ IS((dm,); (tin,.,)) - S((d,);(tj))I < 2L(C)c.
(t j))}' 1 is a Cauchy sequence; we denote its limit Thus we see that by a. Then it follows that T1 - TO
1(d)I <
IS((d);(fi)) - at < L(C)E.
2n
That is,
a = lim
I(d)I- 0
S((d);
We call a the curvilinear integral of f along C. The limit a is invariant under the change of parameter of C. We write (3.2.8) 1c,
f(z)dz =
or
= 1(
o
S((d); (f,))
If fl and f2 are continuous functions on D, and if a1 and a2 are constants, then
f(oifi(z) + a2f2(z))dz = al This might be clear. Taking the absolute value of the right side of (3.2.7), we set S = 6i osup
I(d)I < b j=1
< max{I f o 01) L(C) < oo.
In fact, we show that 1
(3.2.9)
S= It E if o
IO(tj) - O(tj-1)I
For an arbitrary E>0wetake a60>0sothat
{It)_t'I< e,
It - t,I < 60
If a m(t) - f o (t ')I < E.
Letting 60 > 0 be smaller if necessary, we have for I(d)I < bo r
S + E > E if 0 j=1
10(tj) - m(tj-1)l
3.2. CURVILINEAR INTEGRALS
59
On the other hand, it follows from the definition that there are (0) : To = To < < Tn = T1 and (k E [Tk_1i Tk] with I(is)I < bo such that 7-1 < S-E<
if ° 0(Sk)I . Im(Tk) - OTk-01 k=1
Define b1 = min{Tk - Tk_1i1 < k < n}. Then 0 < b1 < bo. We take a partition (d) of [To, T1] and (a,) with I(d)] < b1. Let (d) be the partition formed by all partition points of (d) and (A).
FIGURE 23
If an interval [tS-1, ii] contains a E._,, we set i = tj; otherwise, we take £i E (ti-1i ti]. So, we have (ci), and by the choice we get n
t
0(ti-1)1-
if ° i=1
j=l
If ° m(oI.10(ti) -
0(tj-1)1
<3nmax{Ifo01} e; n
FL
If
0(ti-1)I - i If o 0((k)I Im(Tk) -
°.O(Si)I.10(ti) -
k=1
i=1
> -E
-EL(C}. i=1
Summarizing the above, we obtain
S + e > i If 0 0(ti)I - 10(t,) -O(t;-1)1 1=1
> S-e{L(C)+3nmax{If o0I}+1}. Hence we have shown (3.2.9). We write
S = f If(z)IIdzI C
It follows that (3.2.10)
ifc
f(z)dzl s f
c
If(z)IIdz1
m(Tk-1)I
3. HOLOMORPHIC FUNCTIONS
60
Let fn, n = 0, 1,... , be continuous functions on D. If {fn}n 0
or F " fn converges uniformly on compact subsets of D, then by (3.2.10) we have respectively fn(z)dz, lim fn(z)dz = Jim n-oo JC J n--oo 00
(3.2.11)
/
J
fn(z)dz =
J fn(z)dx.
n=0 C Let 0(t) = 01(t)+i02(t) be piecewise continuously differentiable. As in (3.2.2) it follows from (3.2.7) that n=O
+e1i(ti - ti-1))
S((d);
=1
+i¢z(ti-1 +02i(ti - ti-1))}(ti - ti-0
(0<01i<1, 0
tt
/
f °
i2(W)(ti - ti-1)
i=1
+o(1)(ti -ti_y). Here, as I (d) I
0, o(1) - 0 uniformly. Hence we get
Ic f (z)dz = JToTt f o 0(t)¢'(t)dt.
(3.2.12)
In the same way, we deduce that T
J If(z)IIdzI = C
IT.
If o0(t)II0'(t)Idt.
To
It is clear that L(-C) = L(C), and for the curvilinear integral we have
ff(z)dz = - j cf(z)dz.
(3.2.13)
For curves C1 and C2 in D such that the terminal point of C1 is the initial point of C2 (3.2.14)
J , +C,
lf(z)dz = J f(z)dz + J f(z)dz. C,
We deal only with a general curve C(O : I -+ D) such that L(CIJ) < 00 for every bounded closed interval J C I. Let f be a continuous function on D. Let {Jn}n 0 be an increasing sequence of bounded closed intervals such that 0 0 J n C Jn+1 C I with the interior J,,+1 of Jn+i, and J .0 Jn = I. Assume that the sequence
J f(z)dz, n=0,1,...,
3.2. CURVILINEAR INTEGRALS
61
converges, and that the limit is independent of the choice of such {1n }no. Then we write for the limit
lc f (z)dz' and thus the curvilinear integral of f along C is defined.
Let f : D - C be a holomorphic function, and assume that the derivative f' is continuous. (The continuity of f' will be proved in Remark (3.5.9), but is assumed here for a moment.) Let C(q : [To, TI] -+ D) be a curve with finite length. We compute the length L(C'(f o 0)) of the curve C'(f o 0). Let < t = TI be a partition of [To,T1]. For sufficiently small (d) : To = to < t1 I (d) we have by Exercise 5, § 1
If o 0(t)) - f o 0(tj-1)I j=1 n
_ F, If' O 0(tj-1)(O(tj) - 0(tj-1)) + OW(0(tj) - 0(tj-1))UI j-1 I
n
n
j=1
If' o -O(tj-1)I IO(tj) - 0(tj-1)I + o(1) E 1(0(tj) - 0(tj-1)I j=1
0. Then o(1) - 0, and so
Let I (d) I
(3.2.15)
L(C) = I If'(z)IIdzI. c
When 0 is piecewise continuously differentiable,
L(C') =
(3.2.16)
f
T1
if, o m(t)IIO'(t)Idt.
To
Let g be a continuous function on a domain D' containing f (D). In the same way as above, we deduce that
Egof 00(tj-1)(f oO(t,) -f o0(tj-1)) j=1 n
_ D9 o f 00(t3- 0 . f' o O(tj-1) +00)00)) -f o 0(tj-1)) j=1
Letting I(d)I - 0, we obtain (3.2.17)
J
c
g(w)dw =
r
Jc
g o f(z) f'(z)dz.
When 0 is piecewise continuously differentiable, (3.2.18)
fc,
g(w)dw =
Tt
To
g o f o 0(t)
o 0(t) $'(t)dt.
3. HOLOMORPHIC FUNCTIONS
62
For a function f on D a holomorphic function F such that F' = f is called a primitive function of f . It follows from Theorem (3.1.7) that if the primitive function exists, it is unique up to a constant term.
EXERCISE 4. Show that a primitive function of z', n E Z \ {-1}, is of the form zn+1/(n+ 1) + constant. (3.2.19) THEOREM. Let f be a continuous function on D, and let F be a primitive function of f . Let C be a curve with finite length with initial point zo and terminal point zl. Then
j f (z)dz = F(zi) - F(zo). In particular, if C is a closed curve,
Jc
f (z)dz = 0.
D. As in the proof of (3.2.15), we PROOF. Let C be given by 0 : [0,1] have for a partition (d) : To = to < tl . . < to = T1 with small I(d)s
F o 0(ti) - F o 0(ti-1) = (f o O(ti-1) +o(1))(O(ti) - 0(ti-1)) Therefore
=
F(zi) - F(zo)
n
E f o 0(ti-1)(m{ti) - 0(t;-0) j_1
+0(1)
n
D4(t;) - 0(ti-1)) 1=1
As I (d)
0, we get the desired equation. 0
EXERCISE 5. Integrate the function f (z) = f along the circle C(0; r) (with anti-clockwise orientation). EXERCISE 6. Integrate the function f (z) = 1/z along the circle C(0; r). EXERCISE 7. Integrate the function f (z) = Re z = (z+z)/2 along the contour of the square {z = x + iy; Ixt < 1, lye < 1) with anti-clockwise orientation.
3.3. Homotopy of Curves We summarize the necessary facts on homotopy of curves. Let D C C be a D and ¢2 : [So, St] -+ D domain of the Riemann sphere C. Let 01 : [To, T1]
3.3. HOMOTOPY OF CURVES
63
be curves with the same initial points and the same terminal points.
FIGURE 24
We say that ¢1 and 02 are homotopic (more precisely, homotopic in D), or
that ¢1 is homotopic to 02 (in D), if for 4'1(t) = ¢1(To + t(Ti - To)) and '2(t) = 02(So+t(S1 - So)) there is a continuous mapping 0: [0,1] x [0,1] -+ D satisfying
i) t ( 0 , 8)
ii) 4'(t,o) _ 4'1(t), (t,1) = 42(t), 0 < t s 1. The mapping 45 is called a homotopy connecting 4S1 and '02. The relation of being homotopic is an equivalence relation. : [To, T1 ] --+ D and ¢2 : [So, SjJ --+ D have the same initial point, and be a change of parameter of each other. Then 01 and 02 are
(3.3.1) LEMMA. Let 01 homotopic.
PROOF. We may assume that [To, T1J = [So, Sl) = [0, 1]. Then there is a strictly monotone increasing continuous function r : [0,1] -+ [0,1] such that 01(t) = 02 (74)), 0 < t < 1. Set
$(t, s) _ 01((1 - s)-r(t) + st),
0:5 t:5 1,
0:5 s:5 1.
Then 0 is a homotopy connecting 01 and 02. 0
Two curves C1 and C2 in D are said to be homotopic if C, are given by OJ : [0,1] -. D, j = 1, 2, such that ¢1 and 02 are homotopic. We see by Lemma (3.3.1) that this is an equivalence relation. We denote by {C} the equivalence class of a curve C in D, and call it the homotopy class of C. When C is a closed curve and is homotopic to a constant curve, C is said to be homotopic to a point. If every closed curve in D is homotopic to a point, D is said to be simply connected EXERCISE 1. We say that a domain D C C is star-shaped, if there is a point zo E D such that for every point z E D the line segment connecting z and zo is contained in D. In this case, show that D is simply connected. In particular, a convex domain and C \ [0, +oo) = {z E C; z is not a real non-negative number} are star-shaped. Where can zo be taken for these domains?
3. HOLOMORPHIC FUNCTIONS
64
If C1 and C2 have the same initial and terminal points, we set
C1 - C2 = C1 + (-Ca), which is a closed curve.
(3.3.2) LEMMA. For an arbitrary curve C in D, C - C is homotopic to a point. PROOF. Let C be given by ¢(0,1] --r D. Define 0{(0,1) 0111, 2)
: (0, 2] -+ D by
(0,1] 9 t --' t(t) E D,
0(2 - t) E D,
: [1,21 a t
which gives the curve C - C. Define
: 10, 21 x (0,1] - D by
' I([0,1] x 10,1])(t,s) = 0(st), $I([1, 21 x 10,1])(t, s) _ 0(2s - at).
Then 0 is continuous, 4? (t, 1) = fi(t), and 4,(t,0) - 0(0). Hence, C - C is homotopic to a point. 0
FICuRE 25
(3.3.3) THEOREM. Let Co, C1 be two curves in D which are homotopic. Then Co - C1 is homotopic to a point.
D be a homotopy connecting Co and C1. [0, 1) x [0,1] PROOF. Let a). Then io = +(0, s) (resp., zI =- 4(1, s)) is Let C. be curves given by the initial (resp., terminal) point of every C,. We define a continuous mapping :
9:[0,2)x[0,1/2)-.Dby (F1[0,1) x [0,1/21)(1, s) = 6 (t, a),
('I'I[1, 2) x (0,1/2))(t, s) _ +(2 - t,1 - s). 1
i 2
0
1
FIGURE 26
2
3.3. HOMOTOPY OF CURVES
65
It follows that Co - C1 is homotopic to C1/2 - C1/2. Then by Lemma (3.3.2) Co - C1 is homotopic to a point.
By making use of the complex coordinate z (= 1/z) about oo E C, the curves connecting zo E C' and co given by
ztzo, z(1-t)zo,
05t<_1, 0 =
where zo = 11zo, are called line segments connecting zo and oo. z
a'
z',=I/z. 0
FIGURE 27
A piecewise linear curve in C is defined as a finite sum of line segments in C and line segments connecting points of C' and oo. A constant curve is considered as a special case of a piecewise linear curve. (3.3.4) LEMMA. Every curve C(0 : [0, 1) - D) is homotopic to a piecewise linear curve in D. PROOF. By making use of the uniform continuity of 0, we may take a partition
0 = to < tl < . . . < t = 1 satisfying the following: (3.3.5)
For every j,1 5 j < n, there is a disk neighborhood Aj in C or a disk neighborhood Aj of oo such that (5([tj_1,tj1) C Aj or m([tj_1itjJ) C Oj. Here, when fa([tj_,,tj]) C Oj, 0(tj_1) or !¢(tj) is assumed to be oc.
FIGURE 28
Connecting 0(to), 0(t1), ... , 0(tn) in order by line segments, we have a piecewise
linear curve C. We show that C and 0 are homotopic. It is sufficient to show
3. HOLOMORPHIC FUNCTIONS
66
that C1 [t j _ 1, t j 1 and the line segment from 0(t, _ 1) to 0(t j) are homotopic. If
CI[tj_1,ti1 C Oj, we set
4(t,s) _ (1
+sS t, t4 t't 0(ti-1)+ t-, - t 11 1
tj-1t
ti ] C A., we form a homotopy
$ by making use of z in the same way as above. 0 (3.3.6) THEOREM. There are at most countably many homotopy classes of curves connecting two arbitrarily given points of D.
PROOF. Let C be a curve connecting the given points of D. We use the same notation as in the proof of Lemma (3.3.4). If .0(ti) with 1 < j < n - 1 is not oo, we take a rational point zj sufficiently close to QS(ti) so that the piecewise linear curve C' connecting zo = 0(0), zl,.... z., = 4(1) in order by line segments is homotopic to C. The set of rational points is countable, and hence so is the set of such curves C. Therefore the set of homotopy classes is at most countable. The following lemma will be used in the next section. (3.3.7) LEMMA. Let 0 : [0,11 x [0,11
D be a continuous mapping. Then there
areab>0,apartition
disks Oj,1<j=< n,inD
(some of them may be disks about oo) such that 0([tj-1,tj] x [0,61) C Oj. PROOF. Applying (3.3.5) to 0(t,0), we may take 6 > 0 so that
0([tj-i,tj1 x (0,61) C oj,
1 <_ j < n.
The next theorem is called Jordan's theorem. We will not use it, but it will be helpful to know that such a theorem holds. (3.3.8) THEOREM. A Jordan closed curve in C divides C into two domains. One of them is a bounded simply connected domain.
EXERCISE 2. Show that if a closed curve C in a domain D is homotopic to a point, then so is nC with n E Z. Here OC is considered as a constant curve mapped to the initial point of C. EXERCISE 3. Let D = C \ (0, 11. Set C1 = C(0;1/2) and C2 = C(1;1/2), let the initial (=terminal) points of them be 1/2, and let the orientation be anti-clockwise. Show that CI + C2 is homotopic to Cs + C1.
3.4. CAUCHY'S INTEGRAL THEOREM
67
3.4. Cauchy's Integral Theorem Let D be a domain of C, and let f be a holomorphic function on D. Let Co denote the perimeter of a closed triangle Eo with vertices, zz, 1 <= j < 3, which is a closed piecewise linear curve formed by the sides of Eo f (z )dz = 0.
(3.4.1) LEMMA. If Eo C D, then
PRooF. Let Co be endowed with the anti-clockwise orientation, and let z1, z2 and z3 be ordered anti-clockwise, too. Let Cj be the side from z., to z,+1
(1<_j<3, z4=z1)
FIGURE 29
Then CO = Cl + C2 + C3.
Taking the middle point of each C. and connecting them by line segments, we divide EO into four closed triangles, Ell, E12, E13, E14. Let Clkl, 1 < I < 3, denote the sides of Elk as in the case of C;. The closed piecewise linear curves
1:5 k:5 4,
CIA, = Clkl + Clk2 + C1k3,
form the perimeter of Elk. It follows from (3.2.13) and (3.2.14) that (3.4.2)
f f (z)dz = E
ok=1 C1k f (z)dz.
Repeating this process n times, we have 4" closed triangles Enk (1 < k < 4n) and their perimeter Cnk, and as in (3.4.2) we obtain (3.4.3)
I
f(z)dz = F
JCo
k=1
f.f
(z )dz.
k
Now we assume that a = L1.0 f (z )dz>
Then it can be deduced from (3.4.2) that for some Clk,
k
f(z)dzi > 4 k,
3. HOLOMORPHIC FUNCTIONS
68
Applying the same argument to Elk, and Ciki, we have Elk, and C2k, such that E2k2 C Elk, and
a
IC3!.2
f(z)dz > 4z
with perimeter Cnk,,, so that
We have inductively Enk C
f (z)dzl > a = 4n' For the sake of simplicity we write E', = Enk and C, = Ckn, and so (3.4.4)
fc f(z)dz
4n ,
n = 1, 2, ...
.
Letting M = max{L(C,);1 < j < 3}, we have
L(CC)<2M.
(3.4.5)
The diameter R of E;, (R.1 = max{ jz - wI; z, w E E'n }) satisfies
M
Rn=2n
(3.4.6)
Since E;,, n = 1, 2, ... , are decreasing compact sets, nO=1 E,, 0 (cf. Exercise 7, Chapter 1). Let a E nnO E. It follows from (3.4.6) that R. - 0 (n -- oo), and so f °° E,, = {a}. Since f is complex differentiable at a, for an arbitrary 1
1
e>0there isab>0such that (3.4.7)
If (z) - If (a) + f'(a)(z - a)}1 < ejz - al,
z E 0(a; b).
Take no so that M/2n° < b. Then, for n ? no we have by (3.4.5)(3.4.7) and (3.2.10) (3.4.8)
f f(z) - j{f(a)
< j Iz-aldzI
+ f'(a)(z -
a)}dz;,
L(C)
The function If (a) + f'(a)(z - a)} has a primitive function f(a)z + f 2a)(z - a),,
and so Theorem (3.2.19) implies
Jc^ If (a) + f'(a)(z - a)}dz = 0.
.
3.4. CAUCHY'S INTEGRAL THEOREM
69
It follows from (3.4.8) that f(z)dzeM2. I
'
This with (3.4.4) yields a/4" < 3&M2 /4n, and hence a < 3M2e. Since e > 0 is arbitrary, we have a contradiction. 0 REMARK. A good point of the above proof is that it is not necessary to assume
the continuity nor the integrability of the derivative f'(z) of f(z). (3.4.9) LEMMA. On an arbitrary disk 0(a; r) C D, f has a primitive function. PROOF. For a point z E A(a; r) we define a curve CZ by
0:tE(0,11-+a+t(z-a)EO(a;r), and set
F(z) = fc* f ( ()d(. We show that F(z) is a primitive function of f (z).
FIGURE 30
Fix an arbitrary point zo E 0(a; r). For an arbitrary e > 0 there is a 0 < 6 < r - Ial such that (3.4.10)
If (zo + h) - f (zo) I < e,
IhI < 6.
Let rh denote the line segment from zo to zo + h. It follows from Lemma (3.4.1) and (3.4.10) that
F(zo + h) - F(zo) h
=
I
_ f (zo)
f f(z)dz-f(zo)I =lh,h(f(z)-f(zo))dz h
< IhI f If(z) - f(zo)IIdzl < I' L(I'h) = C ti
Therefore F is holomorphic, and F' = f .
0
3. HOLOMORPHIC FUNCTIONS
70
Let C(V : [0; 1] -' D) be a curve. As in the proof of Lemma (3.3.4), there is < t,, = I of [0,1] fine enough so that (3.3.5) holds. a partition (d) : 0 = to < Then the piecewise linear curve C(d) connecting W(to), ..., and p(t,,) in order by line segments is homotopic to C. (3.4.11) LEMMA. The curvilinear integral fC(d) f(z)dz is independent of (d), provided only that (3.3.5) is satisfied. PROOF. It is sufficient to prove that for a refinement (d) of (d) (3.4.12)
Jc{d
f (z)dz =
f (z)dz.
For we see by (3.4.12) that for two partitions (dl) and (d2)
J
G'(f(z)dz = (
fd3)
f(z)dz = f(dz) f(z)dz,
d1)
where (d3) is the partition formed by all partition points of (dl) and (d2). To show (3.4.12) it is sufficient to deal with the cage where one partition point tJ is
added to (t,,t,fl). at,)
r
Pu")
FIGURE 31
In this case we denote by E the triangle with vertices cp(t,), :p(t;), fp(tj+1), and by r the perimeter of E. Lemma (3.4.1) implies
Id.) f (z)dz - fe(d) f (z)dz = / f (z)dz = 0.
0
By Lemma (3.4.11) we can define the curvilinear integral off along C by (3.4.13)
f f(z)dz = J (d) f(z)dz,
where (d) satisfies (3.3.5). The next theorem is called Cauchy's integral theorem: (3.4.14) THEOREM. Let f be a holomorphic function on D, and let C be a closed curve in D which is homotopic to a point. Then
jf(z)d.z=0.
3.4. CAUCHY'S INTEGRAL THEOREM
71
PROOF. Let C be given by p : [0,1[ - . D, and set zo = ap(0) = v(1). There is a homctopy
45:[0,1]x[0,1]
D
connecting W and zo. Then ap(t) = 45(t, 0), and 45(t,1) _- zo. Denoting by C. the curve given by 45,(t) = 45(t, s), we define
A=sup{8E[0,1]; J f(z)dz=J f(z)dz, 0Sa'5s} c
llC.,
ll
As in Lemma (3.3.7) we take 6 > 0 and 0 = to <
JJJ
< to = 1. Let r, denote
the piecewise linear curve connecting 0(t,, 0), 4s(ti, s), 0(ti+1, 45(ti, 0) in that order.
0), and
e(t,. o)
(t,- 0)x- -:r(t,_1, a)
FIGURE 32
We infer from the definition of r;, Lemma (3.4.9), and Theorem (3.2.19) that
f f(z)dz = 0. The line segment connecting 45(t3, 0) and #(t a) has the opposite orientation in
r,_1 and r,, so that n-1
fc.
f(z)dz -
fc
f(z)dz = E f f(z)dz = 0. , j =O
Thus A ? 6. If A < 1, there would be a small 6' > 0 such that, as in Lemma (3.3.7), 0([ti_1i t,I x [a - 6, A + 6'J) is contained in a disk. Then f(z)dz = fc f(z)dz.
f(z)dz = fCA-$1
Ca+s,
Therefore A > A+6', which is a contradiction. We have A = 1. Taker > 0 so that
A(zo;r) CD. There is a6">0such that C,CO(zo;r)forall 1-6"<_s<1. Take C,(d) C s(zo;r) as in (3.3.5). We infer from Lemma (3.4.9) and Theorem (3.2.19) that f(z)dz =
1C. Hence fc f(z)dz = 0.
C.(d)
f(z)dz = 0.
3. HOLOMORPHIC FUNCTIONS
72
By Cauchy's integral theorem we find that the curvilinear integral along a curve C in D,
Jf(z)dzdepends only on the homotopy class of C in D. In fact, if C, and C2 are mutually homotopic, then Theorem (3.3.3) implies that Ci - C2 is homotopic to a point, and that
J
f(z)dz = J
f(z)dz -
JC,
,
f(z)dz = 0.
C, -C,
In particular, if D is simply connected, the curvilinear integral fe f (z)dz is determined only by the initial point zo and the terminal point z1 of C. Thus we write , f (z)d-- = / f (z)dz.
/
C
Fixing zo E D, we define
F(z) =
r
J
sl f (z)dz.
We see by Lemma (3.4.9) that F is holomorphic and a primitive function of f. We call f Z' f (z)dz the indefinite integral of f . Therefore we obtain the following:
(3.4.15) THEOREM. A holomorphic function f on a simply connected domain D has a primitive function F; moreover, for affixed point zo E D
: F(z) = F(zo) + / f (z)dz. EXERCISE 1. Let f (z) = 1/(z + i)2, and R > 0. Show that fjo Rj f(z)dz and f[-Ro f(z)dz converge as R -. +oo, and that fR f (z) = Rlim
it- R,R]
f (z)dz = 0.
EXERCISE 2. On A(1), find a primitive function of f (z) = (z2 -1)-2 in power series.
3.5. Cauchy's Integral Formula Let C(a; r) be the perimeter circle of the disk 0(a; r) given by 0: 9 E [0, 2a] -+ a + te'B E C(a; r).
(3.5.1)
f
(3.5.2) THEOREM. J
C(a:r) z - a
dz = 2iri.
PROOF. Set z = 0(9) = a + re'°. Since qV(O) = riese, we have by (3.2.12) r 1 r2* rie; e O J dz= 1 d6=2rri. C(a:r) z - a 0 77C.6
3.5. CAUCHY'S INTEGRAL FORMULA
73
Let D C C be a domain, and a E D. We define d(a; 8D) = inf { Ia - wI; w E 8D} < +oo.
If d(a; 8D) < +oo, there is a w E 8D with d(a; 8D) = is -wl. The next theorem is called Cauchy's integral formula (3.5.3) THEOREM. For an arbitrary a E D and 0 < r < d(a; OD)
f (z) =
1
f (S) dC,
.
C(a;r) l; - z
27ft
z E 0(a; r).
PROOF. For the sake of simplicity we may let a = 0. Let z = Izle'BO E A(r). Take a 6 > 0 with 0(z; 6) C i(r). Let Cl = C(0; r), let C2 be the line segment from z+6e`BO to rei00, and let C3 = C(z;6). We may assume that the initial and terminal point of C, is reie0, and that of C3 is z+6e'e0. Set C = Cl -C2 -C3+C2. Then C is homotopic to the curve C given in Figure 33.
FIGURE 33
Let Cbe given by ¢: 10, 1] - D, and let 0: (t, s) E [0, 11 x [0,1[ - (1 - s).3(t) - sreieO E D \ {z}
be a homotopy connecting C and the point -re'B° in D \ {z). Hence C is homotopic to a point in D \ {z}. Since the function f in D \ {z}, it follows from Theorem (3.4.14) that
z) is holomorphic
Jid(=0. C - z
f C)=f f.
One obtains
J
,
C-z
C, ( - z
Since
f
f2x
f(C)
3C-z
o
f (z + bee)
(z+beie)-z
f2w
=
f (z
+ 6eie)id8
2xif(z)
(6
0),
3. HOLOMORPHIC FUNCTIONS
74
one gets
J
27ri
Fix z E 0(a; r). Ash
J h
, C
("
0
d(= f (z).
0, the following convergence is uniform in ( E C(a; r):
_ f(C) I
f(C)
(-(z+h) (-z
f(()
_4
Thus we deduce the integral formula for the derivative off from Theorem (3.5.3):
f,(z) =
1 f(C) 2;i- 1C(o;.) (C - Z)2 4,
z E 0(a;r).
Repeating this n times, we find that the n-th derivative f (") (z) is expressed by f(n)(z)
(3.5.4)
f(() r = n) 17r, z)"+, dC C(a;r) ((-
zE0(a;r), n=0,1,.... E C(a; r) one has
For z E A(a; r) and 1
_
1
1
_
00
1
fz-a
(-any
a
Here, note that
{ z-a l <
(-a
r
l
<
Therefore the above power series converges uniformly in ( E C(a; r). It follows from (3.2.11) that 00
f(z) = E
(3.5.5)
a)",
n-0
an =
(3. 5 . 6)
21ri
((
f)
>+t
1
27rrn o
Since the right side of (3.5.5) converges for z E A(a;r), its radius R of convergence is not less than r. Letting r / d(a; 8D), we get R > d(a; 8D). Hence we have the following:
(3.5.7) THEOREM. A holomorphic function f on D is analytic, and about an arbitrary point a E D f (z) is expanded to a power series 00
A z) = E an(z - a)n,
z E 0(a; d(a; 8D)),
n=0
whose radius of convergence is not less than d(a;BD).
3.5. CAUCHY'S INTEGRAL FORMULA
75
One sees by (3.5.6) and (3.1.13) that (3.5.8)
f(")(a) = n! JC(a;r)
(( - a)
= n!a".
(3.5.9) REMARK. We see by Theorem (3.5.7) and Corollary (3.1.12) that holomorphic functions and analytic functions are the same. Therefore theorems for analytic functions also hold for holomorphic functions. It follows from (3.5.4) and also from Theorem (3.1.10) that holomorphic functions are complex differentiable arbitrarily many times, so that the derivative of a holomorphic function is again holomorphic. The identity Theorem (2.4.14) is valid for holomorphic functions. Conversely, the theorems for holomorphic functions hold for analytic
functions. Thus the composite of analytic functions is analytic (cf. Theorem (3.1.9)).
(3.5.10) THEOREM. Let f and g be holomorphic functions on D. If f(")(a) _ g(") (a), n = 0, 1, ... , at a point a E D, then f - g on D.
PROOF. Theorem (3.5.7) and (3.5.8) imply that f - g on 0(a; d(a; OD)). One deduces from Theorem (2.4.14) that f - g on D. 0 Now we return to the logarithmic function log z, which was defined only for z E A(1;1) by (2.5.4). Let C be a curve from 0 to z 36 0, not passing through 0, and define (3.5.11)
Fo(z) =
1
do c
If D is a simply connected domain containing 1, but not 0, and if C C D, then Fo(z) restricted to D gives rise to a holomorphic function on D. A direct computation yields Fo") (1) = log(") (1), n = 0,1, ... , so that by Theorem (3.5.10) Fo(z) _- logz on A (l; 1). Therefore (3.5.12)
eF°(`) = z,
z E A(1;1).
Hence we call the function Fo(z) defined by (3.5.11) the logarithmic function, and denote it by log z. The logarithmic function log z is defined for z ,- 0, but not a function such as those we have dealt with before. It is a so-called multivalued function (in contrast, the functions dealt with before are called 1-valued functions). One deduces from (3.5.12) and identity Theorem (2.4.14) that (3.5.13)
e1Og = = Z.
Therefore log z is determined modulo 2ai, and
logz=log(zj+iargz.
3. HOLOMORPHIC FUNCTIONS
76
Let D C C' be a simply connected domain, and fix a point a E D. Take a curve CI from 1 to a. For an arbitrary point z we take a curve C2 in D from a to z. Then d( log; z= +J d( J, +c, C Cl C c2 ( When CI is fixed, this integration is independent of the choice of C2, and one may write
_(
logz=f(z)=
JCl
'+J I
I
This defines a 1-valued function on D, which is called a branch of log z. (The general notion of branch will be given in Chapter 5, §2.) If CI is changed, one gets another branch g(z). Since ef(=) = e08f, there is an n E Z such that f (z) -- g(z) = 2mri,
z E D.
That is, the difference of two branches of log z is an integral multiple of 27ri. Let a E C, and let C be a curve not passing through a. The rotation number of C around a is defined by
n(a;C) = 2xi 1 c C-a
E Z.
(3.5.14) THEOREM. A curve C in C \ {a} is homotopic to a point if and only if n(a; C) = 0.
PROOF. We may assume that a = 0, and by Lemma (3.3.4) and Theorem (3.4.14) that C is a piecewise linear curve. The "only if" part is clear. Assume that n(0; C) = 0. Let C be given by p : (0,1] - C'. We may assume p(0) = 1. Set
I
E (0,11 x [0,1] -. (1- 8)jP(t) +
FIGURE 34
IwtI
8 E C.
FIGURE 35
Then W is homotopic to 0(0 : (0,11 - C(0;1)) with 0(t) = $(t,1). Since the piecewise linear curve C may be changed within its homotopy class, one may
assume that IMO(t1) lm*(t2), tI < to < t2, in a neighborhood of to with 0(to) = I. L e t 0 = to < tl < .. < tt = 1 be all the points t E (0,11 such
3.5. CAUCHY'S INTEGRAL FORMULA
77
that ra(t) = 1. Let e > 0 be taken small, so that 01[tj_1,tj) does not intersect {x E R; x > 0). Since C \ {x E R; x > 0) is simply connected, a branch of log z is determined. Then '!S
= logV/(tj - e) - log '(t1_1 + e)
(e - 0).
-+ 0, ±2iri
Thus
[e,) <
=0,±27ri. By the condition we
have
t
=0.
(3.5.15)
0, then Ct[tj-1,tj] Let Co = C(0;1) be as defined by (3.5.1). If 0I[tj_1,tjI is = 27ri (reap., -21ri), then is homotopic to a point. If fcl[t, I't') homotopic to Co (resp., -C0). By (3.5.15) the numbers of C1 [tjtj] homotopic to Co and to -Co are equal. Since Co - Co is homotopic to a point, so is C. p l om now on, let D C C be a general domain. (3.5.16) THEOREM. Let f be a holomorphic function on D. Let C be a curve in D homotopic to a point, and let z E D not be on C. Then n(z; C)f (z) = 27ri PROOF. Set
9(4} -
A0
J
_ !(z)
(()
.
(ED\{z}.
Expand f (() in a neighborhood 0(z; r) C D of z: 00
!(() _
an(( - z)',
ao = !(z)
n=o
Then one has 00
(3.5.17)
E A(z;r) \ {z}.
g(() = fan(( - z)"-1, n=1
Since this right side converges in A(z;r), g(() defined by (3.5.17) on A(z;r) is holomorphic on D. Hence
Jcj$C -J c
i(zzcK
=Lg(()d(=0.
By the definition of n(z; C) one gets the desired formula. 0 The integral formula in the above theorem is called Cauchy's integral formula,
too. The next gives the inverse of Cauchy's integral Theorem (3.4.14), and is called Morera's theorem:
3. HOLOMORPHIC FUNCTIONS
78
(3.5.18) THEOREM. Let f be a continuous function on D. If for the perimeter C of an arbitrary closed triangle contained in D
Ic f (z) = 0, then f is holomorphic. PROOF. Let a E D. It is sufficient to show that f is holomorphic in A(a; ro) with ro = d(a; 8D).
FIGURE 36
For an arbitrary point z E 0(a; ro) we take the line segment C,z from a to z and set
F(z) = f f (()d(+
By By making use of the assumption one sees as in the proof of Lemma (3.4.9) that
F is holomorphic. Since F' = f, f is holomorphic. (3.5.19) THEOREM. Let {f,}0 be a sequence of holomorphic functions on D which converges uniformly on compact subsets of D. Then the limit function f is holomorphic and f(k)(z) = lira f(k) (z), ,1 -*
zED, kEZ+.
Here the convergence is uniform on compact subsets of D.
PROOF. Let C be a curve as in Theorem (3.5.18). Then jf(z)dz =
limf,(z)dz = lim J fn(z)dz = 0.
Jc
Thus by the same theorem f is holomorphic.
c
3.5. CAUCHY'S INTEGRAL FORMULA
79
Let a E D and 0 < rl < r < d(a; 8D). It follows from (3.5.4) that for z E 0(a;rl) f(k)
(Z)
)
= k!
f (C)
27rt JC(a:r) (C - Z)k+l k!
27ri
f
n (0
d( lim (a;r) n-oo (C - Z)k+l
k!
lim n--- 27ri
! J (a;r) ((- Z)k+1 d f. (C)
lim fnk) (Z).
n-oo
Here the convergence is uniform in z E A(a; r1). 0 (3.5.20) THEOREM. Let f (z) _ >n°_0 anz" be a holomorphic function on A(R). For 0 < r < R we have2*
i) Ianl < 21rr"
If(re's)Ide,
00
2a
ii) E
Ianl2r2n
n=0
=
f If (re'a)2Id6. 27r
PROOF. i) This follows immediately from (3.5.6). ii) Noting that E 0 Ian 1r" converges, we have 2x
1
j2,
+mi(nm)9
If (Te e)I2d9 = 1 f
0
n.m=o
°O
= E andm n,m-0
rn+m f2x 21r
ei(n_m)ed9.
o
If n 96 m, f0" ei("-m)Bd@ = 0, and so the desired equality follows. 0 The next theorem is called the maximum principle.
(3.5.21) THEOREM. If a holomorphic function f on D takes the maximum at a point of D, then f is constant. PROOF. Assume that f takes the maximum at z0 E D. Let 0 < r < d(z0; 8D). Expand f (z) = En_0 an(-- - z0)", z E A(z0; 8D). Since la0l = If (zo)I, it follows from Theorem (3.5.20), ii) that IaoI2 +
0C
Ianl2r2n < Iaol2.
n=1
Hence an = 0, n >_ 1, so that f is constant by Theorem (3.5.10). 0 The next is called Liouville's theorem: (3.5.22) THEOREM. A bounded holomorphic function f on C is constant.
3. HOLOMORPHIC FUNCTIONS
80
PROOF. Expand f (z) about 0 to a power series:
x f(z) = F, anzn n=0
Then the radius of convergence is +oc. There is an M > 0 such that if (z) I < M for z E C. For an arbitrary R > 0 one has by Theorem (3.5.20), i)
la"I
n=0,1,....
Letting R -+ +oo, one gets a" = 0 for n > 1. Thus f is constant. p We get the following fundamental theorem of algebra due to Gauss as an application of the above theorem. a,1zn necessarily has a zero if it is not constant; i.e., there is a z0 E C with f (zo) = 0. (3.5.23) THEOREM. A polynomial f (z)
PROOF. Suppose that f (z) does not have a zero on C. Then g(z) = 1/ If (z) is holomorphic on C. We may assume ON 0 0, N >_ 1. Taking a sufficiently large
R0, we have forIzI>Ro {
aN +
aN-1 z
+ ... +
ap ZN
I
IaNI
z +... + -I aN-'
ao I ZN
1
2 IaNI.
Thus, for Izi > Ro, I9(z)I < 2/IaNIIzIN < 2/IaNI4. Since g(z) is bounded on A(Ro), g is bounded on C, and so constant. This is a contradiction. 0 EXERCISE 1. Let f be a continuous function on 0(R) which is holomorphic in A(R). Show that f (z) =
1 J f d(, 21ri C(o;R) (- z
z E 0(R).
EXERCISE 2. Let (f}`=0 be a sequence of continuous functions on 0(R) which are holomorphic on A(R). Assume that {fn} ' converges uniformly on the circle C(0; R). Show that {f,,}'0 converges uniformly on A(R). EXERCISE 3. Let R > 0, R 0 1, and let C denote a curve from R to -R on C(0; R) with anti-clockwise orientation. Show that
r
Ic
1
2 arctan R
1+z2dz- 7r-2arctanR
(0 < R< 1), (R> 1).
EXERCISE 4. Let f (z) be a holomorphic function on C satisfying If W1 <_ MIzI",
IzI > R
where M > 0 and R > 0 are constants. Show that f (z) is a polynomial of degree at most n.
3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS
81
EXERCISE 5. Let f be a continuous function on A(1) which is holomorphic in A(1). Assume that f (eie) is constant on an arc {eie; 0 < 81 < 0 < 02 5 27r}. Then, show that f (z) is constant on A(1). EXERCISE 6. Let f be a holomorphic function on D. For an arbitrary point w E f (D), show that the inverse image f -1(w) is either the whole of D, or a discrete subset of D.
3.6. Mean Value Theorem and Harmonic )Functions Let D C C be a domain, let f be a holomorphic function on D, and let a E D. By Theorem (3.5.3) we have f2w (3.6.1)
f(a) =
0 < r < d(a;BD).
f(a + reie)dO,
2w
This is called the mean value theorem for holomorphic functions. We set a = 0 and r = 1 for the sake of simplicity in what follows. The value f (O) of f at the center 0 of the disk A(1) is expressed by the integration of f over the boundary circle C(0;1) of A(1). We consider a similar expression for a point of A(1) other than the center. Let z E A(1) and take O_(() =
< E A(1).
z(+1 '
and 0j 1 are injective and surjective mappings As seen in Theorem (2.8.11), between A(1), and holomorphic in A(1/IzI), a neighborhood of A(1). Therefore f o 0.1(() is holomorphic in a neighborhood of A(1), and fo.=1(w)dw.
fo0=1(0)= 1f 2ari
W
C(o;l)
Changing the parameter by w = z((), C = eie, 0 < 0 5 tar, one gets from (3.2.18)
j2* 1
(3.6.2)
f(e,e),O:(e'e)ieed9 4,/ (e1e)
2ari Jo Tar
f
2x
f(e'e) We
IzZI2 dB.
(3.6.3) THEOREM. Let f be a continuous function on A(a; R) which is holomor-
phic in A(a; R). Then
f2,r P z) = tar
f (a + Re`s) I
2 _I(z
aa)12
'
z E A(a; R).
3. HOLOMORPHIC FUNCTIONS
82
PROOF. By the change of variable w = (z - a) /R, it is reduced to the case where R = 1 and a = 0. Let z E A(1) be an arbitrary point, and take 0 < r < 1. Applying (3.6.2) to g(C) = f (r(), one obtains r2A f(reis)I
f(rz) = 2_
IZ12
d$.
Letting r -+ 1, we have the desired formula. 0 The function of ( and z
K((' z) -
2_ z2
I
IC
- ZI2
C # z,
> 0'
is called the Poisson kernel. The formula of Theorem (3.6.3) is written as (3.6.4)
f(z) =
1f 2R
f(C)K(C - a,z - a)dO. (a;R)
This is called the Poisson integral.
Let a = 0 and set f (z) = u(z) + iv(z). Then it follows from (3.6.4) that (3.6.5)
1 u(C)K(C,z)d8. Za f (a;R)
u(z) =
The same holds for v(z). Since
K((,z)=ReC+i,
(3.6.6)
the function u(Z)
2,r C(O;R)
U(()
C+z
C-z is holomorphic on A(R), and Reu(z) = u(z). The above integral is called the complex Poisson integral. Thus we have (3.6.7)
f (z) = u(z) + i Im f (0).
The second order differential operator (3.6.8)
A=(+) =4&&=4&8:
is called the Laplacian. Since 3.f = 0, we have the so-called Laplace equation (3.6.9)
4u=0.
A real-valued C2-function u(z) = u(x, y) which satisfies the partial defferential equation (3.6.9) is called a harmonic function. The real and imaginary parts of a holomorphic function are harmonic. Conversely, for a given harmonic function u(z, y) we will construct a holomorphic function f with real part u.
3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS
83
Let D be a simply connected domain. Let u(x, y) be a harmonic function on D. Then 8=u is holomorphic by (3.6.8) and (3.6.9). Fixing a point a E D, by Theorem (3.4.15) we have a holomorphic function
f(z)=J
=28=udz+u(a), a
which is a primitive function of 8,u. Let 0 < r < d(a; 8D). Then f (a + reie) = / r 28=u(a + te'9)e'9dt + u(a)
=/ (cos8+sm9)dt 0
+
Jr( sine-
clu
cose)dt+u(a).
Setting a = al + ia2, one gets
Ref (a + rei8) = f
u(aI + t cos 0, a2 + t sin e)dt + u(a)
= u(al + r cos e, a2 + r sin 9) - u(al, a2) + u(a)
= u(a + reie)
Set D' = {z E D; Ref (z) = u(z)}. Since D' a, D' # 0. Clearly, D' is closed. To conclude that D' = D, it is sufficient to show that D' is open. Let b E D' be an arbitrary point. For z E 0(b; d(b; 8D)) f (z)
n
28,udz + u(a) z
rb
a
28,udz +
20,u(z)dz + u(a). b
The above computation and the assumption b E D' yield
Ref (z) = Re f.b 28,
+ u(a) + Re
rz
28,u(z)dz
= u(b) + u(z) - u(b) = u(z). Hence A(b; 8D) C D', and so D' is open. (3.6.10) THEOREM. i) For a harmonic function u on a simply connected domain
D, there exists a holomorphic function f = u + iu' on D with real part u, which is unique up to a purely imaginary constant. ii) Let u be a harmonic function on a domain D. If ulU _ 0 on a non-empty open subset U C D, then u = 0 on D.
3. HOLOMORPHIC FUNCTIONS
84
iii) Let u and D be as in ii). Let a E D, and 0 < r < d(a; 8D). Then for z E A (a; r)
u(z) =2ir1 J
u(()K(( - a, z - a)dO,
(= a + reiB
( ax)
In particular,
u(a) = 2n f,(a;r) u(a + reiB)d6
(the mean value theorem).
iv) Let u be as in ii). If u is not constant, then u does not attain the maximum nor the minimum in D.
PROOF. i) It remains to show the uniqueness; this follows from Theorem (3.1.8).
ii) Let D' denote the set of points z E D such that there is a neighborhood V C D of z with ujV - 0. By definition, D' is open. Since D' D U, D' # 0. Let a E D' n D. There is a point b E D' such that a E 0(b; d(b; 8D)).
FIGURE 37
Let f be a holomorphic function on i(b;d(b;BD)) such that Ref = u. Since b E D', f is constant, so that uIA(b; d(b; 8D)) = 0. Thus a E D'. Since D is connected, D' = D. iii) This follows from i) and (3.6.5).
iv) Assume that a attains the maximum value at a E D. (For the minimum, it suffices to take -u.) For 0 < r < d(a; 8D) 1
2r.
2n
f(u(a) - u(a + reio))dO 2'
1
= u(a)
-j
u(a + reiB)dO = 0.
0
Since u(a) - u(a + rei°) is continuous in 0 and non-negative, u(a + reiB) = u(a),
0 < 0 < 27r.
Therefore uI0(a; d(a; 8D)) _- u(a), and it follows from ii) that u - u(a) on D. This is a contradiction.
3.6. MEAN VALUE THEOREM AND HARMONIC FUNCTIONS
85
The function u' in i) is called the adjoint harmonic function of u, and it is unique up to a constant term. EXAMPLE. a) The adjoint harmonic function of x is y. b) The adjoint harmonic function of log jzj is arg z.
EXERCISE I. Show that x2 - y2 is harmonic. What is its adjoint harmonic function?
The importance of harmonic functions for the study of holomorphic functions is made clear by the existence of adjoint harmonic functions. Generally speaking,
it is easier to construct a harmonic function than to construct a holomorphic function. The problem of finding a harmonic function on a domain D with prescribed boundary values on 8D is called the Dirichlet problem. We solve it for a disk:
(3.6.11) THEOREM. Let h be a continuous real valued function on C(0;1). Then the real valued function defined by u(z) = j27r h(eae)K(e:B, z)dO
2rr
is harmonic on A(1), and for every point e" E C(0;1)
lim, u(z) = h(e't). PROOF. It follows from (3.6.6) that u(z) is the real part of the holomorphic function
f(z) =
1 27r
f
h(eie)e
+zdd
e
(o:1)
ie,
z
and hence harmonic in A(1). Applying (3.6.5) to u = 1, one gets
f2r 2>r
h(e`O)K(e`$,z)d6.
Noting that K(eie, z) > 0, we have for a small 6 > 0 Ju(z) - h(e't)1
If 2L
s,+
{h(eie)
o
<_ 1
- h(e't)}K(e'6, z)d9IIII
I
rt+6 Ih(e'e) - h(e:t)jK(e,e z)dO
27r t-
rt+xJ
+J + I (ft-6 -r
I Ih(e'B) - h(eit)IK(e'e,z)d6.
e+6
For an arbitrary e > 0 there is a small 6 > 0 such that
jh(eie)-h(e't)l <e,
t-6<_0
3. HOLOMORPHIC FUNCTIONS
86
FIGURE 38
6 and I argz - tI < 6/2
On the other hand, for 10 - tI
K(eie z) =
1 - I=12
(1 - IzI)2 +4lzlsine 8-2 1-IZ12 4IzI sine
Thus it follows that for I arg z - tI < 6/2 Iu(z) - h(e'`) I <
2n fe t ab K(e*, z)d9 +
ma)'
IhI
4IzI
sinee6 (2a - 2b) 4
<e+1(maxlhl) 1-IzI2
Izlsin +'
2
The above second term is less thane for IzI sufficiently close to 1. Therefore there is a 6' > 0 such that Ju(z) - h(e")I < 2e, so that limx_e,i u(z) =
h(e").
z E a(1) fl A(e,'; b'),
0
REMARK. By the above proof one sees that if h is bounded, h(e';) holds at a point e" where h is continuous. EXERCISE 2.
u(z) _
Let D and D' be domains of C, and let f : D -+ D' be a
holomorphic function with values in D' (which is called a holomorphic mapping from D to D'). Show that if u' is a harmonic function on D', so is the composite
u of on D. EXERCISE 3 (Harnack inequality). Let u(z) be a harmonic function on A(R) such that u(z) >-- 0. Show that
R - IzI u(0) < u(z) R + IzI S u(0) R- IzI R+ IzI EXERCISE 4. Show that if u is a harmonic function on C, and u ? 0, then u is constant.
3.7. HOLOMORPHIC FUNCTIONS ON THE RIEMANN SPHERE
87
EXERCISE 5. Express a harmonic function u on A(1) by the Poisson integral which has boundary values 1 on C(0;1)n{Imz > 0} and -1 on C(0;1)n{Imz < 0}.
3.7. Holomorphic Functions on the Riemann Sphere Let D be a connected open subset of the Riemann sphere C. Then D is called a domain of C, as in the case of C. On the Riemann sphere, there are complex coordinates z on 6\ {oo} and i on a\ {0}. They are related on C' = 6\ {0, oo} by
On C', z = 1/i is a holomorphic function of i, and the converse holds, too.
Let f be a function on D. On C', there are two coordinates, z and i. If f is holomorphic as a function of z on D n C', then by Theorem (3.1.9) it is holomorphic with respect to i; the converse holds, too. We define f to be complex differentiable at oc, provided that oo E D, if f is complex differentiable
as a function of i at i = 0. We say that f is holomorphic on D if f is complex differentiable at every point of D. We know that the complex coordinates z and i are related by zi = 1, and represent the same point of the Riemann sphere C. When f is considered as a function of z (resp., i), we write f (z) (reap., f o i). For example, we have
foi=f
(z),
f0i(0)=f(oo)
The various properties of holomorphic functions on domains of C remain valid for those on domains of C. We need, however, the following notion of differential to deal with curvilinear integrals. We set
Uo=DnC\{oc},
U1 =Dn{0}.
A pair w = {rlo(z),171(i)} of holomorphic functions, rlo(z) on Uo and rl1(i) on U1 is called a differential on D if on Uoo n( U1 the following holds: (3.7.1)
r10(z) = r11
(I) `-z2) _ III
To represent this change of variables, we write (3.7.2)
w = rlo(z)dz.
Here dz is considered to be transformed to dz = (-1/i2 )di. Then, on Uo n U1, w = r1o(z)dz =171(i)di.
Let C(¢ : I - D) be a curve in D. Assume that C C D n C \ {0} and it has a finite length. Then frio(z)dz =
jt7i(i)d2
3. HOLOMORPHIC FUNCTIONS
88
where C in the right side is a curve given by i o 4(t) = 1/0(t). Thus we may define the curvilinear integral of w along C by
/w=
(3.7.3)
j
go(z)dz =
j m(i)di.
c c For a general curve C(q5 : [To, T1 J -e D), if there is a partition (d) : To = to < < t = T1 such that 0(It;-1it)]) C Uo or ¢([tj-1,tj]) C U1, and such t1 < that (3.7.4)
JCI[tj-.f)
IdzI < +oo,
or J Iits-i til Idi[ < +oo,
we define the curvilinear integral of w along C by (3.7.5)
f=>f
w.
If it exists, it is independent of the choice of the partition (d). The curvilinear integral fC f (z)dz with C C C is considered as a curvilinear integral of the differential f (z)dz along C. A differential w = {rlo(z),r)1(i)} on D is called a holomorphic differential if rb(z) (reap., r,1(i)) is holomorphic in z (reap., z). A holomorphic function F on D is called a primitive function of w if
dzF(z) = qo(z),
z E Uo,
d Foz=rtl(z),
zEU1.
We write dF = w for this fact, and call it the differential of F. If C is another primitive function of w, then the difference F - G is constant. All properties of holomorphic functions and curvilinear integrals of holomorphic functions on domains of C which have been proved up to the last section also hold for holomorphic functions and for holomorphic differentials on domains of d which may contain the infinite point oo. For example, let w be a holomorphic differential on a domain D C C and let C be a curve in D. Take a piecewise linear curve C for C as in Lemma (3.3.4). Since a piecewise linear curve always satisfies (3.7.4), one may define the curvilinear integral of w along C by Jcw=
jew.
As in the proof of Lemma (3.4.11), the above right side is independent of the choice of C so long as C satisfies (3.3.5), and Theorem (3.2.19) extends to (3.7.6) THEOREM. Let w and D be as above. Let F be a primitive function of w, and let C be a curve in D with initial point zo and terminal point z1. Then
jw = F(z1) - F(zo).
PROBLEMS
89
Cauchy's integral Theorem (3.4.14) extends to (3.7.7) THEOREM. Let w and D be as above. Let C be a closed curve in D which is homotopic to a point. Then w = 0.
IC (3.7.8) THEOREM. Any function holomorphic on the whole C is constant.
PROOF. Let f be a holomorphic function on C. Since a is compact, if I attains the maximum at some point a E C. If a 0 oo, Theorem (3.5.21) implies that f is constant; if a = oo, then the function f o z, holomorphic in z, gives rise
to a constant. Thus f is constant. 0 If we deal with non-constant holomorphic functions on D, by this theorem we may exclude the case of D = C. If D 0 C, we take b E C \ D. If b = oo, then D C C; if b # oo, we may reduce the case to D C C through the transformation z 1/(z - b).
Problems 1. Show that f (z) = x2 + iy2 satisfies the Cauchy-Riemann equations at the origin 0, but is not holomorphic in any neighborhood of 0. 2. Let f (z) be a holomorphic function on a domain D. Prove that f (z) is a holomorphic function on D' _ {z; z E D}. 3. Let a E C` and define f (z) _ (1 + z)° = e° °s(1+;i. Take a branch of f (z) on 0(1), and expand it to a Taylor series about 0. 4. Let C(O : [a, b] -+ C) be a curve. Show that L(C) = 1cdlim0 L(C; (d))
(5 +oo),
where (d) is a partition of [a, b].
5. Let C be the perimeter of the square {(x, y); -r 5 x 5 2r, -r 5 y 5 2r} (r > 0) with anti-clockwise orientation. Compute the curvilinear integral of f (z) = x2 + iy2 along C. 6. i) Let f (z) = (z - a)/(z -)8) with Q i3 (E C). Show that f is holomorphic in a neighborhood of oo.
ii) Let L denote the line segment from cr to J3. Show that D = C \ L is simply connected.
7. Let f be a holomorphic function on 0(R) such that If (z) 15 M for all z E A(R). Prove that MRn! = (R - IzI)n+1
I f(n)(z)I <
1
z E A(R),
n=1,2,....
3. HOLOMORPHIC FUNCTIONS
90
8. Let f be a holomorphic function on a bounded domain D C C which is continuously extended over the closure D. Show that if f does not take the value 0 on D, and If I is constant on 8D, then f is constant. What happens if f may take the value 0 on D?
9. Let f be a holomorphic function on C. Show that if Ref > 0, then f is constant. 10. (Hadamard's three circles theorem) Let f be a holomorphic function on a
domain {z E Q RI < Izl < R2} with 0 < Rl < R2, which is called an annulus. Set M(r) = max{I f (z)I; IzI = r} for r E (R,, ,R2). Show that logM(r) is a convex function with respect to logr: that is, for Rl < r, < r2 < r3 < R2
logM(r2) <
log r3 - log r2 log r3 - log r,
logM(ri)+
log r2 - log rl logM(r3). log r3 - log rl
11. Let f be a holomorphic function of a neighborhood of A(1). Show that
f(z) dz - f(0)
1
z-a
27ri
12. Show that
12,r Jo
7(0
(lal < 1),
- f (a
zn
(1al > 1).
(lal < 1),
log la - e`sld9 = 10
log lal
(lal > 1).
13. Find the adjoint harmonic function of the harmonic function x+y+x2 -y2. 14. Show that u (x, y) = ez (x cos y - y sin y) is harmonic, and find a holomorphic function f (z) with real part u(x, y).
15. Let {un(x, y)}n o be a sequence of harmonic functions on a domain D, converging to u(x, y) uniformly on compact subsets. Prove that u(x, y) is harmonic.
16. Let f (z) be a bounded function, holomorphic on H, and continuous up to R. Prove that f (z) = r J_ 00
(t z) - t2
I 1
Ref (t)dt + Imf (i).
17. Let h(t) be a bounded continuous function on R. Show that the function u(x, y) =
!J
y h(t)dt z (t-x)2+N2
is harmonic on H. Moreover, as x -+ to E R and y -+ +0, u(x, y) - h(to).
CHAPTER 4
Residue Theorem
We begin in this chapter with Laurent series, and then explain meromorphic functions and the residue theorem. By making use of it, we will prove the Prinzip von der Gebietstreue (open mapping theorem), and the inverse function theorem. In the last section we explain methods for applying the residue theorem to calculate various kinds of definite integrals. These are direct consequences derived from Cauchy's integral theorem proved in Chapter 3; nevertheless, the reader may sense the depth of content of the theorem.
4.1. Laurent Series
For aECand O
which is called a ring domain or an annulus. In the case of a = 0 we write R(rl, r2) = R(O; rl, r2)
For the sake of simplicity we assume a = 0, and let f (z) be a holomorphic function on R(rj,r2). Let z E R(r1,r2) be an arbitrary point, and take rl < r2 so that
rl
the half-circle of C(z;r) from (jzj - r)e181$z to (Izl + r)e'"g2 with clockwise orientation, let C4 be the line segment from (Izl +r)e'" ' to rte'"`gz, let C5 be the circle C(0; r2) with initial point rte' "rg z and with anti-clockwise orientation, and finally let CO be the half circle of C(z; r) from (Izj +r)e'"gz to (Izj-r)e'"`gz with clockwise orientation. Then, in the domain R(r1, r2) \ {z} the curve (4.1.1)
C= -C1 +C2+C3+C4 +C5 -C4 +Cg - C2 91
4. RESIDUE THEOREM
92
FIGURE 39
is homotopic to a point, so that fc fd(= 0. It follows that f (z)
2irz
fC,+C4f(-(()z d(
-I
=
Jc1
(- z
d(+
1
f (d(
/
27rz Jas (- z
Since K(/zl < 1 on C1, we have
-1 2na
f(C)d(=
1c, (- z
2zri
C,
/ f(C) d(
1
c, 1 - 4
21ri
f (() , (cyd(.
=
1
z
n=1
C,
f ((nd( zn.
The last power series of the above expression is absolutely convergent, and is written as -O0
1
2ni n=-1
ff
(()(-n-`d( zn.
i
Similarly we have
f(()
I 27ri
Since f
(()(-n-1
f Szd(-
1
n=0
27ri
f f(()(
n-ld(
is holomorphic in R(r1iT2), one infers in the same way as above
that id=
21-
1
Jc5
This shows that for n E Z (4.1.2)
zn
cS
an
=
f(()
JC(O,r)
(n+1
r1 < T G r2,
4.1. LAURENT SERIES
93
is independent of the choice of r. Therefore we have the expression -00
f (z) _
(4.1.3)
00
a,,zn + >2 anzn =
ri
n=-1
anzn. n=-0o
En00 The power series , an zn in 1/z converges absolutely and uniformly on compact subsets of R(r1, oo), and so does anzn in i(r2). Hence, °O n=-oo anzn converges absolutely and uniformly on compact subsets of R(rl, r2), and is called
the Laurent series of f. As in the case of the Taylor series, the derived function f'(z) is written as
fl(z) _ E nanzn-1.
(4.1.4)
n=-a
Thus, if a_1 = 0, the function f has the primitive function F, which is written as
a F(z)
(4.1.5)
1
zn+1 + cont.
n=-00
n*-1
As in Theorem (3.5.20), ii), for r1 < r < r2 we have 00
E
(4.1.6)
IanI2r2n 27r
fIf(r&8)I2d9.
In the annulus R(a; r1, r2) we have the following, called the Laurent series of
f about a: 00
f(z) = E an(z - a)n,
(4.1.7)
r1 < IzI < r2-
n=-00
We consider the case of rl = 0 (R(a; 0, r2) = 0(a; r2) \ {a}), and assume that f is not constant. If there are infinitely many coefficients an # 0 with n < 0, then a is called an isolated essential singularity of f. If there are only finitely many an # 0 with n < 0, f is said to have at most a pole at a, and f is expressed as f (z)
= (za- a)m + (za- a)m-1 + ... + ao + ... ,
a-,,, 0 0.
If m > 0, a is called a pole of order m of f , and one writes f (a) = oo. If an = 0 for all n < 0, then f is holomorphic in A(a; r2), and f(z)=a,n(z-a)m+a,n+,(z-a)m+l+...,
If m > 0, a is called a zero of order m of f .
am #0,
m>_0.
4. RESIDUE THEOREM
94
For a = oo, we define the notion of zeros and poles in the same way as above by making use of the complex variable z. Note that
{i;0 < lil < R} = S z; IzI >
l .
T1 The relation between the Laurent series of f with respect to i and that with respect to z is given by o0
f(z) = >00 anzn = E a_nZ = f o i.
(4.1.8)
n=-oo
n=-oo
Therefore we have the following: (4.1.9)
f (z) = E'n=-. anzn is holomorphic at oo if and only if an = 0 for all n ? 1;
(4.1.10)
the differential
_
(n
f
°°
1 anzn ] dz = oc
a_nzn
/
)
(_.) dz
is holomorphic at oo if and only if an = 0 for all n >_ -1. EXERCISE 1. Obtain the Laurent series of f (z) = el/= about z = 0. EXERCISE 2. Obtain the Laurent series of f (z) = 25(11+ z) about z = 0.
4.2. Meromorphic Functions and Residue Theorem Let D be a domain in C. A function on D which has at most a pole at every point of D is called a meromorphc function. Let f be a meromorphic function on D. If a E D is a pole of f, the function
fuo f(z)=
1
f (z)
is holomorphic in a neighborhood of a. In this sense f uniquely defines a continuous mapping (4.2.1)
f:D-.C.
For example, a rational function P(z)/Q(z), defined as a ratio of two polynomials P(z) and Q(z) with Q(z) $ 0, is meromorphic on C. By definition, the set E of all poles off is a discrete subset of D. If E = 0, then, of course, f is holomorphic
in D. Two meromorphic functions f and g on D are said to be equal if there is a discrete subset F C D such that f and g are holomorphic on D \ F, and if f = g on D \ F. This property is independent of the choice of F by the identity Theorem (2.4.14). If f 0, l/f is meromorphic, and the sum and product of two meromorphic functions are meromorphic. Thus the set of all meromorphic
4.2. MEROMORPHIC FUNCTIONS AND RESIDUE THEOREM
95
functions on D forms a field. For meromorphic functions we have the following identity theorem.
(4.2.2) THEOREM. Let fl and f2 be meromorphic functions on D. Let A C D be a subset with an accumulation point in D. If fl = f2 on A, then f, = f2 on D. PROOF. We may assume that f2 = 0. Let zo E D be an accumulation point of A, and E be the set of all poles of f . If zo ¢ E, Theorem (2.4.14) implies that f1 - 0 on D \ E, so that f1 = 0. Assume that zo E E. If zo 36 oo, we take
0
gi(z) =
fe(z)' Then gl is a holomorphic function on i (zo; r). Therefore f, does not take the value 0 on 0(zo; r). This contradicts the fact that zo is an accumulation point of A. When zo = oo, by making use of the complex coordinate z we make the same argument as above about z = 0. Let a E C and let f be a holomorphic function on 0(a; r2) \ {a} expanded as (4.1.7). We define the residue of f at a by (4.2.3)
Res(a; f) = a- i =
2rri C(n:r) f (()dC
(0 < r < r2).
When a = oo, we consider a holomorphic function f on (z E C; 0 < (il < R} _ {z E C; Izi > 1/R}. Then we have the Laurent series x x
f(z) = E anzn = 1: a-nzn = f o1, n=-x n=-x and we define the residue of f at oo by (4.2.4)
7ra f Res(oo;f)=-a-1=2-
foi
{I=1=r}
-1-dz',
T
0
By this definition we we that the residue is a notion not for the holomorphic function f , but for the differential w = f (z)dz. That is, no matter whether a is oc or not, taking a small circle C around a with anti-clockwise orientation, we set (4.2.5)
Res(a; w) = tai j w.
c
It follows that Res(a; w) = Res(a; f) for all a E C.
In general, let D C C be a domain, and let w= f (z)dz = g(i)di be a differential on D. If the coefficient f (z) (resp., g(i)) is a meromorphic function of z (resp., z), w is called a meromorphic differential on D. We say that w has
4. RESIDUE THEOREM
96
a zero (resp., pole) at a E D if a E C and f (a) = 0 (resp., oo), or if a = oo and g(O) = 0 (resp., oo). The set of zeros and poles of w is discrete in D. Let C be a closed curve in D which is homotopic to a point in D. Let
be the homotopy connecting C to a point. Assume that there is a point b E C \0(10, 11 x [0, 1]). For a E D \ C we define the rotation number nb(a; C) with respect to b by (4.2.6) i) The case of a # oc: If b # oo,
nb(a; C) =
tai
j(L ;
If b= cc, 1
2Ai
1a
1 b/
ik.
/ c (-a 1
ii) The case of a = oe:
In the case of b = cc we set n(a; C) = no,(a; C).
Note that if a # oo and b # oo, the differential (1/(( - a) - 1/(( - b))d( is holomorphic at oc. EXERCISE 1.
Let a,b E C \ C(0;1). Obtain nb(a;C(0;1)). Here C(0;1) is
endowed with the anti-clockwise orientation.
(4.2.7) THEOREM. Let D, C, and b be as above. Let E = {a,,; v = 1, 2.... } be a discrete subset of D such that E ft C = 0, and let w be a holomorphic differential
onD\E. Then 00 1
2w* c
w = E nb(av; C)Res(av; w) L-1
Here the number of a,, with nb(a,,; C) # 0 is finite.
PROOF. If a ¢ 0((0,1J x (0,11), then the differential to define nb(a,,;C) in (4.2.6) is holomorphic in a neighborhood of 0((0,11 x (0,11). Thus, by Theorem
(3.7.7), nb(a,,; C) = 0. Changing the order of numbering of the a,,, we may assume that
Efo$(10,1)x[0,11)={a,,;1
05vo<+oo.
4.2. MEROMORPHIC FUNCTIONS AND RESIDUE THEOREM
97
Fix a domain D' C D such that b ¢ D' and D' fl E _ {a,,;1 5 v S vo}. About each a we write w = f (z)dz (or g(i)di), and expand f (z) (or g(i)) to a Laurent series. Suppose first that a = oo and b # oo. We set -1
oc
an (z -
f(z) _
an (z - a,,)",
n=-oo _2 17,
n=O
_
(
\
l
n=-oo
1
-
1
b
l/
dz.
If b = oo, we set (z - b)" = 0 for n < 0 in the above equations. Note that n converges absolutely and uniformly on compact subsets of C \ {a,,, b}(D D'), and hence is holomorphic there. The first term of nY has a primitive function _2
E
n+1{(z-a
v)n+1-(z-b)n+1).
n=-oo
By Theorem (3.7.6) the curvilinear integral of the first term of nY along C is zero. It follows from this and the definition that (4.2.8)
tai I q = nb(a.,; C)Res(a,,;w).
In the case of a,, = oo, we similarly define q by making use of g(z), and deduce that
( 4.2.9 )
1
tai
rl =
w)
The differential w - E; 0_1 q is holomorphic in D'. Since C is homotopic to a point in D', it follows from Theorem (3.7.7) that V0
Jt7v1=0. C
Thus, (4.2.8) and (4.2.9) imply the required formula. 0 We explain a practical method to calculate residues. If f (z) has a pole of order 1 at a E C, then (4.2.10)
Res(a; f) =Z-a lim(z -a)f(z).
If a E C is a pole of order m of f, (4.2.11)
Res(a; f) =
1-1
(m - 1)! dz*"-1
EXERCISE 2. Prove (4.2.10) and (4.2.11).
(z - a)- f (z). x=o
4. RESIDUE THEOREM
98
If a = oo, the equations similar to (4.2.10) and (4.2.11) hold in terms of z. That is, if f is holomorphic in a neighborhood of oo and has a zero at oo, then
Res(oo; f) = lim -z f (z).
(4.2.12)
If f has a pole of order m at oo, (4.2.13)
Res(oo; f) = lim
(-1)"` zm+2 f(m+1)(z)
I.,-. (m+ 1)!
In the case where f is holomorphic at oc, this formula is valid with m = 0. EXERCISE 3. Prove (4.2.12) and (4.2.13).
Let D C C be a bounded domain surrounded by finitely many Jordan closed curves, Ci, 1 < i < k, as in Figure 40 below. When a description depending on geometric intuition is used, every Ci is assumed to be endowed with orientation so that D is lying on the left.
FIGURE 40
Let the numbering of the C, be as in Figure 40. Then C1 is called the outer boundary of D, and C2, ... , Ck are called the inner boundaries of D. Let D' be a domain containing the closure DU (uk1 Ci), and let f be a meromorphic function on Y. Connect C; (2 < i < k) and C1 by suitable curves Li. Then, which is as in (4.1.1) we obtain a curve Ci+L2+C2-L2+C'I homotopic to a point in D', where C1 = Cl' + + C;k-1) Assume that f has
no pole on Uk1 Ci, and take the Li so that f has no pole on them, either. Then Theorem (4.2.7) takes the following form: k
(4.2.14)
1
i_1 2ni C
f (z)dz =
Res(a; f ). aED
In the above, the description was based on geometric intuition. We will, however, rigorously check the conditions when we use it. We will summarize in
the last section of this chapter how to apply the residue theorem to calculate various definite integrals.
4.3. ARGUMENT PRINCIPLE
99
EXERCISE 4. Let E be a finite set of C, and let w be a holomorphic differential
on C \ E. Show that
ERes(a;w)=0. a meromorphic function on C is a rational function.
EXERCISE 6 (Fractional expansion of a rational function) Let F (z) =
P(z)
Q(z) be a rational function, where P(z) and Q(z) are coprime. Let ai E C, 1 5 i 5 k, be zeros of Q(z), and expand f (z) to a Laurent series about ai: P.
z
au
!(z) _
i)',
j=1
15iSk.
j=0
Let the Laurent series of f (z) about oo be -00
PO
!(z) = Eaojz' + E NUj j=-1
j=o
Then, show that k
PO
P.
!(z)=Eao;z'F, E(z j=0
i=1 j=1
EXERCISE 7. What are the residues of f (z) = z/(1 + z2) at poles? EXERCISE 8. What are the residues of f (z) = (1 + z)/(a2 + z2)2 at poles? EXERCISE 9. What are the residues of f (z) = ez+1/= at z = 0, oo?
4.3. Argument Principle Let D be a domain of C, and assume that D function in D. We call (4.3.1)
w
f(z)
dz
)dz
C. Let f be a meromorphic
f of di zdi
the logarithmic differential of f . If a E D is not a zero nor a pole of f , then w is holomorphic in its neighborhood. Let a E D be a zero of order m of f. If
aiA oo,wehave
!(z) = a,,,(z - a)m + a,,,+, (z - a)m+t + ... ,
an
dz(z) =a,,,m(z-a)m-t +a,,,+1(m+1)(z-a)"'+ , amm1+a"
Q; =
_
Tm+t (z-a)+...
a,,,(z-a)tl+a(z-a)+ m G- a
+ 4 + ai(z - a) + -
dz.
dz
0, m > 1,
4. RESIDUE THEOREM
100
Thus Res(a; w) = in.
(4.3.2)
If a = oo, by making use of z, we similarly deduce that
foz= c(fo
di
az +
a,,, # 0, m ? 1,
,
= ammi -I + ...
so that Res(oo;w) = m. Suppose that a is a pole of order p of f. If a # oo, we have
f (Z) _ (za- a)P + -Pa-P
f1(Z) _ (z - a)P+I w
I
(za
a-P 36 O,P > 1,
P+'
+ -(P - 1)a_v+l + ... (z - a)P
-P z-a
Thus Res(a; w) = -p.
(4.3.3)
This holds also for a = oo. The following theorem is called the argument principle.
be a domain, and take c E C \ D. Let f be a
(4.3.4) THEOREM. Let D
meromorphic function in D, and let w be its logarithmic differentiaL Let C be a closed curve in D such that f has no zero nor pole on C and it is homotopic to a point in D. Let a,, E D, v = 1, 2, ... (reap., 6 µ E D, P = 1, 2, ... ), be zeros (resp., poles) of order m (t esp., of f . Then
C)Pµ tai f w = > Ync(a,,; C)m - E nc(b,u; µ PROOF. This easily follows from (4.3.2), (4.3.3), and Theorem (4.2.7). 0
Let D c C be a bounded domain as in (4.2.14), and let 8D = UkI C, and c = oo. Let f be a meromorphic function in a neighborhood of or poles on 8D. Then Theorem (4.3.4) takes the form (4.3.5)
without zeros
k 1 Jidzmv-P :
f(z)
He re the first term on the right represents the total number of zeros off counting
multiplicities, and the second that of poles. Note that the curvilinear integral 1
2ai fc,
L 7 (z)
dz
4.3. ARGUMENT PRINCIPLE
101
is the rotation number of the curve f (C,) around 0. Therefore, if C is a closed curve in D such that f has no zero nor pole on C and the image f (C) is contained in a simply connected domain D' inside C \ {0}, then the logarithmic differential w of f has a primitive function log f, once we choose a branch of log on D'. Then, it follows from Theorem (3.7.6) that (z)dz = 0. lc f (z) dz If in Theorem (4.3.4) D is a bounded domain of C, c = oo, and f is holomorphic, then the sum F_,, n(a,,; C)m is called the number of zeros off surrounded by C. The next theorem is called Rouche's theorem. (4.3.6)
(4.3.7) THEOREM. Let D c C be a bounded domain, let f be a holomorphic function on D, and let C be a curve in D. Assume that C is homotopic to a point, and that there is no zero off on C. Let g be a holomorphic function on D such that Ig(z) I < If (z) I for z E C. Then the number of zeros off surrounded by C is equal to that off + g.
PROOF. Let M be the number of zeros of f surrounded by C. Then, that of
f+9is 1
M
2ni
+9'
c f + 9 dz
_
1
2ai
(9/f)'
f + 1 + 9/f c (hi
dz
)I
(91f dz. 21ri Jc 1 + 9/f Since Ig/f I < 1 on C, 1 + 9(z)/f (z) E A(1;1) for z E C. Taking a branch of log on the simply connected domain A(1;1) (! 0), we have __M +
1
(g/f)'
1+g/f
dz
log1+9 ` f
Hence, it follows from (4.3.6) that I
/f
J 1 +9/f d= 0,
and that M = M'. Let D and CJ be as in (4.2.14), and let f and g be holomorphic functions on D with 19(z)I < If (z) I on C. Then
(4.3.8) the number of zeros of f + g in D equals that of f in D. EXERCISE 1. Set f j (z) = z" + a1 z"' 1 + + an, and f2(z) = z". Show by making use of Rouchd's theorem that for a sufficient large R > 0 the number of zeros of f1 surrounded by C(0; R) equals that of f2, i.e., n.
We give some applications of the above theorem. The first is Huruntz's theorem.
4. RESIDUE THEOREM
102
(4.3.9) THEOREM. Let D C C be a domain, and let { f,}n o be a sequence of holomorphic functions in D converging uniformly on compact subsets of D to f . I f fn, n = 0,1, ... , do not take the value zero, then either f has no zero, or
f e0. PROOF. First of all, f is holomorphic (Theorem (3.5.19)). Assume that f -$ 0.
Take a E D. Suppose a E C; otherwise, use the coordinate i about oc. Since the set of zeros of f is discrete, there is a b > 0 such that A(a; b) C D, and (A (a; b) \ {a}) n {f = o} = 0.
Here, {f = 0} stands for (z E D; f (z) = 0}. For C(a; b) with anti-clockwise orientation 1
f
) dz >= 1.
i JC(a;b) f (Z)
Since {f,, } converges uniformly on A(a; b), there is an no such that t
f (z)I < min{I f (w)j; w E C(a;b)},
z E C(a; b), n ? no.
It follows from Theorem (4.3.7) that the number of zeros of f in A(a; 6) equals
that off + (f - f) = f (n ? no). Thus fn has a zero in i(a; b); this is a contradiction.
Let D C C be a domain. A meromorphic function f in D_is said to be univalent if f is injective (one-to-one) as a mapping from D into C. (4.3.10) THEOREM. Let {fn}n o be a sequence of holomorphic functions in a domain D C C such that it converges uniformly on compact subsets of D to f. If all fn are univalent, then f is either univalent or constant.
PROOF. Assume that f is not constant, nor univalent. Then there are two points z1 i z2 E D with f (z1) = f (Z2) = wo. Take neighborhoods U; of z i = 1, 2, with U1 n U2 = 0. Since {f,, - wo} 0 converges to f - wo uniformly on compact subsets of D, it follows from Theorem (4.3.9) that there are some n = wo. That is, f,, (z') = fn(z2), and this and zi E U;, i = 1, 2, such that f,. is a contradiction.
a point z E D For a meromorphic function f in D and for a point w E such that f (z) = w is called a w-point of f; an oo-point of f is nothing but a pole of f . When w # oo and the order of zero off - w at z is m, m is called the order of the w-point z; the order of the oo-point z of f is defined as that of the pole off at z. (4.3.11) THEOREM. Let f be a meromorphie function in a domain D C C. Let zo E D be a wo-point of order m of f with wo = f (zo). Then there are neighborhoods V of zo and W of wo such that for an arbitrary w E W \ {wo} there are
exactly m distinct w-points, z,... , z,,,, of order 1 off in V.
4.3. ARGUMENT PRINCIPLE
103
PROOF. We assume that zo and w are not oo; otherwise, we argue similarly by using z or tD about oo. By assumption we may expand f to a Taylor series about zo: f(z)=wo+a,,(z-zo)'"+...,
a,,,#0.
There is 0 < b < d(zo; OD) such that A(zo; b) n {f = wo} = {z0}, (4.3.12)
(4.3.13)
0(zo;6)n{f'=o}\{zo}=0, f '(Z)
tai ./C(w;a) f (z) - wo
dz = m.
There is0<e<1such that If (Z) - tool > e,
z E C(zo;6).
Applying Theorem (4.3.7) to f (z) - to = (f (z) - wo) + (wo - w) with to
E
A(wo; e) \ {wo}, we infer that the number of zeros of f (z) - wo in A(zo; b) is equal
to that of f (z) - to for to E A(wo; e). If z E A(zo; 6) and f (z) = to E A(wo; e), then z is a u)-point of order 1 of f by (4.3.12). Therefore there must be exactly m such points z. The following theorem is called the Prinzip von der Gebietatreue (open mapping theorem). (4.3.14) COROLLARY. Let f be a non-constant meromorphic function in a domain D C C. Then, as f is considered to be a continuous mapping from D into C, the image f (D) is a domain of C.
PROOF. By Theorem (4.3.11) f (D) is an open subset. Take two arbitrary points f (z1), f (z2) E f (D). There is a curve C(45) connecting z1 and z2 in D. Then f o 0 is a curve connecting f (z1) and f (z2). Thus, f (D) is connected. We give one more application of Theorem (4.3.11), which is called the inverse function theorem.
(4.3.15) THEOREM. Let f be a holomorphic function in a neighborhood of zp E C, and set wo = f (zo). If f'(zo) 0, there are neighborhoods U of zo and W
of" such that f (U) =W and the restriction f I U : U- W off to U has a holomorphic inverse function g : W -, U of f JU with 9(wo) = zo.
PROOF. Take neighborhoods U of zo and W of wo as in Theorem (4.3.11). Since f'(zo) 96 0, m = 1, so that for an arbitrary to E W there is a unique point
4. RESIDUE THEOREM
104
z E U with f (z) = ur. Write g(w) = z. We first check the continuity of g. Take an arbitrary w' E W, and set z' = g(w'). In a neighborhood V' of z', we have (4.3.16)
f (Z) - f (Z') = al (z - z') + a2(z - z')2 + ...
(a, # 0)
= (z - z')(al + a2(z - z') + -) = (z - z')h(z).
Since h(z') = a, 34 0, after taking V' smaller if necessary, there is a positive number b such that (4.3.17)
z E V.
lh(z)I >_ 6,
Then f(V') = W' is an open neighborhood of w'. It follows from (4.3.16) and (4.3.17) that
619(w)-9(0I 5 1w-ti1. Hence g(w) - g(w) -' 0 as w -+ w'; that is, g is continuous. Again by (4.3.16) we have for wE W' with w76 w'
g(w) - g(w') w - w'
_
I h(g(w))
The continuity of g implies that lim,,,_,, h(g(w)) = h(g(w')) = h(z) = f'(z), and so lira g(w) - 9(&) w-.w' w - w'
= g'(w') =
1
P(z')
Thus, g is complex differentiable in W'; i.e., g is holomorphic. Then we may set
U=9(W)=f-'(W)- o (4.3.18) THEOREM. Let f be a univalent meromorphic function in a domain D C C. Then there is a meromorphic function g in the domain W = f (D) C C such that the mapping g : W -. C is the inverse mapping off : D - W.
PRooF. By Theorem (4.3.14), the image W = f (D) C C is a domain, and we have the inverse mapping g : W Doff since f is univalent. Take an arbitrary point zo E D, and set wo = f(zo). It follows from Theorem (4.3.11) and the univalence off that zo is always a we-point of order 1. Therefore, in the case of zo 0 oc and wo 96 oo, f'(zo) 96 0. By Theorem (4.3.15) g is holomorphic in a neighborhood of wo. In the case of zo = oo and wo 0 oo, we have that f o i/di(0) 0. Thus z o g(w) = 1/g(w) is meromorphic in a neighborhood of ", and hence g(w) is meromorphic there. In the case of zo = oo and " = oo, we see that riw o f = 1/f is holomorphic in a neighborhood of i = 0. Thus, for the same reason as above, g is meromorphic in a neighborhood of oo. 0
4.3. ARGUMENT PRINCIPLE
105
It should be noted that Theorem (4.3.18) does not hold for real differentiable functions, nor for real analytic functions. For example, take f (t) = t3 for t E R. Then f : R R is injective, but the inverse function ' t E R is not differentiable
at t = 0. Let f be a holomorphic function in a neighborhood of zo E C. If f'(zo) # 0, then by Theorem (4.3.15) f has an inverse function g defined in a neighborhood of wo = f (zo). In what follows, we consider the case of f'(zo) = 0. For the sake of simplicity, we assume that zo = wo = 0. Let m be the order of zero of f at 0. Then m > 1, and we have
f (z) =amzm +a,.+ l = amzm
zm+l + .. .
+ ...1 \(1 +'-+1 am
= a,,, z'h l (z)
(a,,, # 0, h l (0) = 1).
Taking 6 > 0 small, we have
hi(z) E A(1;1),
z E 0(6).
Fixing one branch of log in 0(1;1), we get h1(z) = elg hi (s), and set
h2(z)=expl mlogh1(z)). we set
Taking one value of
h3(z) =/zh2(z). Then f (z) = (h3(z))m, and h3(z) _ /z + ... # 0.
h3(0) =
It follows that v(z) = h3(z) has the inverse function z = gi(v) with gl (0) = 0. Let the Taylor series expansion of g1 be
91(v) = clv + c2v2 + ..., Since w (4.3.19)
f(z)=vm,wewrite v= i,andset 92 (W)
= C1
VW-
+ c2 ("vf_)2 + .. .
Then f(g2(w)) = w, and so one may consider g2(w) = f-1(w). But, 92(w) is a holomorphic function of v = and for the variable w, 92 (w) = g1( w-) is a multi-valued function with m values if w # 0. In general a series such as (4.3.19) in the powers of Vw- is called a Puiseux series, and in the present
4. RESIDUE THEOREM
106
case, in particular, (4.3.19) is called the Puiseux series expansion of the inverse
function f -'(w). 4
0
0001.7.
V"
FIGURE 41
EXERCISE 2. Show that the numbers of zeros and poles of a rational function counting multiplicities over the Riemann sphere d are equal. EXERCISE 3. Set f (z) = z2/(1 + z). Obtain the Puiseux series expansion of f -1(w) about w = 0.
4.4. Residue Calculus In this section we explain bow to calculate the values of some definite integrals
as applications of residue Theorem (4.2.7) or (4.2.14), classifying the definite integrals into several types. i) Let R(x, y) be a rational function of two variables x and y such that it has no pole on the circle C(0;1). We want to calculate the value of !2w
1= /
R(cos t, sin t)dt.
Jo
For z=e'twehave Cost
=Z(z+ZI
slat=
///
dt =
2ilz-z \\ //
1 dz. sz
It follows from Theorem (/4.2.7) that
I=J
C(0;1)
=2w
R(1 Iz+lZ 1, 1
zW ldz
` Res(a;R(2 (+)2i
nE0(1)
(z-
zl)l
4.4. RESIDUE CALCULUS
107
102 a
EXAMPLE. We calculate I = 1
C(o;1) a+ 2 (z + _) 2
_
tzdz
a + cost
, a > 1. Then
_2
1 z2+2az+1dz
C(o:i) 1
C(o;I) {z - (-a - a -1)}{z-(-a+ a -1)}
? 27ri Res
(-a + `
dz
1
a2 - 1;
{z - (-a - a 1)}{z-(-a+ a -1)} {z - (-a + a --1)j = 47r lim z-.-a+ a -1 {z - (-a - /a2 - 1)} {z - (-a+ a --I)) _
27r
e
a -1*
In the latter half we used (4.2.10) to obtain the value of the residue.
FIGURE 42
EXERCISE 1. Calculate the following definite integrals. dB
)1"
i
ii) f iii)
(0
1-2acosO+a2
I
(a > 0).
a + sine 8 2-
a+sc
9do
(a > 1).
1i) Let R(z) be a rational function of z. We want to get (4.4.1)
I =1
00
R(x)dx = oc
10
o
R(z)dz.
1-10
Express R(z) = P(z)/Q(z) with coprime polynomials P(z) and Q(z). For the convergence of the integral (4.4.1) at ±oo the following condition is necessary and sufficient: (4.4.2)
degP-degQ <-2.
4. RESIDUE THEOREM
108
If Q(z) = 0 has a real root, the integral (4.4.1) does not exist. We add the following condition:
(4.4.3) The orders of poles of R(z) on the real axis are at most 1.
Nonetheless, if R(z) has a pole b E It, the integral (4.4.1) does not. exist in the ordinary sense; we have to take the principal value of integration defined by P.V.
f
b+6
rb
R(x)dx
b -6
'E
Um +0
J
b-6 + rb+
6
R(x)dx
Here 6 > 0 is chosen so that b is the only pole of R(z) contained in the interval [b - 6, b+ 61 . The existence of the above limit will be proved below. We take the integral (4.4.1) in this way, call it the principal value of integration, and write
I = p.v.
r
'Q R(x)dx. J o0
Let a; E C, Im a, > 0, 1 < i S k, be all the roots of Q(z) = 0 in the upper half-plane. Let b, E R, 1 < j 51, be all the real roots of Q(z) = 0. Then we are going to show that k
(4.4.4) p.v.
for,
R(x)dx = 2xi
1
Res(a:; Rz)) + Sri E
R(z)).
j_1
FIGURE 43
Take r > ro > maxi; {la, I, Ib; 1), and small e > 0 so that A(a e) do not intersect It, and 0(ai; e) and 0(bj; e) are mutually disjoint. Let C(,.) denote the half circle from r to -r in the upper half plane, and let C(r,,) denote the curve from -r to r on the real axis where about every bJ the curve goes around the half circle C- (b,; e) of C(b,; e) in the lower half plane (see Figure 43). Then (4.4.5)
/
r
R(z)d.- = 2ri
Res(ai; R(z)) +
R,es(b;; R(z))}
.
By condition (4.4.2) there are an M > 0 and a ro > 0 such that for all r > ro IR(Te`s)I S
ra
IN
4.4. RESIDUE CALCULUS
and hence
M -rdz = < = Icf, r2 r
R(z)dz
JC()
y0
(r --+ oo).
Expand R(z) about every b., to a Laurent series:
R(z) = C-I + co + cl (z - b,) + -
z-bj C-I
z-b3 We get
f
C-I z
bj
+ h(z) dz -+ rric_I
(e - 0).
Since c_1 = Res(b,; R(z)), we obtain (4.4.4) by letting r - oc and e -b 0 in (4.4.5).
EXAMPLE. We calculate 1 = p.v. r x41
1
dx = 2 P.V.
j x41
1 dx.
Since
Q(z) = z4 - 1 = (z - 1)(z + 1)(z - i)(z + i), conditions (4.4.2) and (4.4.3) are satisfied. We have
_
1
Res (1' Q(z)) Therefore
4
Res (-1; Q(z))
1 = 2 [2viRes (i; =
__1
1
1
7ri
{Res
Res
i
4
(1;
1
Q(z))
_ i 4
Res (-1; Q(z) )}]
22ai 4) _ 4
EXERCISE 2. Calculate the following integrals. °°
i)
p.v.1. 2 P.V. r J ao
1
X2
x
0o
1
ii) 1
dx.
+
x4-+-X21 2 dx.
iv) 1o
x4 + 1 dx.
x + 1 )°
dx (n = 1, 2,... ).
( 2
Hi) We let R(z) = P(z)/Q(z) be as in ii), but relax condition (4.4.2) to
deg P - deg Q < -1.
(4.4.6)
We keep condition (4.4.3). Then the following holds: (4.4.7)
p.v.
foo
R(x)e'=dx = tai E R s(a; R(z)e':) + si E Res(b; R(z)e'=). Ima>O
Imb=O
4. RESIDUE THEOREM
110
The proof is the same as in ii); the only difference is to show that
fC)
R(z)e"dz - 0
(r -. oo)
under condition (4.4.6). Set z = re' (0 5 0:5 ir). Then Ie'r there is some M > 0 such that JR(re'BW< Mr. Therefore
sI
= e-r°in9, and
a-redo.
R(z)e'= < M 0 To
If one knows Lebesgue integrals, it is clear that the above right side converges to 0 as r - oo. Relying only on Riemann integrals, one deduces the same from the following estimate:
r
a-r sin Bd0
=
a-r sin e W
2
0
<
Jo
7r
r
7re_-r
r
-' 0
x/z +r 2 f _ A e-2rB/J L
2r
0
(r - oo).
F1cuiE 44 !
EXAMPLE. Set I = I
sinx dx = Ip.v. /OD sinxdx.
Note that (sin x)/xJJis continuous at 0 if the value there is defined to be 1. Since (cosx)/x is an odd function, we have the principal value of integral, p.v.
l
ooe x
Thus
x
dx = 0.
00 ei:
21 = p.v.
f -dz. iz 00
Since R(z) = 1/iz satisfies conditions (4.4.6) and (4.4.3), it follows from (4.4.7)
that esi
21 = ,rRes (o;
Hence, I = 7r/2.
) _ it.
4.4. RESIDUE CALCULUS
REMARK 1. The integral I =
111
r' sin x dx is not absolutely convergent; that x
is,
f/'
o
sin z dz=oo. x
R(x)e-'=dx, one has to take a curve
REMARK 2. To calculate I' = p.v. f in the lower half plane as in Figure 45.
FIGURE 45
Then
I' = -21ri
Res(a; R(z)e") - ai
Res(b; R(z)e'2).
Imo
Ima
EXERCISE 3. Calculate the following integrals. s
P.V.
i)
iii)
00
ii)
foo °O cos x dx. Jo x2 + 1
iv)
P.V.
Jsin
r°a sin3 x 1
x
zdx (a E R).
dz.
iv) Let R(z) = P(z)/Q(z) be a rational function which has no pole at 0 nor on the positive real axis, and satisfies (4.4.6). Let 0 < a < 1. We calculate
x It is clear that the integral converges at 0 and oo. Set D = C\{z E R; z > 0), and take a branch of log z on D so that lim, I log z = 0. Consider a meromorphic function on D,
f(z) = R(z) = R(z)e oboe:.
4. RESIDUE THEOREM
112
Let r > 0 be sufficiently large so that A(r) contains all the poles of R(z). Take sufficiently small f > 0 and c' > 0 to form a curve C(c, c, r) as in Figure 46:
FIGURE 46
Then we have f(z)dz = tai R s(a;f). C(e,e', r) aED
(4.4.8)
Letting e' - 0, (4.4.8) is decomposed to (4.4.9)
J
C(O;e)
+f
f(z)dz+Jlim+0 f(x+it')dx+J
f(z)dz
C(O;r)
lira f (x - iE )dx = 2 ri FRes(a; f (z)). aSD
ForxER,x>0, lim f (x + iE )dx - lim f (x - it) = e'+0 e,-+O
R(x) ZQ
(1- e-2a*') .
There areM>0andM'>0such that
J
C(O;e)
Jc(0r)
f (z)dz1 <
2we = 2xMc'-Q -+ 0
f (z)dzl < rme 2ar
2aM' r°
0
(r-.oo).
It follows from (4.4.9) that
I-
R(°)
E
1- a-20- aE D
Rs(a;R(z)e-agog:).
4.4. RESIDUE CALCULUS
EXAMPLE. Set I = -a log z
1+z
1
J0
(1 + x)xa
113
dx, 0 < a < 1. The only pole of f (z) _
is -1 and its order is 1. The residue there is
Res(-1;f(z)) = -a log(- 1) = -ari Therefore,
I
_ _ 27rie-ar' 2iri 1 - e-2ar, -ears - e-art - sin air 717
EXERCISE 4. Calculate the following integrals. i)
( 1 + x2)xa
Jo 00
ii)
L
iii)
0
(1+x)^xa
dx
(0 < a < 1).
dx
(0 < or < 1,n = 1, 2, ... ).
or,
(a>0,0
1
(x2+a2)xadx
v) Let R(z) = P(z)/Q(z) be a rational function without pole at 0 nor on the positive real axis, satisfying (4.4.2). Assume moreover that all the coefficients of P(z) and Q(z) are real. We calculate
I = I R(x)logxdx. 0
The convergence of the integral at 0 and oo may be clear. Let D and C(e, e', r) be as in iv), and take a branch of log as in iv). Set f (z) = R(z) (log z)2. Then !C(c,t',r)
f (z) = tai
Res(a; f (z)). aED
Letting' -. 0, we get
J
l
c(ore)f
f(z)dz + J r R(x){(logx)2 - (logx + 2ri)2)dx Res(a; f (z)).
f (z)dz = 2iri
+ (O;r)
aED
There are M, M' > 0 such that R(z)(log z)2dz <= M((loge)2 + 4ir2)2Ae
(e --i 0),
J-C(O;c)
ko;r)
R(z)(logz)2dzl <
M,((logr)2+4ir2)2rr
(r - oo).
4. RESIDUE THEOREM
114
f
Therefore, 00
- 41ri
R(x) log xdx + 4x2
o 0
R(x)dx
0
Res(a; R(z)(log z)2).
= 21r aE D
EXAMPLE. Calculate I =
nogx2dx, a > 0. Note that R(z) = 1/(a + z2) 0
has no pole at 0 nor on the positive real axis, and satisfies (4.4.2). Its poles are z = ± fai, of order 1. One has (log z)2
Res + fa
(
Res
\-
= (log fa + a /2)2
2fi
a+z2 }
f i; l
2) =
(log Vra +
/2)2
-2v
2
Hence,
I=-2Re{-7 (logfa+si)}= mar. EXERCISE 5. Calculate the following integrals. i)
r°Q (x l+og x
x
Jo
°O
(a > 0).
dx
log x
Jo
(a + x)(b + x) dx
Jo
(1 + x2)2
log x
(a, b > 0
).
dx.
Problems 1. Obtain the Laurent series of the following functions. e1/s z 1) 1-z (about z = 0). ii) sin z 1 (about z = 1). z
(about z = oo).
in) 1
+3Z2
2. What are the poles and residues of the following functions on C or C? i) z" (n = ±1,:k2,.. . ). ii) tan z (on C). iii)
z
iv )
a 1/=
z2+5z+6'
3. Calculate the following curvilinear integrals. 1
`)
JC(0;1)
z(z2 + 2) ex
ii) C(0;1)
1
2n
dz
+ a2
dz.
(n = 1, 2,. .. ).
v)
1 z+z+1
3
Il
PROBLEMS
z"(z - 1)mdz
115
(n, m E Z).
IC(O;2) iv)
(O;1)
(z - a)^ (z - b)-
dz
(IaI<1
I z2dz (C={z=a+iy;-oo
V)
4. i) How many roots of the equation, z6 - 2z5 + 7z4 + z3 -Z+ 1 = 0, are there in A(1)? ii) How many roots of the equation, z4 - 6z + 3 = 0, are there in the annulus R(1, 2)?
5. Let f (z) be a holomorphic function on i(R) such that If (z) I < OR (0 < 0 < 1). Show that there is a z E 0(R) with z = f (z). 6. Let f,g be holomorphic functions in a neighborhood of A(1) such that If (z) I > tg(z)I on C(0;1). Then, show that zf (z) + g(z) has a zero in A(1). 7. Let {f,} ,0°_ 1 be a sequence of holomorphic functions in a domain D C C
which converges uniformly on compact subsets to f with f # 0. Then, show
that a zero of f is a zero of f, n = 1, 2, ... , or an accumulation point of zeros of f,,, n = 1, 2, ... . 8. Calculate the following integrals. i)
0082 z dx Jo a+sin x 0o
iii)
p.v.100
ii)
f
Cos
(a E R).
xZdx X2 +a
o
2W
2
iv) J
x3
2a2 dx
V) I
(a > 0).
cosxdx
a
(a E R).
0
J00
vii)p.v.
1 o o
viii)
Jo
(1 - x2)x° 2
(x2+
1
dx.
dx
(0
/
log
1
dx.
(a> 1).
CHAPTER 5
Analytic Continuation
The analytic continuation is defined, and the fundamental extension
theorem of Riemann is proved. After the reflection principle and the monodromy theorem are proved, the construction of universal coverings is described, and then the notion of Riemann surface is introduced. We will not go into function theory on Riemann surfaces, but readers will comprehend the necessity of this notion.
5.1. Analytic Continuation _ Let D C C be a domain, and let f be a meromorphic function in D. Let D' C C be another domain, and assume that D' fl D 54 0. If there is a meromorphic function g in D' such that f = g on a connected component U of D' ft D, we define a function f on DUD' so that f = f on D and f = g on D'. The function f thus defined is not necessarily one valued, for the values of f and g on other components of D' fl D may be different. We call f the analytic continuation of f from D to D U D' (or sometimes simply D').
FIGURE 47
In general, we call it the analytic continuation extending a meromorphic func-
tion to one defined on a larger domain, allowing the resulting function to be multi-valued. The above meromorphic function g is unique if it exists and if it coincides with f on U (Theorem (4.2.2)). This property is called the uniqueness of analytic continuation. The following theorem is called Riemann's extension theorem.
(5.1.1) THEOREM. Let f be a holomorphic function in 0(r) \ {0}. If there is a 117
5. ANALYTIC CONTINUATION
118
0 < r' < r such that f is bounded on,&(r') \ {0}, then f is uniquely extended to a holomorphic function on 0(r). PROOF. Expand f to a Laurent series: 00
f(z) _
n=-x
az"
By the assumption, there is an M > 0 such that If (r"e'8)1 <_ M for all 0 < r" < r' and 0 S 0:5 2,r. It follows from (4.1.6) that 0C
21r
jf(r-e'8)12d0 < M2.
1
o
n=- 3C
Letting r" \ 0, we have that an = 0 for n < 0, and that 00
anzn
f (z) _ n=O
The right hand side defines a bolomorphic function in A(r). The uniqueness is clear.
As in Theorem (5.1.1), let f be a holomorphic function defined in a neighbor-
hood of a point a except at a itself. If f has an analytic continuation so that it is holomorphic about a, a is called a removable singularity. As a corollary of Theorem (5.1.1) we have Casorati-Weierstraas' theorem:
(5.1.2) THEOREM. Let f be a holomorphic function in A(r) \ {0} such that it has 0 as an isolated essential singularity. Then the image f (0(r) \ {0}) is dense in d; that is, f (A(r) \ {0}) = C.
PROOF. Assume that f (i(r) \ {0}) # C. Then there are a point p E C and a neighborhood U of p such that f (0(r) \ {0) n U = 0. Suppose that p = oo. Take 6 > 0 such that {tu; IwI < 6} c U. Since unit = 1,
f (A(r) \ {0}) c A(1/6).
Hence, f is a bounded holomorphic function in 0(r) \ {0}. It follows from Theorem (5.1.1) that f is analytically continued about 0; this is a contradiction. If p E C, we consider g(z) = 1/(f (z) - p). We deduce from the above that g is analytically continued about 0, so that f (z) = p + 1/g(z) has at most a pole at 0; this is again a contradiction.
In the above theorem, it can actually be proved that the image f (0(r) \ (0)) contains all points of d except for at most two points. This is called the big Picard theorem, which will be proved in Chapter 6. A holomorphic function on all of C is called an entire function. Let f be an entire function. Then one of the following three cases holds: i) f is a constant function.
5.1. ANALYTIC CONTINUATION
119
ii) f has a pole at oo. iii) f has an isolated essential singularity at oo. These correspond respectively to the following cases, if f is expanded to a Taylor series
x
f(z) = >anzn n=0
i) an=0,n>1. ii) There is some number no such that an = 0, n > no; i.e., f (z) is a polynomial.
iii) There are infinitely many n with an 54 0. In the last case iii), f is said to be transcendental.
(5.1.3) COROLLARY. The image f (C) of a non-constant entire function f is dense in C.
PROOF. If f is a polynomial, the equation f (z) - w = 0 has a root for every w E C (Theorem (3.5.23)). Thus f (C) = C, and f (C) = C. If f is transcendental, it suffices to apply Theorem (5.1.2) for the function f o z in
z#0. In the above corollary, if f is transcendental, then in fact it will be proved in Chapter 6 that f takes every point of a infinitely many times except for at most two points; this is called the little Picard theorem. The next theorem is called Schwarz' reflection principle.
(5.1.4) THEOREM. Let D C C be a domain such that the boundary 8D contains
an open interval L on the real axis, and OD \ L does not accumulate at any point of L. Let f be a continuous function on D U L which is holomorphic in D. Assume that f takes real values on L. Then f is analytically continued to
D=DU{zEC;zEDUL}.
FIGURE 48
PROOF. Set D' = {z' E C; z' E D}, and
g(z') = f (Y),
z' E DU L.
5. ANALYTIC CONTINUATION
120
On L, g = f. We expand f about an arbitrary point a' with a' E D' to a Taylor series:
00
f (z) = E an (z - d')',
z E 0(a ; r) C D.
n=0
Then
g(z') _
an(-' - ii)"
an(z' - a')",
x E A(a'; r).
Hence g is analytic in D', i.e., holomorphic there. Define a function f on D =
D U L U D' so that f = f on D U L, and f = g on D' U L. The function f is holomorphic in D U D', and continuous on D. It remains to prove the analyticity of f about an arbitrary point a E L. Take a small 6 > 0 with 0(a; b) C D. To prove that f is holomorphic in 0(a; b) it suffices by Morera's Theorem (3.5.18) to show that for an arbitrary triangle contour C (with the anti-clockwise orientation) inside A(a; 6) f(z)dz = 0.
(5.1.5)
Jc
If C C D or C C D', the interior E of C is also contained in D or D', and so Cauchy's integral Theorem (3.4.14) implies (5.1.5). Assume that C fl L # 0. Taking a sufficiently small c > 0, we set
EE ={zEE;Imz>E},
FIGURE 49
EE ={zEE;Imz>-E}.
Suppose that E,1- 0 0 and E, 0 0. Let C} be the closed piecewise linear curves of the contours of Et with anti-clockwise orientation. Then f f(z)dz
= 0.
Moreover,f + f(z)dz --,
J rGn{Imz>0} J
J f(z)dz -. J
Cn{Im z_U}
f(z)dz + J
f(z)dz
(E
0),
rLfE
f (z)dz -
JLn E f (z)dz
Here one notes that in the second terms of the above right sides the orientation of the closed interval L fl E C R is the natural one, compatible with the order of the real number R. Thus we get (5.1.5). If one of Eis empty, say E= 0, fCf (z)dz
Therefore, in either case, (5.1.5) holds.
J/(z)dz.
5.1. ANALYTIC CONTINUATION
121
A curve 0 : [To, T1 ] - i C is called an analytic curve if for an arbitrary point to E (To, T1) with zo = 0(to) 54 oo, there area 6 > 0 with (to -b, to+b) C (Ta, T1) and real analytic functions 01, 02 on (to - b, to + b) such that
0(t) _ 01(t) + io(t),
to - b < t < to + b;
when zo = oo, the same property as above with respect to the coordinate z is to be satisfied. In particular, if for any t E (To, T1) 0'(t) $ 0, or gz o 0(t) & 0, then ¢ is called a non-singular real analytic curve. Let D C C be a domain. Assume that a part of the boundary 8D is represented by a non-singular analytic curve C(O : (To,T1] -+ 8D). Set O
C = C \ {4(To),m(T1)}.
Assume that the set OD \ C does not accumulate at any point of C. Let to E (To, T1) be an arbitrary point, and set zo = 0(to). Assume for a
moment that zo 0 oo. Taking a sufficiently small b > 0, we have a Taylor expansion of 0 about to: For t E (to - b, to + b) 30
c E C.
fi(t) = zo + E n=1
(5.1.6)
Introduce a new complex coordinate
1;=t+is,
t,sER.
It follows from (5.1.6) that oc
OW = zo +
cn(t - to)"
if - tol < b,
n=1
is a convergent power series, and defines a holomorphic function. Since 0'(to) = c1 96 0, by the inverse function Theorem (4.3.15) 0(1:) may be assumed to be injective on A (to; 6). Set
u+ = A(to; b) n {Iml; >o}, u- = L V
= (t0-6,t0+5), = 00(to; b)),
A(to; b) n {Imp
= O(L), V± = O(U±). Co
By the assumption we may suppose, after taking a smaller b > 0 if necessary, that
OD n v = Co. It does not lose generality to assume that V+ C D (if V- C D, one may replace s by -a). Then C E U+ and Z E U- are the reflection points of
5. ANALYTIC CONTINUATION
172
each other with respect to L. We say that O(C) and O(Z) are the reflection points of each other with respect to Co.
FIGURE 50
We show that this concept of reflection does no depend on the choice of OD be a change of parameter of 0, which parameters of y. Let 4 : [To,T1I gives C and is non-singular and analytic. Take 4 with 0(io) = 0(to), and define A(bo; b), Ut, L, V. ..., etc., as above. Let b > 0 be sufficiently small so that the values of the holomorphic function 0' o t on A(io; 6) are contained in A(to; 6). Set h = 0-1 o I,1(to; 6) : A(to; 6) --' A(to; 6).
The holomorphic function h takes real values on L: h(L) C L. Therefore it follows from Theorem (5.1.4) and the uniqueness of the analytic continuation that for f E A(to; 6), h(t) = h(1). Setting t = h(t), we have 0w _ k)T
0(7 =0(0-
Hence, two points which are reflection points with respect to Co are also reflection
points with respect to Co. In the case of zo = oo, we use the coordinate i to define the notion of reflection points as well.
EXERCISE 1. Show that if C is a circle (the case of a line is included), the notion of reflection points defined in Chapter 2, §8 coincides with one defined here.
(5.1.7) THEOREM. Let D; C C, i = 1, 2, be domains such that a part C, of 0 8D, is a simple non-singular analytic curve and 8D, \ C, does not accumulate 0
at any point of C.,._Let f be a meromorphic function on D1 with values in D2 as a mapping into C. Assume that f is extended to a continuous mapping over 0 0 0 D1 U C1 so that f (Cl) C C2. Then, if z E DI is sufficiently close to C1, to its 0 reflection point z' with respect to Ci there corresponds the reflection point f (z)' 0 of f (z) with respect to C2, so that f is analytically continued, crossing C1, to a 0 domain containing C1 as a meromorphic function.
5.1. ANALYTIC CONTINUATION
123
FIGURE 51
0
0
PROOF. Take arbitrarily a point z1 E C1, and set z2 = f(z1) E C2. We suppose that in a neighborhood of z,, C, is represented by a simple non-singular analytic curve O.: (-b;, b;) - C; with b; > 0. As described above, we extend 4;
over a complex variable {, E 0(b;). We may assume that O; : 0(b;) - MA00) is injective. Let U, denote the part of A(b;) in the upper half plane. Take b1 > 0 sufficiently small. Then the composite function 021 o f o 01 : Ui - U2 is holomorphic, extends continuously over the real axis (-bl, b1), and takes real values on it. By Schwarz' reflection principle, Theorem (5.1.4), 021 o f o 01 extends to a holomorphic function g on 0(b1) with values in i (b2). Set h1 = 02 o g o 0,' on V1 = 01 (A(61)). Then f is analytically continued to h1 on V1. e V1(zj ). Then W is a domain, We write V1(zl) for this V1, and set W = U 0
z,EC,
C1, and there is a meromorphic function h such that h = f on W n D1. Thus f is analytically continued to D1 U W. 0 W
REMARK. When C1(0 : [To, T1] - 8D) is a closed curve in Theorem (5.1.7), there are two possible cases:
i) There is a change m : [To,TI] - OD of parameter of
such that
is
non-singular analytic and for some io E (To, T1), ¢(to) _ O(To) ii) There is no such change of parameter of 0. If C2 is not of case i) at f oO(To), f cannot be analytically continued to a domain containing i(To) no matter whether C1 is of case ii) or i) (see Figure 52, below). If C1 (reap., C2) is of case i) at 0(To) (reap., f o O(To)), then f is analytically
124
5. ANALYTIC CONTINUATION
continued to a domain containing C1.
f(Z) =
FIGURE 52
EXERCISE 2. Let f (z) be a continuous function on 0(R) that is holomorphic
in 0(R), and let If (z)I - K (constant) on the boundary C(0; R). Show that f (z) is a polynomial. EXERCISE 3. Show that if an entire function f (z) takes real values on the
real axis, and purely imaginary values on the imaginary axis, then f (z) is an odd function.
5.2. Monodromy Theorem A meromorphic (resp., holomorphic) function element about a point a E C is defined as a pair (Fa, Ua) of a meromorphic (resp., holomorphic) function in a disk neighborhood Ua of a, and U. is called the disk of definition. We sometimes abbreviate (Fa, Ua) by Fa, calling it a meromorphic or holomorphic function element about a. Let C(O : [0, 1] -+ C) be a curve. Assume that a meromorphic function element (F,(t), Ut) is given for every t E [0, 11. We write Ft = Fo(t). We say that {(Ft, Ut)}tE(o,11 is an analytic continuation (of (Fo, Uo)) along C if for an arbitrary to E [0, 11 Fto - F t
on Uta fl Ut,
t E (t1, t2),
where t1 = inf{t E [0, to]
q5([t,to]) C Uto},
t2 = sup{t E [to, 11 : O([to, t]) C U10 }.
FIGURE 53
FIGURE 54
5.2. MONODROMY THEOREM
125
In this case, the disk Ut of definition of every Ft is written as 0(t) j4 oo,
Ut = {(t);rt),
{z E C; IzI < rt},
4(t) = oo,
and by increasing rt if necessary we may assume (cf. Figures 53 and 54) that (5.2.1)
inf{rt; t E [0, 1]} > 0.
(5.2.2) THEOREM. Let {(Ft, Ut)}tE[o.1! and {(Gt, Vt)}tE[o,1] be two analytic con-
tinuations along a curve C(t¢ : [0,1] C). If Fo = Go in a neighborhood of ¢(0), then F1 = G1 in a neighborhood of 0(1).
PROOF. Set E _ {t E [0, 1]; Ft = Gt in a neighborhood of 0(t)}. By the assumption, E # 0, and E is clearly open in [0,1]. Take an arbitrary to E E. Take a connected neighborhood Ut, of q(to) so that it is contained in the disks of definition of Ft,) and Gto. If t' E E is sufficiently close to to, O(t') E Ut,. Since Fta and Gt,, coincide on a neighborhood of O(t'), Ft. = Gto on Ut0 by Theorem
(4.2.2). Thus to E E, and so E is closed in [0, 1]. We have that E = [0,1]. 0 Let C} (O, : [0,1] x [0, 1] --+ C), j = 0, 1, be two curves which are homotopic to each other. Let the homotopy connecting them be
4':[0,1]x[0,1]-4C, 4'(t, 0) = 00 (t),
4'(t,1)
(t).
Then we have the following monodromy theorem
(5.2.3) THEOREM. Let CJ, j = 1, 2, and 4 be as above. Suppose that for every (t, s) E [0, 1] x [0,1] a meromorphic function element F(t,,) about 4i(t, a) is given so that for each fixed a E [0, 1] {F(t.,)}t gives rise to an analytic continuation along C,(0. = If F(o,o) = F(o,8) in a neighborhood of the initial point P = t>5o(0) (0 < s < 1), then F(1.o) = F(1,1) in a neighborhood of the terminal point Q = 0o(1)
PROOF. Set S = {s E (0, 1]; F(1,.) = F(1,o) in a neighborhood of Q). Since 0 E S, S 0 0. We write U(t,,) for the disk of definition of F(t.9). Note that U(t,3) is connected. Fix arbitrarily so E S. We may assume (5.2.1) for {F(t,,0) }t. By the uniform continuity of 4 there is 6 > 0 such that for every a E (so -b, so+b)fi [0,11 (5.2.4)
0 (t, s) E Ut.,,,
0 < t < 1.
126
5. ANALYTIC CONTINUATION
FIGURE 55 We
FIGURE 56
fixsE(so-b,so+6)fl[0,1],andset T = it E (0, 1); F(t,,) = F(t,,,) on U(t.,) fl U(e.,o) }.
Since 0 E T, T # 0. In the same way as in the proof of Theorem (5.2.2) one sees that T is open and closed in [0, 1]. Hence T = 10, 11, and so in a neighborhood of Q, F(1,,) = F(,.,o) = F(i,o). It follows that S is open. In the above arguments, without assuming so E S, we infer that if there is an s which satisfies (5.2.4) and belongs to S, then so E S. Hence S is also closed, and then S = [0, 1].
Let f be a meromorphic function on a domain D C C. Let P E D, and consider the meromorphic function element obtained by the restriction of f to a disk neighborhood of P in D. We consider meromorphic function elements (FQ, UQ) produced by all possible analytic continuations of f along curves in C with initial point P, and set b = U UQ. Then b is a domain containing D, and we have a multi-valued meromorphic function 1(Q) = Fq(Q). It follows from Theorems (3.3.6) and (5.2.3) that (5.2.5) THEOREM. Let f and b be as above. i) Let Q E D. Identify the meromorphic function elements (FQ, UQ) and (Gq, Vq) if Fq and Gq are identical in a neighborhood of Q contained in Uq fl VQ. Then the number of meromorphic function elements of f defined about Q is at most countable. ii) Let Do c D be a simply connected domain such that for a point Q E Do a meromorphic function element (Fq, Uq) about Q can be analytically continued along any curve in Do with initial point Q. Then f defines a one-valued meromorphic function fo on Do.
The above fo in ii) is called a branch of f. The identification of meromorphic function elements used in the above i) induces an equivalence relation in {(FQ, Uq)}. Let [Fq, Uq] denote the equivalence class of (FQ, Uq). We denote by X the set of all [FQ, UQJ. We introduce a topology on X as follows. It is sufficient to define neighborhoods of [FQ, Uq] E X. For every P E Uq, we take a
5.2. MONODROMY THEOREM
127
disk neighborhood Up C UQ and the restriction Fp = FQJUp of FQ to Up. We define the system of neighborhoods of (Fq, Uq] by {[Fp, Up]; P E V},
where V runs over all neighborhoods of Q in UQ. Thus X is a topological space, which is called a Riemann surface defined by f. Defining f ([FQ, Uq]) = Fq(Q), f gives rise to a one-valued function on X. (5.2.6) EXAMPLE. The Riemann surface defined by the multi-valued meromorphic function
f(z)=logz, zED=C\{0}, is obtained as follows. Take countably many complex planes Cj, -oo < j < co. Let L,1 (resp., L,) denote the real pos-
=C
itive axis in CJ \ {0} as accumulation points of the upper (resp., lower) halfplane, and distinguish between Lt and L7. Then we identify L with L,_1 of
CJ_1 \ {0} for all -oo < j < oo (see Figure 57). Let X denote the resulted
topological space. Defining f(i) = 0 at 1 E Co\ {0}, we obtain a one-valued func-
tion f on X.
FIGURE 57
Making use of g(w) = ew, we identify injectively G; = {z E C; 2ja < Im z < 2(j + 1)ir} with (C3 \ {0}) U L+:
g1G.' :Gj-+(C.\{0})UL?. G; = C, X is identified with C. The logarithmic function w = log z is the inverse function of the holomorphic function z = z(w) = e', and so the above Riemann surface X is also called the Riemann surface of the inverse function of z(w) = e'°. Since trJ,°
FIGURE 58
5. ANALYTIC CONTINUATION
128
(5.2.7) EXAMPLE. The Riemann surface X of the inverse function w = f of z(w) = w2 is obtained by gluing (identifying) two copies of C along the nonnegative real axis (see Figure 58). Let Do be a simply connected domain in
C \ {0}, and take a branch fo(z) of f on Do. The other branch of f is f, (z) = - fo(z). Set G; = f,(Do), j = 0,1. Then G; is a domain of C \ {0}, the mapping
fi : Do -. Gi is injective, and Go n G, = 0.
FIGURE 59
In general, if the function fin Theorem (5.2.5) is expanded to a Puiseux series with an integer p > 2,
f(z)
_
an(
z - a)n
(a
oo,0 < N < oo),
n>-N
foz
=
an(v z)n
(a = oo, 0 < N < oo),
n>-N and if p cannot be reduced to a smaller one by another Puiseux series expansion,
then a is called a brunch point, and p - 1 is called the order of the branch. If in a neighborhood of a f is expressed as a meromorphic function of log(z - a) or log z, and is an infinitely-many-valued function, then a is called a logarithmic branch point. For instance, if a E R is irrational, then
f(z) = z° has logarithmic branch points at z = 0 and coo.
5.2. MONODROMY THEOREM
129
(5.2.8) EXAMPLE. We consider the Riemann surface X of the inverse cosine function w = cos-1 z. 00
00
w = C08-1 z
FIGURE 60
Let L 1 denote the half line Re w = -ir, Im w >_ 0 in the w-plane from oo to -7r. Let L2 denote the line segment of the real axis from -7r to 0, and let L3 denote the half line Re w = 0, Im w >_ 0 from 0 to oo. Let Go be the domain surrounded by them. Let Cj be the image of Lj mapped by z = cos w with orientation for j = 1, 2, 3. Then C1 runs from oo to -1 on the negative real axis, C2 from -1 to 1 on the real axis, and C3 from 1 to oo on the positive real axis. Since a holomorphic function preserves the orientation except at points where the derivative vanishes, Go is injectively mapped to the upper half plane Do. This mapping is analytically continued by the reflection principle, Theorem (5.1.7) or Theorem (5.1.4). The reflection of Do with respect to C1 is the lower half plane D1, and its image by cos-1 z is the reflection G1 of Go with respect to L1. We carry out this continuation process with respect to Cj, j = 1, 2, 3, as far as possible. The branch points of cos' -z are -1 and 1, and their values are 2nir and (2n + 1)ir (n E C), respectively. The orders of branch are 1. Thus the Riemann surface X is identified with C. EXERCISE 1. Construct the Riemann surface of sin-1 Z. EXERCISE 2. What is the Riemann surface of 1 - cos z? Show that in fact it defines two 1-valued entire function on C.
Let F(z1i... , zn) be a complex function of n variables, z1, ... , z,,. Set zj = xj + iyj, 1 5 j 5 n. If F is continuously differentiable in the variables x., and 1 5 j 5 n, and if for every j the Cauchy-Riemann equations (5.2.9)
1
8
2 \ 8X,
1 8 )F=O, i 8yj
1<j:n,
hold, then F is said to be holomorphic in (z1, ... ,
Let fi (z),1 5 j 5 n, be n holomorphic functions of z, and assume that the composite F(f1(z),... , fn(z)) is defined. Then, if F is holomorphic, so is Let C(O : (0,1] -+ C) be a curve and let {fjt)1i 1 5 j 5 n, F(f1(z),... , be analytic continuations of holomorphic functions along C. Let F(w1, ... , w)
5. ANALYTIC CONTINUATION
130
be a holomorphic function such that all composites F(flt(z),... , f,,t(z)) are defined. Then it follows from the identity Theorem (2.4.14) that (5.2.10)
F(f11(z),... ,f",(z)) ° 0.
F(f1o(z),... ,f,,o(z)) =-O
This is referred as the principle of the permanence of the functional relation of analytic continuation.
5.3. Universal Covering and Riemann Surface Let D C C be a domain and fix a point P E D. Let 7r1(D)p denote the set of all homotopy classes (Cl of closed curves C in D with the initial and terminal point P. A multiplication in 7r1(D)p is defined by
{Cl} {C2} _ {C1 +C2}. Here it is sometimes written as {C1 } + {C2 }, while the multiplication is not commutative in general. Let e denote the homotopy class {C} E 7r1(D)p of C which are homotopic to the point P. We set {C}-1 = {-C) for {C} E 7r1(D)p. Then 7r1(D)p forms a group with unit element e, and is called the fundamental group of D with base point P. For another point Q E D we have a group isomorphism (5.3.1)
tQ : 7r1(D)p 9 {C}
{CQ + C - CQ} E 7r1(D)Q,
where CQ is a curve from Q to P in D.
FIGURE 61
Thus the groups irl (D)p and 7r1(D)Q are mutually isomorphic as abstract groups.
We simply denote it by 7r, (D), and call it the fundamental group of D. Let D' be another domain, and f : D D' be a continuous mapping. Then we have a group homomorphism (5.3.2)
f.: 7r1(D)p E) {C} -+ {f(C)} E 7r1(D')fiQi.
Here f (C) is defined by f o 0 with C = C(O).
The continuous mapping f : D -* D is called a covering mapping, and D is called the covering of D' if (5.3.3) i) f (D) = D';
5.3. UNIVERSAL COVERING AND RIEMANN SURFACE
131
ii) for every point w E D' there is a neighborhood V C D' of w such that for any connected component U of f -'(V), the restriction off to U,
flu: U- V, is a homeomorphism (a surjective and injective continuous mapping with continuous inverse mapping).
EXAMPLE. 1) f : C 3 z - ez E C' is a covering mapping. 2) f : C' 3 z z2 E C' is a covering mapping. EXERCISE 1. Let f : D -+ D' be a covering mapping. Show that f (w) has no accumulation point for every w E D'. EXERCISE 2. Show that the fundamental group of C' is isomorphic to Z.
Now, let D be as above, and fix a point Po E D. Let X be the set of all homotopy classes ap of curves from Po to P in D, where P moves over D. Take an arbitrary point P E D and a disk neighborhood Up. For a point Q E Up we take a curve L from P to Q in Up. Then the homotopy class ap + {Cp } is independent of the choice of Cp , since Up is simply connected.
FIGURE 63
FIGURE 62
Set
V(ap) = {aP + {C }; Q E UP}.
If ap 0,3p, then (5.3.4)
V(ap) fl V(13p) = 0.
Let co E X be the homotopy class of the constant curve P0, and let (5.3.5)
7r:X 3ap-+PED
be the natural projection. We define a topology of X by taking neighborhoods V(ap) for each point ap E X; that is, a point sequence {an}°°_0 of X converges to ap E X if and only if for an arbitrary neighborhood V(ap) of ap there is a number no such that an E V(ap) for all n > no. The restriction mapping (5.3.6)
7rIV(ap) : V(ap) -+ Up
5. ANALYTIC CONTINUATION
132
is a homeomorphism. Take two distinct points ap 31 fig E X. If P 36 Q, then, taking Up and UQ with Up n UQ = 0, we define V(ap) and V(13q) so that
V(ap)nV(fQ)=0; if P = Q, the same holds by (5.3.4). Therefore, X is a so-called Hausdorff topological space. For ap = {C(Q+ : [0,1] -+ D)} we have a continuous mapping (5.3.7)
0 : [0,1] 30 t
am(p) _ {C(0J[0, t])) E X,
and 0(0) = co and O(1) = ap. Thus X is arcwise connected. While we do not know if X is a domain of C, the mapping a : X - D defined by (5.3.5) satisfies the properties of (5.3.3), i), ii). We also call this a covering. If we choose another
Po E D, we have another X, which is homeomorphic to the original one, and so we may identify them as a topological space. We call this X the universal covering of D, and it : X --; D the universal covering mapping. EXERCISE 3. Let Po E D and X be as above. We take another point FO E D, and then obtain another X' as X. Let CP° be a curve from Po to Po. Show that (5.3.8)
: X' E) apo
{CPo } + apo E X
is a homeomorphism. An element -y E 1r1 (D) = art (D)p,, of the fundamental group defines a homeomorphism (5.3.9)
ry:X3a-+ry+aEX.
For two elements -r1. y2 E a1(D) (5.3.10)
'Y1 C ry2 = 71
'72 (= 71 +'Y2),
where the left side is the composition of the mappings and the right side is the multiplication in the group ir1(D). In this case, we say that the group a1(D) acts on X. The set {ry(a)-,1y E ir1(D)} with an element a E X is called the ir1(D)-orbit of a. It follows that {ry(ap);'y E ir1(D)} = {y(aQ);7 E 1r1(D)}
FIGURE 64
P = Q.
5.3. UNIVERSAL COVERING AND RIEMANN SURFACE
133
Considering a r1 (D)-orbit as one point, we get a quotient space X/r1 (D) of X by the action of rl(D). We have
X/r1(D) = D.
(5.3.11)
When an element 7 of r1 (D) is considered as a homeomorphism defined by (5.3.9), r1(D) is called the deck transformation group or the covering transformation group. For distinct ap, /3p E X with P E D, set 7 = Op - ap E r1 (D). It follows from (5.3.9) that -y: X
X satisfies
7(V(ap)) = V0p)
(5.3.12)
In general, let Y be a closed interval of R, an open interval of R, a domain of C, or some finite product of them. Let f : Y --. D be a continuous mapping.
A continuous mapping f: Y - X is called a lifting of f if f= r o f:
D
Y f
For a curve C(O : [0,11
D) with the initial point P E D and ap E X we set
1 : [0,11 9 t - ap + {CI[0, t1} E X.
(5.3.13)
Then r o fi(t) = '(t), so that tai is a lifting of ' and gives a curve CQP, called a lifting of C. We see by (5.3.6) that a lifting of C is unique if the initial point ap is given. Therefore, for a lifting Cpp of C with the initial point /3p 0 ap we have
CQP nOp. =0.
(5.3.14) ca,
x j
p
D C
FIGURE 65
ar
D
C p
FIGURE 66
If C is closed, then -y = ap + {C} - ap E r1 (D) and it follows from (5.3.9) that (5.3.15) i) -y(ap) = the terminal point of CQP.
5. ANALYTIC CONTINUATION
134
ii) {C} is homotopic to a point if and only if y(ap) = op. Since X is arcwise connected, the notion of curves, their homotopy, and simple connectedness is defined for X in the same way as in the case of domains of C. (5.3.16) LEMMA. Let it : X - D be as above. Let 4) : [0,1] x [0,1] - D be a
continuous mapping, and set P = 0(0,0). For an arbitrary op E X there exists a unique lifting 4: [0,1] x (0,11 - X with 4)(0, 0) = op. PROOF. Take a lifting 4(., 0) of 0) : [0,1] 3 t --+ 4(t, 0) E D with 4)(0,0) = op. By making use of (5.3.6) we see that there exist a a > 0 and a continuous mapping 4: [0, 1] x [0, or]
X such that 4)(0, 0) = op and tr o 4 = 0.
Let S be the set of all such o E [0,11. In the same way as in the proof of the monodromy Theorem (5.2.3) we infer that S is open and closed, so that S= [0, 1]. To show the uniqueness, we take a continuous mapping 'P : [0,1] x [0,11 - X such that W(0, 0) = op and it o is = 4). Set E = {(t, s) E [0,11 x [0,1J; 'Y(t, 8) _ +$(t, s)}.
Then E is clearly closed, and E 3 (0, 0). It follows from (5.3.6) that E is open in [0,1] x [0, 1]. Thus E = [0,11 x [0, 1], and 1i = 4. Let 4' be another lifting of 4) in Lemma (5.3.16). Then, by (5.3.12) there is a y E 7r, (D) such that (5.3.17)
6'=yo4.
(5.3.18) THEOREM. Let D and X be as above. i) X is simply connected. ii) Let Do C C be an arbitrary simply connected domain, and let f : Do -. D be a continuous mapping. Then, for an arbitrary of(.,) E X with
zo E Do there is a unique lifting f : Do - X such that f (zo) = of(...) and n o f = f. If j' is another lifting of f, then there is a y E Ai(D) such that j' = y o f . PROOF. i) Take a closed curve e in X with the initial point op E X. Then C = ir(C) is a curve in D, and C is a lifting of C with the initial point ap. It follows from (5.3.15) that C must be homotopic to a point. Let fi : [0,11 x [0,11 ---
D be a homotopy connecting C and P. By Lemma (5.3.16) there is a lifting 4) : [0, 1] x [0, 1] - X with 4)(0, 0) = op. It follows that $ is a homotopy connecting C and op. ii) Take an arbitrary point z E Do and a curve C from zo to z. Let C be
the lifting of f (C) with the initial point afand denote the terminal point by j(z). We see by Lemma (5.3.16) that %(z) is independent of the choice of C. For every point z' E Do we take a sufficiently small neighborhood W C Do so
5.3. UNIVERSAL COVERING AND RIEMANN SURFACE
135
that for neighborhoods Uf(z,) of f (z') and V(f (z')) of j (z') satisfying (5.3.6), f (W) C Uf(z,), and zEW. f(z)EV(f(z')), Since it o j (z) = f (z) is continuous and 7rlV (f (z')) is a homeomorphism, f is continuous. The latter half follows from (5.3.17). EXERCISE 4. Let ire : X3 -. D3, j = 1, 2, be universal covering mappings,
X2 be a lifting of and let f : D1 - D2 be a continuous mapping. Let f : X1 f o 7r1i and let f.: n1(Dl) -+ Tri (D2) be as in (5.3.2). Then, show that
Jo. =f.(7)°f,
7Enl(D1)
EXERCISE 5. Set Dj = C', j = 1, 2, and A : D1 3 z -- z" E D2. 1 + t E D2. i) Obtain all liftings of the curve C, given by 01 [0,1] 3 t e2'"'t E D2. ii) Obtain all liftings of the curve C2 given by 02 : [0, 1] 3 t EXERCISE 6. The mapping it : C (= D1) 3 z -+ e2' E C' (= D2) is a universal covering mapping. In this case, answer the questions similar to the above i) and ii).
Let X be as above. Then every point ap E X has a neighborhood V(ap) which is homeomorphic to a disk neighborhood Up (c D) of P. We call V(ap) a disk neighborhood of ap. There are a complex coordinate z in Up for P 34 oo, and z in U,,,. Through the restriction irIV(ap), the points of V(ap) are identified with the points of Up, so that we introduce a complex coordinate zv(Qp) in V(cep) by making use of z or z. The mapping (5.3.19)
zv(ap) : V(ap)
zv(ap)(V((P))(C C)
is homeomorphic, and if V (ap) n V (,OQ) 0 0, (5.3.20)
ozvI P) : zv(ap)(V(ap) nV(AQ)) -' zv(RQ)(V(aP) nV(fQ)) is a homeomorphism, and a holomorphic function as well. We call zv(ap) a ZV(RQ)
holomorphic local coordinate in V(ap) of X. In general, if every point of a connected Hausdorff topological space has a neighborhood endowed with a complex function satisfying (5.3.19) and (5.3.20), then the space is called a Riemann surface, and the complex function is called a holomorphic local coordinate. The Riemann sphere C is a Riemann surface with holomorphic local coordinates z and z, which is compact and simply connected. The Riemann surfaces of a multi-valued meromorphic function and of the inverse function dealt with in the last section are Riemann surfaces in this sense. The above X gives rise to a simply connected Riemann surface.
EXAMPLE. The mapping it : C (= D1) 3 z - e2x`z E C- (= D2) is a universal covering mapping, and C is the Riemann surface of the inverse function
136
5. ANALYTIC CONTINUATION
(I/2wri) log w of the holomorphic function w = e2*'z. The fundamental group of
C' is Z, and acts on C by
C3z-.z+nEC,
nEZ.
REMARK. We have a "global' holomorphic Coordinate z in C or in its domain D. For multi-valued meromorphic functions on D or for the universal covering of D, the coordinate z is no longer "global". Here it becomes necessary to introduce the notion of Riemann surfaces. For Riemann surfaces the notion of holomorphic functions and holomorphic mappings is defined in what follows, and all theory which has been described can be developed with suitable generalizations. It can
be shown that a simply connected Riemann surface must be one of C, C, or A(1). This fact is known as the unifortnization of Riemann surfaces, and the proof requires considerable preparations of function theory on Riemann surfaces. We will prove this fact for domains of C in Chapter 6.
In general, let X1 and X2 be Riemann surfaces. A continuous mapping f : X2 is defined to be holomorphic if for every point P E X1 the function w o f (z) is holomorphic in z, where z (resp., w) is a holomorphic local coordinate in a disk neighborhood V (reap., W) of P (resp., f (P)) with f (V) C W. By (5.3.20) this definition is independent of the choice of holomorphic local coordinates z and w. Setting X2 = C, we call a holomorphic mapping f : X1 C a holomorphic function on X1. The mapping f : D C (cf. (4.2.1)) defined by a meromorphic function on D is a holomorphic mapping. Conversely, a holomorphic mapping f : D C with f ; oo is defined by a meromorphic function. Hence, a meromorphic function on D is a holomorphic mapping from D into C which is not constantly oo. X1
EXERCISE 7. Let X be a Riemann surface. Show that the followings are
defined independently by the choice of holomorphic local coordinates.
i) A function u : X -+ R (or C) is defined to be of class Ck (k 5 oo) if in a disk neighborhood V of an arbitrary point P E X with a holomorphic local coordinate zv(Q) = z = x + iy (Q E V), u o zV_ I is of class Ck as a function in (x, y). [To, TI] -, X is said to be of class Ck if ii) A continuous mapping for every point to E [To, Tl ] the function zv o O(t) is of class Ck in a :
neighborhood of to, where zv is a holomorphic local coordinate in a disk neighborhood V of 0(to). The notion of piecewise class Ck is defined in the same way as above. Let 7r : X -+ D be the universal covering defined in (5.3.5), and consider X and D as Riemann surfaces. Of course, the mapping n is holomorphic. Let Do C C be a simply connected domain, and let f : Do - D be a holomorphic
mapping. The following is clear by definition:
5.3. UNIVERSAL COVERING AND RIEMANN SURFACE
137
(5.3.21) A lifting f : Do -. X of f is holomorphic. For a holomorphic mapping between Riemann surfaces, the Prinzip von der Gebietstreue holds as in Corollary (4.3.14). As in Theorem (4.3.18), the inverse mapping of a univalent (injective) holomorphic mapping f : Xl - X2 between Riemann surfaces X1 and X2 is called a biholomorphic mapping. In this case, by Theorem (4.3.18) the inverse f -1 is holomorphic, too, and X 1 and X2 are said to be mutually biholomorphic. In the special case of X1 = X2, a biholomorphic mapping f : X1 - X1 is called a holomorphic transformation, and the set of all of them is denoted by Aut(Xi ), which forms a group under the law of composition. We call Aut(X1) the holomorphic transformation group of X1. Assume that for every point P E X2 a disk neighborhood Wp with a holomorphic local coordinate wp and a function hp(wp) in wp are assigned. We call the family h = {hp(wp)}pErz an hermitian (reap., pseudo-) metric or a conformal (reap., pseudo-) metric if the following conditions are satisfied:
(5.3.22) i) hp(wp) > 0 (reap., > 0), and hp(wp) is of class C°°. ii) If Wp n Wq 0, -2
hp(wp(a))
'W (wp(a))
= hg(wQ(a)),
a E Wp n WQ.
Because of the above equation we write h = hpdwp dwp = hp(dwp(2. We consider dwp dip = idwp 12 to be transformed as dwp dwQ
2
dwQ dwQ.
X2, the pull-back f'h of h by f is
For a holomorphic mapping f : X1
defined as follows. Take a disk neighborhood V (reap., W) of X1 (reap., X2) with holomorphic local coordinate z (reap., w) so that f (V) C W, and h = h(w)(dw(2
in W. Set (5.3.23)
f'h = h o f
dw o f dz
(dz(2
It is clear that f'h is independent of the choice of holomorphic local coordinates. If dw o f/dz 54 0, then f'h defines an hermitian metric there. If f is biholomorphic, then f'h is an hermitian metric. Let 0 : (To, T1 ] -, X2 be a piecewise continuously differentiable curve. For a t E (To, T1] at which 0 is differentiable, take a disk neighborhood W containing 0(t) and a holomorphic local coordinate to in W as in (5.3.23). Then the function dw
h(O(t)) I
t(t) I
d
(>
0)
5. ANALYTIC CONTINUATION
138
is independent of the choice of w by (5.3.22), ii). We define the length Lh(C) of the curve C = C(0) with respect to h by T
(5.3.24)
Lh(C) =
h(O(t))
! To
Idw
d0(t) I
dt.
For a subset E of X2 we define the area Ah(E) with respect to h by (5.3.25)
Ah(E) = J h(w)dudv,
w = u + iv.
Here it is easily checked in the same way as in the case of length that the right side of the above equation is locally well-defined independently from holomorphic
local coordinates and hence globally, provided that the integration exists. Idzj2. On a\ 101 EXERCISE 8. i) On C\ {oo} = C, we set h = 4(l +1212)-2 C, we set h = 4(1 + IzI2)-2Jdz"I2. Show that these define an hermitian metric on the Riemann sphere C, which is called the Fe+bini-Study metric. ii) Obtain the length of the real axis R with respect to the above Fubini-Study metric. iii) Obtain the area of C with respect to the Flibini-Study metric. EXERCISE 9. Assume that the composition 9o f of two holomorphic mappings f and g is defined. Let h be an hermitian metric defined on a domain containing the image of g. Show that (g o f'(g'h).
Problems 1. Construct the Riemann surface of the multi-valued function z(z -1). What are the branch points and the branch orders? 2. Construct. the Riemann surface of the inverse function of f (z) = z2 - 2z (z E C). What are the branch points and the branch orders? 3. Let X be a Riemann surface and let u be a real valued function of class C2. We define u to be harmonic if for an arbitrary point P E X and for a disk neighborhood V of P with holomorphic local coordinate zy = z = x + iy, the function u o zy 1(x, y) is harmonic in (x, y). Show that this definition is independent of the choice of a holomorphic local coordinate.
4. Let D C C be a simply connected Riemann surface and let f be a nonvanishing holomorphic function in D. Show that there is a holomorphic function g in D such that f (z) = es(z).. 5. Let f be a holomorphic function on a Riemann surface X. Show that if III takes a maximum at a point of X, then f is constant. 6. Let u be a harmonic function on a Riemann surface X. Show that if u takes the maximum or the minimum at a point of X, then u is constant. 7. On A(R)(0 < R < oo) we define an hermitian metric 2 4RzI2)2 Jdz12,
h= (R2
139
PROBLEMS
which is called the Poincar# metric. Obtain the length of the curve Ct,,el(0 :
[0,r] 3 t te'B E .(R)) (0 < r < R), and the area of the disk A(r) with respect to h. 8. Let h = 2a(z)ldzI2 (local expression) be an hermitian metric on a Riemann surface X. Show that the function Kh(z) _
44 loga(z) _
.9. 5. log a(z)
is defined independently of the choice of the local holomorphic coordinate z. We call Kh the Gaussian curvature of h. 9. i) Let hi be the Fabini-Study metric on C (see §3, Exercise 8). Show that
Kh, =1. ii) Show that for the hermitian metric h2 = jdzj2 (the Euclidean metric) on
C, Kh, = 0. iii) Show that for the Poincari metric h3 (cf. problem 6) on 0(R) with
0
CHAPTER 6
Holomorphic Mappings
The aim of this chapter is to prove the Riemann mapping theorem and Picard's theorem, which are the two greatest theorems in the theory of complex functions of one variable. The former is a very specific and essential property of the one variable case, and cannot be extended to the several variable case as it is. It is an open
and deep problem to generalize it in the higher dimensional case. On the other hand, Picard's theorem furnished the starting point for modern complex function theory, and was later developed to a quantitative theory, called the Nevanlinna theory (1925). Furthermore, introducing a geometric method, L. Ahlfors proved that the number 2 of the so-called exceptional values is the Euler number of C, which is a topological invariant of the 2-dimensional sphere (1935). For this work Ahlfors was awarded the first Fields medal in 1936. Picard's theorem and the Nevanlinna theory have been extended to the several variable case in a number of ways, and interesting results have been obtained.
6.1. Linear Transformations As defined in Chapter 2, §8, a linear transformation f is represented by f (z) = (az + b)/(cz + d), ad - be # 0, which is a meromorphic function on a, and also a holomorphic mapping f : C - C.
(6.1.1) THEOREM. i) A holomorphic transformation f of C is a linear transformation. ii) A holomorphic transformation g of C is a linear transformation represented by g(z) = az + b,a 0. PROOF. i) Set c = f (oo). If c 0 oo, we consider 1/(f (z) - c); if this is a linear transformation, so is f itself. Thus we may assume that c = oo. Restricting f to C, we have a bibolomorphic mapping f IC : C - C. Hence, it suffices to prove
141
6. HOLOMORPIUC MAPPINGS
142
ii) Set g(O) = b. Since g-1 is continuous, g-'(i(b;1)) is abounded and closed subset of C. Take R > 0 so that C A(0;R). Let 0C
9(z) = Ea. Zn n=O
be the Taylor expansion of g. The Laurent expansion of h(z) = goi = g(1/z), z E c.(1/R) \ {0}, is given by 0
00
h(i) _ E anz-n = E a_nzn.
-a
n=0
Note that h restricted to A(1/R) \ {0} does not take any value in i(b;1). By Casorati-Weierstrass' Theorem (5.1.2) i = 0 is not an isolated essential singularity of h; that is, z = 0 is at most a pole of h. Therefore, there is a number no such that ano 0 0 and an = 0 for all n > no, and so g(z) is a polynomial no
0"00.
g(z) = b + E anzn, n=1
Since g is biholomorphic, g(z) has no zero. The fundamental theorem of algebra,
Theorem (3.5.23), implies that no = 1, and hence g(z) = a1z+b,g'(z) = a1 96 0.
The next result is called Schwarz' lemma:
(6.1.2) LEMMA. Let f (z) be a holomorphic function on e(1) such that f (0) = 0
and If(z)I < 1. Then
lf(z)I s Iz
If'(0)l <_ 1.
,
The first equality holds for some z i4 0, or the second holds if and only if
0 E R.
f (Z) = e'0z,
PROOF. Set f (z) = En 1 anzn. Then the function 00
9(z) = f zz)
=E
an+1.Zn
n=0
is defined on A(1) and holomorphic there. The assumption implies Um 19(z)I < I.
jZI
1
It follows from the maximum principle (Theorem (3.5.21)) that unless g is a constant such that lgI = 1, Ig(z)I < 1 for z E A(l). Since g(0) = f'(0), If (z)I < Iz1, 1f'(0)1 < 1.
0 < IzI < 1,
If .q= eie (0 E R), then f (z) = eiez; in thias case, If(z)l = Izl and I f'(z)l . 1
6.1. LINEAR TRANSFORMATIONS
143
Let f (z) be a holomorphic function on i(R) such that f (0) = 0 and if (z)l < M. Applying Lemma (6.1.2) to g(z) = M
f
Izl
(Rz),
<
1,
we get (6.1.3)
If W1 <
Izl,
If'(0)I < R
R
(6.1.4) THEOREM. i) A holomorphic transformation of a disk /(a; R) is a linear transformation. ii) A holomorphic transformation of the upper half plane H is a linear transformation.
PROOF. i)Wemayassume that a=OandR=1. Let f:A(1)-A(1)be a holomorphic transformation, and set a = f (0) E A(1). As in (2.8.10) we set
z-a 0°(Z)
-az + 1
It follows from Theorem (2.8.11) that 0a is a holomorphic transformation of 0(1), and 0 1 is a linear transformation, too. It suffices to show that g = Oa o f is a linear transformation. Since g(0) = 0 and Ig(z)l < 1, Lemma (6.1.2) implies (6.1.5)
I9 (0)I < 1.
Here, the equality holds if and only if g(z) = e'8z, 0 E R; it is a linear transformation. Applying Lemma (6.1.2) to g-1, we have 191(0) < 1.
It follows from this and (6.1.5) that Ig'(0)l = 1; thus, g(z) = e'6Z. ii) Let f : H --+ H be a holomorphic transformation. As in (2.8.12), we set
z+i' which defines a biholomorphic mapping ti : H -. 0(1) . Then lli o f o 10-1 is a holomorphic transformation of 0(1), and hence a linear transformation by i). Therefore f is a linear transformation. By the above Theorems (6.1.1) and (6.1.4) the notations, Aut(C), Aut(O(1)), and Aut(H) used in Chapter 2, §8 may be taken for holomorphic transformation groups in the sense defined in Chapter 5, §3. EXERCISE 1. Let Di (i = 1, 2) be a domain defined as the inside of a circle of C or one side of a line of C. Show that a biholomorphic mapping f : D1 -+ Dz is a linear transformation.
6. HOLOMORPHIC MAPPINGS
144
6.2. Poincare Metric We define an hermitian metric gR on A(R) by
aR(z) =
(6.2.1
2R2 (R2 - 1212)2
9R = 2aR(z)Idzl2
This is called the Poincard metric on E(R). Taking a biholomorphic mapping
fR:A(1):) z- RzEA(R), we have by (5.3.23) (6.2.2)
fRgR = 91,
and hence we may normalize, R = 1. The next theorem plays a basic role. (6.2.3) THEOREM. The Poincare metric g1 on A(1) is Aut(A(1)) -invariant; i.e.,
for every f E Aut(A(1)), f'g1 = 91 PROOF. Since 2
f'g1 =
(6.2.4)
l.f())12)2ldzl2
(1-
for f E Aut(A(1)), it suffices to prove that 1
If'(z)l2
(6 .2.5)
(1-
If(z)12)1
(1- Iz12)2
This clearly holds for f (z) = eiez (0 E R). By Theorem (2.8.11) it is sufficient to prove (6.2.5) for (6.2.6)
a E 0(1).
f (Z) = 0a(z) = -'az + 1'
Since
f,(z) =
-az + 1 + (z - a)a
-
1 - Ia12
(az - 1)2'
(-uz + 1)2
we have IaI2
If'(z)I
1-If(z)I2
z
a
21-az+112
1 - I -az+1
_
1-1a12
1az-112-1z-a12 _
1-1a12
12iz12-71z-a7Z +1-Iz12+za+za-IaI2 1 - 1z12 + 1a121z12 - IaI2
1 - 1z12 -
6.2. POThICARE METRIC
145
Let C(O : [To,T1) - A(1)) be a piecewise continuously differentiable curve in A(1). As in (5.3.24) we define the length L,&(,) (C) = L9, (C) of C with respect to 91 by (6.2.7)
Lo(1)= f
T,
VI-2 To
-
1(t)
2 JTT `
1 - (t)12
2
+
I
(t) I' dt,
where 0(t) = 01(t) + iO2(t). Let f E Aut(A(1)). Then it follows from Theorem (6.2.3) (cf. also (3.2.18)) that (6.2.8)
o(1)(f(C)) =
f
T o
L2
1 - if(0(t))12 I
dt
dt
2
= LA(i)(C). Therefore we see that the length of C with respect to gi is invariant under holomorphic transformations of A(1). We call LA(,)(C) the hyperbolic length of C.
For two arbitrary points z1, z2 E A(1) we define the hyperbolic distance by (6.2.9)
da(1)(zi, z2) = inf{Lo(1)(C); C is a piecewise continuous differentiable curve in 0(1) connecting zi and z2}. In fact, this satisfies the following axioms of metric: (6.2.10) i) do(1)(zl,z2) = do(1)(z2,zi),
ii) dA(1)(zl,22) +dO(1)(z2,z3)
d4(1)(z1,z3),
iii) dn(1)(zl, z2) = 0 b zi = z2. The proofs of i) and ii) are clear, and iii) follows from the fact that aI(z) is a positive continuous function; but it will also be shown in the following argument. It follows from (6.2.8) that (6.2.11)
f E Aut(A(1)).
d&(1)(f(zi),f(z2)) =da(l)(zl,z2),
For this property, we say that do(1) is Aut(0(1))-invariant. Set z - Zi
f(z) = e`6
-'11z+1
,
6. HOLOMORPHIC MAPPINGS
146
so that 4 = P Z2) E R and 4 > 0. Let C((: (0,1] -y A(1)) be a piecewise continuously differentiable curve with initial point 0 and terminal point 4. Set
fi(t) _ 01 (t) + i#s(t),
3(t) _ Since
(t),
j = 1,2.
1 - Io1(t)12,
1
6 2 12
(1 - J0(t)I2)2 = (I - 1(t)2)2 We have a curve C1 = C1(01) given by 431. Then the initial (resp., terminal) point of C1 is 0 (resp., 4), and by (6.2.12) Lo(C1)-
Ln(1)(C)
0. Furthermore, we have
The equality holds if and only if 432(t) L o 1) (C ) = 1
f
21d11(t)J
1
o
(
1
- 01(t)2
dt
> f1
f
o
f
=
1
2
01(t)2
\\1+x 1+ 4 log
+
1 - x2
\ 1 -x111
dz
dx = [l og(l+ x )-log (1- x ) 1o
1-4. It is to be noted that the equality in the above inequality holds if and only if o 1(t) > 0; for such an example, we may take 431(t) = t4. Thus we see that L4%(1)(C) >- log{(1 + 4)/(1 - 4)}, and the equality holds if and only if C is a line segment C(o,,zz) connecting 0 and 4. It follows that (6.2.13)
do(1)(z1, z2) = do(1)(01 zi) = log
l-
_1112+1
Rom this, (6.2.10), iii) follows. In general, a curve C connecting two points 21i z2 E 0(1) is called a hyperbolic geodesic if
do(1)(z1,z2) = LA(1)(C)
The above C(o,Za) is the unique geodesic connecting 0 and 4, and C(,,,,,,) (C(asz)) is the unique geodesic connecting z1 and z2. The mapping f is analytically continued to a neighborhood of 0(1). By the conformality of f (Theorem (3.1.15)) C(s1,s2) is a part of the circle (including the case of a line), passing through z1 and z2 and orthogonally crossing the boundary circle C(0;1) of A(1). We denote by L(z1iz2) the part of that circle contained in A(1). For
6.2. POINCARE METRIC
147
any two points w1, w2 E L(z1, z2) the segment of L(zl, z2) connecting w1 and w2 is the geodesic C(2 ,,,,W2) connecting wl and w2. We also call L(zl, z2) a
hyperbolic geodesic. Take the third point z3 E 0(1) outside L(zl,z2). Then there are infinitely many hyperbolic geodesics passing through z3 and having no
intersection with L(zl, z2) (see Figure 67). To confirm this, it is easier to set z3 = 0 by a linear transformation of Aut(A(1)). If we associate a hyperbolic geodesic to a line in Euclidean geometry, the above property contradicts the axiom of parallels. The geometry in which hyperbolic geodesics in A(l) are considered as lines is called non-Euclidean or hyperbolic.
FIGURE 67
The metric do(1) defines the so-called metric-topology on A(1). That is, a sequence {z} n° o of points zn E A (l) is said to converge to a E A(1) if
slim do(I)(zn,a) = 0. We say that do(1) is complete if any Cauchy sequence {zn}n o with respect to d,%(1) (i.e., for an arbitrary e > 0 there is a number no such that dp(1)(zn, z,n) < e for all n,m > no) converges to a point of A(1). In fact we have the following theorem. (6.2.14) THEOREM. i) A sequence {zn}no of points zn of i(1) converges to a point a E 0(1) with respect to do(1) if and only if it converges to a as a sequence of points of C. ii) dj(1) is a complete metric. PROOF. i) By (6.2.13) d j(1) (z, w) is continuous in the two variables z and w. Therefore llm Izn - al = 0 lira da(1)(z,,,a) = 0.
n-x
nac
Conversely, we assume limn-ao d&(1) (zn, a) = 0. By the transformation 0a E Aut(A(1)) in (6.2.13) we may assume a = 0 without loss of generality. It follows from (6.2.13) that (6.2.15)
1+
dA(1)(zn,0) = log 1 - Iznl
-' 0
(n -. oo).
6. HOLOMORPHIC MAPPINGS
148
Hence, limn Iznl = 0. ii) Let {zn}°°_o be a Cauchy sequence in A(1) with respect to dA(1); that is, for an arbitrary e > 0 there is a number no such that (6.2.16)
da(1)(zn, zm) < E
for all n, m. > no. In particular, da(1)(zn,zn(,) < e, n > no, and so
da(1)(0,zn)
It follows from (6.2.15) that Iznl
e'W - 1 eM _+1 < 1.
Therefore there is a subsequence {z }i_e of (zn)n' o which converges to a point
a E L((eM - 1)/(eM + 1)) C i(1) as a sequence of points of C. It follows from i) that for e > 0 there is a number vo such that da(1)(zn,,,a) < E,
(6.2.17)
v ? ve.
We may assume that n,,,, >_ no. Thus, by (6.2.16) and (6.2.17)
do(1)(zn,a) <2E.
By i) we deduce that it is not necessary to distinguish the convergence with respect to do(1) and the convergence in C. Because of ii) the Poincare metric g1 (and 9R as well) is also said to be complete. The upper half plane H is biholomorphic to 0(1) by (2.8.12): V) (z) _
z_i z+i
H -- A(1). Set It = O'g1. A direct computation yields
(6.2.18) h = (Imz)2Idz12, FIGURE 68
which is called the Poincare metric on H. We can define the length LH(C) of a piecewise continuously differentiable curve C in H, the hyperbolic metric dH, and hyperbolic geodesics of H in the same way as in 0(1). By the correspondence of circle to circle of linear trans-
formations (Chapter 2, §8), a hyperbolic geodesic of H is a part of a circle orthogonally crossing the real axis, contained in H. EXERCISE 1. Show that a hyperbolic geodesic of H is as described above.
6.3. CONTRACTION PRINCIPLE
149
be a sequence of points of a(1). Assume that there EXERCISE 2. Let is a point a E 0(1) such that {do(l) (a, z,))nc=O is bounded. Show that {z.,}n o contains a subsequence which converges to a point of A(1). EXERCISE 3. Compute dH(z1,z2) for points zi,z2 E H.
6.3. Contraction Principle If the universal covering of a domain D C C is biholomorphic to A(1), D is said to be hyperbolic. In this case, the universal covering mapping
zr:A(1)-.D is holomorphic. The deck transformation group Sri (D) consists of holomorphic transformations of Aut(0(1)). Thus a1(D) is a subgroup of Aut(A(l)), and its
elements are linear transformations. For an arbitrary point P E D we take a disk neighborhood Up C D. It follows from (5.3.6) that for every zi E it-i(P) there is a neighborhood Vi of z1 such that the restriction
7rIV1:V1- Up is biholomorphic. Now we define an hermitian metric hu, by (6.3.1)
hu, = ((xtVi)-i)'(91IVi)
Here g, IV, stands for the restriction of the Poincart metric gi to V1. Take another z2 E 1r-I(P) and V2 in the same way. Then, by (5.3.12)
there is an element -y E ir1(D) C Aut(0(1)) such that y(41)
=
'Y (Vi
z2,
CD V2
)=V2
Since 91 is Aut(0(1))-invariant (Theorem (6.2.3)), 7*(g11V2) = g1 I Vi .
Thus it follows from Chap-
-
tr I
I V,
ter 5, §3, Exercise 9 that V1.
'
((irIV1) -1 ) (911V,) FIGURE 69
= ((rtV2)-1)'(911V2)
Therefore hup is defined independently of the choice of Vi, and defines an hermitian metric hD on D. We call hD the hyperbolic metric on D. For a piecewise continuously differentiable curve C in D, we may define the length LD(C) = L60 (C)
with respect to hD, and then the metric dD in the same way as do(,). We call LD(C) the hyperbolic length of C, and dD the hyperbolic metric. Let e be a lifting of C. Then by the definition (6.3.2)
LD(C) = La(1) (C).
150
6. HOLOMORPHIC MAPPINGS
(6.3.3) THEOREM. Let D C C be a hyperbolic domain, and let it : 0(1) -i D be the universal covering. Then we have the following. i) A sequence {zn}n o of points of D converges to a E D as points of a if and only if it converges to a with respect to dD. u) The metric dD is complete. PROOF. i) Let lim zn = a as points of C. Take a disk neighborhood U. of a.
We may assume that zn E U., n = 0, 1, 2, .... Take a point a E it-' (a) and a neighborhood V of a so that zr(V) = U.. Take in E V fl a-'(zn), n = 0, 1, 2, ... . It follows from (6.3.2) and the definition of dD that dD(z.,,a)
(6.3.4)
da(I)(i.,4).
By the choice of the points, limn_,, i = a as points of C. It follows from Theorem (6.2.14) that limn_.,, do(1) (in, a) = 0. By (6.3.4) we have lim dv(zn, a) = 0. Conversely, we assume that lim dD (zn, a) = 0. Take the above U. so that it is relatively compact in D. It follows from the above argument that dD(z,a) is continuous in z E D. Hence
eo=min{dD(z,a);zEBU0}>0. Take an arbitrary point z E D with dD (z, a) < to. There is a piecewise continuously differentiable curve C from a to z such that
dD(z,a) 5 LD(C)
(6.3.5)
If z V U1, then C fl 8Ua 0 0; by the definition of to, co < LD(C). By (6.3.5) we get a contradiction, co < co. Therefore (6.3.6)
{z E D;dD(z,a) < eo} C U0.
By the hypothesis we may assume that z E U. and dD(zn,a) < co, n =
0,1,2,.... Take a piecewise continuously differentiable curve C,, connecting a and zn as in (6.3.5). Then (6.3.6) implies that C C U0. Take a lifting of Cn be its terminal point. We get with the initial point a, and let in E
LD(C,) ? do(1)(I.,a) Letting LD(C,) - dD(zn,a), we see that dD(z,,,a) > do(1)(in,a). Combining this with (6.3.4), we obtain (6.3.7)
dD(zn, a) = do(I)(i.,a)
It follows again from Theorem (6.2.14) that lim i = a as points of C, and that Um z,, = a as points of C. ii) For r > 0 we set
D, = {z E D;dD(z,7r(0))
6.3. CONTRACTION PRINCIPLE
151
convergence of the original sequence with respect to dD follows as in the proof of Theorem (6.2.14), ii). Set
A(1),- = {z E A(1);do(1)(z,0) < r}. For an arbitrary z E Dr we take a piecewise continuously differentiable curve C from ir(0) to z such that
dD(z, lr(0)) < LD(C) < r.
Let C be the lifting of C with the initial point 0. By (6.3.2), Lo(1)(C) < r. Let z be the terminal point of C. Then z E A(1),., rr(z) = z, and Dr C 7r(A(1)*) C 7r(A(1)r.)
By Exercise 2 of §2, 0(1), is compact, and so is 7r(A(1),). Hence Dr is compact
in D. D We deduce the following from the proof of ii): (6.3.8) COROLLARY. Let rr : A(1) - D be as above, and fix a point a E D. Then a sequence {zn}°O_o of points of D has no accumulation point in D (that is, no convergent subsequence) if and only if limn-,,,, dD(zn, a) = oo.
Let D C C be a domain, and let h1, j = 1, 2, be two hermitian pseudo-metrics on D. Write
h; = b,(z)Idzl2 = b;(z)IdiI2,
j = 1,2
i(a) 5 b2(a) bi(0) 5 b2(0)
if a
I .
z
For a E D we define h1 (a) 5 h2(a) if and only if (6.3.9)
i=
oo,
if a = oo.
Ifh1(a) Sh2(a) for all aED,wewrite hl
perbolic metrics hD,. Let f : D1 - D2 be a holomorphic mapping. Then we have
i) f'hD, = hD,; ii) dD1 (a, b) > dD, (f (a), f (b)) for all a, b E D1.
This property of the hyperbolic metrics is called the contraction principle. For the proof we first deal with the case where D1 = D2 = 0(1). The next is called Schwarz-Pick's lemma:
(6.3.11) LEMMA. For an arbitrary holomorphic mapping f : 0(1) - A(1) we have
i) f'g1 5 g1, and if the equality holds at a point, then f E Aut(0(1)); ii) d4,(1)(a,b) ? da(1)(f(a), f(b)) for all a,b E 0(1).
6. HOLOMORPHIC MAPPINGS
152
PROOF. i) We want to show that (6.3.12)
f'91(a)
91(a)
at an arbitrary point a E A(1). By making use of
z+a ox+1 we get 0' ag1(0) = g1 (a) from Theorem (6.2.3). Thus (6.3.12) is equivalent to
f'0-a91(0) < 01a91(0)i
that is, 91(0) > 0-a °f'°0' a91(0) = (0-aofo0a)'g1(0). Therefore it suffices to prove (6.3.12) at a = 0 for an arbitrary holomorphic mapping f : A(1) - A(1): (6.3.13)
f*91(0) < 91(0)
Since f'91(0),
(xf(0))'91(0) =
we may assume that f (0) = 0. By the definition of g1i (6.3.13) with f (0) = 0 is equivalent to (f'(0)I < 1. This was proved by Lemma (6.1.2).
Suppose that there is a point a E 0(1) such that f'g1(a) = g1(a). By the above arguments, we may assume that a = 0 = f (0). Then it follows from Lemma (6.2.1) that f E Aut(A(1)). ii) Take a piecewise continuously differentiable curve C from a to b in 0(1). It follows from i) that Lo(1)(C) ? Lo(1)(f(C)) > do(1)(f(a),f(b))
Thus do(1)(a,b) > do(1)(f(a), f(b)). 0 PROOF OF THEOREM (6.3.10). i) Let irk : A(1) - D,, j = 1, 2, be the universal coverings. Let F : A(1) --+ A (l) be a lifting of f o ir1 : A(1) -s D2:
i(1)
P
1 ir1
D1
A(1) 1 W2
-L
D2
It follows from Lemma (6.3.11) that F'g1 < g1. Hence, the definition (6.3.1) implies that f' hp, < hD,. ii) The proof is similar to that of Lemma (6.3.11), ii). 0 The notion of hyperbolic distance was generalized for general complex manifolds by S. Kobayashi (181, and it has played an important role in complex analysis.
EXERCISE 1. Let D be a hyperbolic domain, and let hD be the hyperbolic metric. Show that f'hD = hD for f E Aut(D).
6.4. THE RIEMANN MAPPING THEOREM
EXERCISE 2.
153
Let D be as above, and let r : A(1) -p D be the universal
covering. Show that for two points wl, w2 E D dD(wl, w2) = min {do(1) (zl, z2); zz E 0(1), r(xi) = w3, 9 = 1, 21.
6.4. The Riemann Mapping Theorem Let D C C be a domain. Let { f } - 0 be a sequence of holomorphic functions on D which converges uniformly on compact subsets to f . Then it follows from Theorem (3.5.19) that f is holomorphic, and that
(6.4.1)
d n
d
dz
dz
4`foi
df of
di
di
(n-oo) onDflC, (n
oo)
on D n C \ {0},
where the convergence is uniform on compact subsets. Now let F be a family of holomorphic functions on D. We say that F is a normal family if any sequence of holomorphic functions in F contains a subsequence which converges uniformly on compact subsets. For normal families the following Montel's theorem is fundamental.
(6.4.2) THEOREM. If a family F of holomorphic functions on D is uniformly bounded, then Y is normal.
PROOF. There is an M > 0 such that for every f E .F If(a)I <_M,
a E D.
Take a E D such that a $ oc. Take a disk neighborhood b.(a; r) with 0 < r < d(a; OD). It follows from (3.5.4) that
x E 0(a; r).
f'(z) = Tai JC(a:,) (C Therefore, for z E A(a; r/2)
If'(z)I = 1
()2M.27rr= 4MFor
arbitrary z, x' E A(a; r/2) Z
If(x)-f(z)I =
J
f'(()d( < 4MIz r - z'I.
Hence F is equicontinuous on A(a; r/2). If a = oo, making use of the complex coordinate i about oo, we infer in the same way as above that.F is equicontinuous on a neighborhood of oo. Thus, F is equicontinuous on compact subsets of D. We see by Theorem (2.2.6) that F is normal.
154
6. HOLOMORPHIC MAPPINGS
It follows from the above argument that Theorem (6.4.2) holds if .F is bounded on every compact subset of D.
EXERCISE 1. Show that a family F of holomorphic functions on D is normal
if and only if for every a E D there is a neighborhood V of a such that the restriction FIV = {f IV; f E .F) of .F to V is normal. (6.4.3) LEMMA. For an arbitrary a E 0(1) there is a unique f E Aut(A(l)) such
that f (0) = a and f'(0) > 0. PROOF. As in (2.8.10), we set
Z+a
f(z) = 0-a(z) = dz+ 1 Then f E Aut(A(1)), and f (0) = a. A simple computation yields
f'(0)=1-1a12>0. Thus we have shown the existence of such f E Aut(.(1)). Let g E Aut(A(l)) satisfy the conditions. Then,
g-1
o f E Aut(©(1)) satisfies
9-1 o f (0) = 0,
(g-1 o f)'(0) > 0.
It follows from Theorems (6.1.4) and (2.8.11) that g-' o f(z) = e'Bz with 0 E R.
Since (g-' o f)'(0) = e'e > 0, e'9 = 1, and henceg-' o f(z) = z; i.e., g = f. 0 Our next theorem is called the Riemann mapping theorem.
(6.4.4) THEOREM. Let D C C be a simply connected domain such that the boundary 3D contains at least two distinct points. Then, for an arbitrarily given a E D there exists a unique bih olomorphic mapping f : D -+ 0(1) such that f (a) = o, CV&
f'(a)>0,
a
z (0) > 0,
a = oo.
oo,
PROOF. We are going to reduce D to a bounded domain of C. Take distinct points bl, bZ E 3D. We may assume without loss of generality that b1 E C. We take a linear transformation
z - bl Here, if b2 = oc, we set z/b2 = 0. Then D is biholomorphic to 01(D) C C \ {0}. We may assume that D C C \ {0}. Since D is simply connected, one may take by Theorem (5.2.5), ii) a branch 02(z) of f on D. As shown in Example (5.2.7), 02 : D 02(D) is biholomorphic, and 02(D) n (-$2(D)) = 0.
6.4. THE RIEMANN MAPPING THEOREM
155
Thus, ¢z(D) C C has an exterior point b3 E C. Set
&(z) =
1
b3
Then, 03o02 : D 03o02 (D) C C is biholomorphic, and 03o02(D) is bounded. Therefore one may assume that D is a bounded domain of C. Let F be the family of all univalent holomorphic functions f on D satisfying
f (a) = 0,
f(a) =1.
Since (z - a) E F, F # 0. Set IIf1100 =sup{If(z)1;zED},
f EF,
p=inf{IlfII ;f E F}.
Since (z-a)EF, p<_max{Iz-al;zEOD}
lim Ilfnll = P
We may assume that II fni10C < p + 1 for all fn. Thus, {fn}n i is uniformly bounded. It follows from Theorem (6.4.2) that {fn}n 1 contains a subsequence converging uniformly on compact subsets of D. We rewrite {f,} n 1 for the subsequence, and set
fo = n--0C hm fn Note that fo(a) = 0, and by (6.4.1) fo(a) = lim fn(a) = 1. Hence, fo is not a constant. Theorem (4.3.10) implies that fo is univalent, and so fo E F. By definition, p < I1foll00 On the other hand, for an arbitrary e > 0 there is a number no such that for all n > no, 11 fn II00 < p + E. Letting n -' oo, we have Ilfo II < p, and hence II fo ll = p. It suffices to show that
fo(D) = o(p).
(6.4.5)
By the maximum principle (Theorem (3.5.21)) it is clear that fo(D) C A(p). Assume that fo(D) 96 0(p). Taking c E 0(p) n 8(fo(D)), we set P(w - c) 04(w) - -ft + p2 The mapping 04 : E(p) - 0(1) is biholomorphic, and 04 (c) = 0. Taking a branch
O4 o fo(z) on D, we set
Os=P 04ofo(z)ED(p),
zED.
It follows that 0s is univalent, and 05(a) = v1--p-c. Moreover, set
06 =
PZ(05(z) - 05(a)) E A(P), -05(a)i5(z) + p2
z E D.
156
6. HOLOMORPHIC MAPPINGS
The mapping 4s is univalent, too, and 06(a) = 0,
110600
P-
A simple computation yields
06(a) = 2 Since 0 < jcl < p, I¢6(a)I > 1. Setting
07(z) = -e(a)46(z),
zED,
we have 07 E F, and 11071130 1
a)I < A
This contradicts the choice of p. Thus, (6.4.5) is proved. To show the uniqueness, we take a biholomorphic mapping g : D
0(1) such
that g(a) = 0 and g'(0) > 0. Then fo9' E Aut(A(1)), fog-1(0) = 0, and (fog-1)'(0) > 0. Lemma (6.4.3) implies that fog-1(z) = z, and so f = g. 0 (6.4.6) THEOREM. An arbitrary simply connected domain D of C is biholomor-
phic to C itself, C, or A(1); furthermore, these three domains are not biholomorphic to each other.
PROOF. If 8D = 0, D = C. If 8D consists of only one point a E C, we may assume a = oc by a linear transformation; hence, D = C. If 8D contains more than one point, it follows from Theorem (6.4.4) that D is biholomorphic to A(1). The latter assertion easily follows from Theorems (3.5.22) and (3.7.8). 0 It is known that the above Theorem (6.4.6) holds in fact for a general simply connected Riemann surface. It is called the xniformization theorem, and was proved by Koebe (1907), and independently by Poincares (1908). The Riemann mapping theorem was first claimed in his doctoral dissertation (1851), and a compact simply connected Riemann surface with boundary was dealt with. Riemann's proof depended on the existence of a solution of a variational problem which had not been established, and was criticized by Weierstrass. Riemann did not answer this criticism, but Hilbert did later (1904). Theorem (6.4.4) was first completely proved in its current form by Osgood (1900). An arbitrary domain or a Riemann surface, even if it is not simply connected, carries the universal covering which is simply connected. It is biholomorphic to C, C or 0(1) by the uniformization theorem. Thus the domain or the Riemann surface is C itself, or is represented as a quotient space of C or z(1) by the deck transformation group (cf. (5.3.11)).
EXERCISE 2. Let f : D - A(1) be a biholomorphic mapping with f(zo) = 0 (zo E D). Let g be a holomorphic function on D such that Ig(z)I < I on D, and g(zo) = 0. Show that I9'(z)1 < 1f'(zo)1.
6.5. BOUNDARY CORRESPONDENCE
157
EXERCISE 3. Show in Theorem (6.4.4) that if a E R and D is symmetric with respect to the real axis, then f (z) = f (z). EXERCISE 4. Obtain a biholomorphic mapping from D = {z = x + iy; y2 > 4c2(x + c2)} (c > 0) to the upper half plane H.
6.5. Boundary Correspondence The boundary of the upper half plane H in a consists of the real axis R and the infinity oo. The two ends, -oo and +oo, of R coincide with oo.
FIGURE 70
We endow 8H with the orientation from -oo to +oo via 0. We take an ordered triple (PI, P2, P3) of distinct points of OR Suppose that it is one of the following:
(6.5.1) i) P3 = oo, P1 < P2;
ii) P2 = co, P3 < P1; iii) P1 = 00, P2 < P3; iv) oo # P1 < P2 < P3 0 00In this case we say that the triple (P1, P2, P3) has the positive orientation. By Theorem (6.1.4), ii) and Theorem (2.8.13) an arbitrary f E Aut(H) is represented by (6.5.2)
f (z) = cz + d'
(c d)
E
SL(2, R).
The restriction
fIOH:OH -iOH is injective and homeomorphic in the sense of the topology induced from C. Since P(Z) -
ad - be (cz + d)2
f'(z) > 0,
1
=
(cz + 1)2
z E R,
it follows that f IR is monotone increasing. Therefore, a triple (P1, P2, P3) of positive orientation is mapped to a triple (f (PI), f (P2), f (P3)) of positive orientation. Hence we say that f preserves the orientation of M.
6. HOLOMORPHIC MAPPINGS
158
Now, cases ii)-iv) of (6.5.1) are reduced to case i) by a linear transformation
f (z) = x - P3 E Aut(H). Moreover, case i) is reduced to (PI1P2, P3) = (0, 1, oo)
(6.5.3)
by a linear transformation
=
f(z)
1
P2 - Pt
(z - P1) E Aut(H).
That is, every triple (P1, P2, P3) of positive orientation is mapped to (0, 1, oc) by the boundary correspondence of some f E Aut(H). Conversely, assume that f E Aut(H) fixes the triple (0, 1, oo):
f(0)=0, f(1)=1, f(oo)=00. As in (6.5.2), write f (z)
= cz + d'
(c
d)EsL(2R).
Since f (oo) = oc, c = 0; f (0) = 0 implies b = 0. Since f (1) = 1. a/d = 1. On the other hand, ad = 1. Thus, a = d = ±1, and so f (z) = z. That is, if f E Aut(H) fixes the triple (0, 1, oo) of positive orientation, then f must be the identity. Summarizing the above, we see that for two arbitrary triples (P1, P2, P3) and (P'1, P2, P3) of positive orientation, there exists a unique f E Aut(H) such that
f(P1)=1'j, f(P2)=P'2,
f(P3)=P.3-
The disk A(1) is biholomorphic to H by (z)
z + i' ty : H - A(1),
0(i) = 0.
By ip the boundary 8H = RU {oo} corresponds to 8A(1) = C(0;1) as follows: ij(oc) = 1,
P(R) = C(0;1) \ {1}.
6.5. BOUNDARY CORRESPONDENCE
159
7
1=v(1) W(P
v(R) 1=v(00)=u'(I',)
FIGURE 71
A triple (P,, P2, P3) of positive orientation with P; E 8H,1 S j < 3, is mapped by ?+/' to a triple (+J'(P, ), z/i(P2), 7/'(P3)) of points of C(0; 1) in anti-
clockwise order. Let (Q,, Q2, Q3) be a triple of points of C(0;1). We say that the triple (Q,, Q2, Q3) of C(0;1) is of positive orientation if they are in anti-clockwise order. A triple (Q1, Q2, Q3) of positive orientation of C(0; 1) is mapped by z[i-1 to a triple (V)- I (Q I ),1' I (Q2), V) - I (Q3)) of positive orientation of 8H. Thus we have proved the following theorem.
(6.5.4) THEOREM. Let D be H or A(1). Then every f E Aut(D) preserves the orientation of OD, and for arbitrary triples (P, i P2, P3) and (Pi, P2, P'1) of positive orientation, there exists a unique f E Aut(D) such that
f(P,)=Pi, f(P2)=P2, f(P3)=P2. Lemma (6.4.3) gives the uniqueness of holomorphic transformations at an interior point, and Theorem (6.5.4) does it by the boundary correspondence. EXERCISE 1. Obtain f E Aut(H), mapping (0, 1, oo) to (-1, 0, 1). EXERCISE 2. Obtain a biholomorphic mapping f : H - A(1), mapping (0, 1,00) to (1,i, -1).
(6.5.5) LEMMA. Let E3 C C, j = 1, 2, be subsets, and let F : E1 , E2 be an injective continuous mapping. If E, is closed, then F is a homeomorphism. PROOF. It suffices to show the continuity of F-I : E2 E,. Let Q E E2 be an arbitrary point, and let be an arbitrary sequence converging to Q. 0 Set P = f `(Q), and P = f - I n = 1, 2,.... Note that El is compact. Let P' be an accumulation point of i°_0. It is sufficient to show that P' = P. Let {P,,,, }v o be a subsequence of {P,,}n o which converges to P. Since F is continuous,
F(P') = Iim F(P,,,,) = lim "-.00Q,,,, = Q = F(P). V-00 The injectivity of F implies P' = P.
6. HOLOMORPHIC MAPPINGS
160
Let D C C be a simply connected domain, and assume that D : C; that is, D has an exterior point. If D is not a bounded domain of C, we take a point c E C\D, and set f (z) = 1/(z-c). Then D is biholomorphic to D1 = f (D) C= C, and f is extended continuously to a homeomorphism between the boundaries:
f:D-,D1.
(6.5.6)
(6.5.7) THEOREM. Let D C C be a simply connected domain such that D # 1b. Assume that the boundary 8D is given by a closed Jordan curve C(O : [0, 1] - C)
(that is, cD = O((0,1])). Then every biholomorphic mapping f : 0(1) -, D is extended to a homeomorphism from E(1) to D.
PROOF. By (6.5.6) we may assume that D is a bounded domain of C. Take an arbitrary point (E C(0; 1). For p > 0 we set OP = a(1) n 0((; p), W, = f(1v), A(p) = the area of we,,
r,
rp = C(() A yP = f(r), L(p) = the length of yp.
f
FIGURE 72
It is sufficient to prove that w, accumulates to one point of C as p 0. Furthermore, it is sufficient to prove it for a sequence {pn},, 1 (p > 0), converging
to 0. Since 1imp-o A(p) = 0, for an arbitrary f > 0 there exists 5(E) > 0 such that A(p) < E,
0 < p < b(E).
By Schwarz' inequality we have
f
e
0L(p)dp = fin
I f'(( + te'B)jtdtdg 6
\ 1/2
(ff
f'(( + te')tdtd9 I
/
i6
<
A(b) b2 9fE,
0 < 6 < b(E).
1 /2
tdtdo)
6.5. BOUNDARY CORRESPONDENCE
161
Therefore, for an arbitrary 0 < 6 < 6(E), there is some p E (0, b) such that
L(p) <
n
Hence, for a positive sequence (e,, I', 1 converging to 0, there exists a monotone decreasing positive sequence 1 converging to 0 such that (6.5.8)
L(pn) <
En,
7rEn.
FIGURE 73
As a point z E rv, approaches on r., to its end points a,,, 0 E tv, f1C(0;1), f (z) E ryo converges to points a,,, b E 7v. fl C, respectively; otherwise, yp should oscillate infinitely times between some positive distance, and hence the length of y., would be infinite. This contradicts (6.5.8). It follows again from (6.5.8) that
Ia -
(6.5.9)
Set C = 4,v fl c (9 a,,, (6.5.10)
?fE.
Then
ac.,', = yP U C,,,
C. D Cn+1
The sets C are closed subsets of C, and connected. Assume that a,,, = b,,1 for some n1.
FIGURE 74
Forn>n1 we have (6.5.11)
C _ {a,,, },
&p. = yp U {a.,).
6. HOLOMORPHIC MAPPINGS
162
Therefore, the diameter of wpn is less than L(pn), and by (6.5.8) n .-- oo.
WP- -{ anl,
This means that for an arbitrary e > 0 there is a number no such that for every n Vi wp.. C A(an,;E). Assume that an bn for any n. We write 91 for C(0;1). There is an injective and surjective continuous mapping
0: S'
8D = C.
It follows from Lemma (6.5.5) that the inverse mapping 0-' : C -. S' is continuous. Hence, 0-' (Cn) = In is a connected closed subset of S1. The points an, bn E C divide C into two curves C;,, C,,. One of them is Cn. Let sn, to be the end points of In. Then it follows from (6.5.9) and (6.5.10) that
Isn - t,,-'0,
InDIn+iJ....
Therefore, f,°°_i In = {r} with r E S', and so
Cn-'c=O(r),
n -'oo.
Since limn- x a. = limn_.n, bn = c, and limn-.x L(-yo,) = 0, we have by the above and (6.5.10) that wp P. -' C,
n - oo.
Thus, f is extended to a continuous mapping f : Y(1) -- D, and f(A(1)) = D. In the above proof, the case of (6.5.11) does not occur. For, if an, = bn,, f takes the value an, constantly on the arc S' f1 A((; pn, ). It follows from Chapter 3, §5, Exercise 5 that f (z) = an z E z(1). This is a contradiction. Therefore we see that f is injective on the boundary, and so by Lemma (6.5.5) f : A(1) -+ D
is a homeomorphism. 0 Let D be as in the above theorem, and let fo : 0(1) mapping. Extend it to a homeomorphism fo A(1)
D be a biholomorphic
D. We say that a
triple (Qi, Q2, Q3) of points of 8D is of positive orientation if (f0 1(Q1), fo 1(Q2), fo 1(Q3)) is of positive orientation. This is a property independent of the choice of fo (Theorem (6.5.4)). We immediately have the following by Theorems (6.4.4), (6.5.4), and (6.5.7)
(6.5.12) THEOREM. Let D C C be a simply connected domain such that D 54
C and the boundary 8D is given by a Jordan closed curve. Let (P1, P2, P3) be a triple of positive orientation of 8A(1), and let (Q1, Q2, Q3) be a triple of positive orientation of 8D. Then there ezists a unique biholomorphic mapping
f :0(1)-yD such that f(Pi)=Q;,1 <_i:3.
6.6. UNIVERSAL COVERING OF C \ {0, 1}
163
REMARK 1. The above theorem holds with 0(1) replaced by H. REMARK 2. If Jordan's Theorem (3.3.8) is admitted, it is not necessary to assume the simple connectedness of D or D C in Theorems (6.5.4) and
(6.5.12).
For an arbitrary simply connected domain D C C, Caratheodory topologically defined the notion of boundary elements, and proved that every boundary element corresponds to one point of 80(1) through a biholomorphic mapping f : 0(1) D; this gives a complete description of the boundary correspondence
of f. EXERCISE 3. Let D be as in Theorem (6.5.7). Let h(z) be an arbitrary realvalued continuous function on 8D. Show that there exists a unique continuous function u(z) on D such that u(z) is harmonic in D and has the boundary value h(z). 6.6. Universal Covering of C \ {0, 1}
As shown in Chapter 5, §3, for an arbitrary domain D C C there exists the universal covering it : X -* D of D. The space X has a structure of Riemann surface such that Tr gives rise to a holomorphic mapping. Here we are going to
construct X for D = C \ (0, 1). Let C1 be the half-line {oo} U (z E C; Re z = 0, Im z >_ 0} from oo to 0, let C2 be the half-circle C(1/2;1/2) n {Im z >_ 0} from 0 to 1, and let C3 be the half-line {z E C; Re z = 1, Im z > 0} U {oo} from 1 to oo. Let L1 be the half-line {oo} U {z E C; Re z < 0, Im z = 0} from oo to 0, let L2 be the real line segment from 0 to 1, and let L3 be the half-line {z E C; Re z >_ 1, Im z = 0} U {oo} from 1 to oc.
H
I., 0
I., I
I
FIGURE 75
Let Go be the domain bounded by C1,C2 and C3. By Theorem (6.5.12) there
exists a unique biholomorphic mapping a : Go - H with the homeomorphic extension zr : Go H such that ir(oo) = oo, 7r(0) = 0, and ir(1) = 1. Therefore we have (6.6.1)
ir(C1) = L1,
ir(C2) = L2,
We shall obtain the universal covering it : H
ir(C3) = L3.
C\{0, 1} by analytic continuation
6. HOLOMORPHIC MAPPINGS
164
of ir. We denote by G1 the domain which is the reflection of Go with respect to C1. By the reflection principle, Theorem (5.1.7), 7r analytically extends, crossing 0
the set C, = {z E C; Re z = 0, Im z > 0} of the interior points of C1, to a holomorphic mapping from G1 to the lower half-plane L = {z E C; Im z < 0}. 0
0
Thus we get a holomorphic mapping it : Go U C1 U G1 - H U L1 U L. We carry out this analytic continuation for the other boundaries, C2 and C3, and again do the same for the resulting images of C;,1 < j 5 3. Let G,,, n = 1, 2, ... , denote all the domains obtained by the above process of reflections of Go. Then, we get
H n=o
and a holomorphic mapping (6.6.2)
A:H--4C\{0,1}.
It is clear that the above A satisfies condition (5.3.3) of a covering mapping. Since H is simply connected, (6.6.2) is the universal covering. Set E = (G1 n H) U Co. Then, the set of the interior points of E forms a domain, A(E) = C \ {O, 11, and for every z E E
a-1(A(z)) n E = {z}. We call such E a fundamental domain of the universal covering (6.6.2). Thus, we see that C \ {0,1 } is a hyperbolic domain. It follows from Theorem (6.3.3) that (6.6.3) C \ (0, 1) carries a hyperbolic metric hc\{o,1}. which is complete.
FicuRE 76 The deck transformation group r1(C\ {0,1}) of the universal covering (6.6.2)
is a subgroup of Aut(H) = PSL(2, R). We determine what it is. Consider the transformation for w E C \ {0, 1} obtained by the reflections with respect to LJ, 1 < j < 3. Then the point w is mapped to itself only by an even number of repetitions of such reflections. Thus we see that an element of
irI(C \ {0,1}) C Aut(H) is a transformation obtained by an even number of
6.6. UNIVERSAL COVERING OF C \ {0,1}
165
repetitions of reflections with respect to Cj,1 5 j =< 3. For example, we take the reflection of Co with respect to C1, and then the reflection with respect to the image C2 of C2, and denote it by T1 (cf. Figure 76):
z = x + iy -i z1 = -x + iy
the reflection with respect to C1
the imageCzofC2
(x+2)2+y2=(z)2,
the reflection with respect to C2
z1
z2,
(zl + 2)
(z2 + 2) _
(2)2.
Therefore we have 1
1
1
z2+2=4 z1+1
_1 4
1
Z'
1
-4 z+2 _
1
2
- z+j 1
1
2
-2z+1
_
1
2z
2
-2z + 1
1
z
-2z + 1
= T1 (z).
Take a point Po E C \ {0,1}; for instance, PO = i. Then, the transformation T1 corresponds to a Jordan closed curve aI C C \ {0, 1) from Po with the antic
O
clockwise orientation, crossing L1 and L2, and then coming back to Po.
FIGURE 77
In the same way, let C3 be the reflection image of C3 with respect to C1, and let T2 denote the composition of the reflections with respect to C1 and C. Then T2(z) = z - 2.
The transformation T2 corresponds to a Jordan curve a2 C C \ {0, 11 from Po 0 0 which crosses L1, goes into L, crosses L3, and comes back to Po. If 03 denotes a O O Jordan curve from P0 which crosses L2 and L3, going around I anti-clockwisely, we have
{a2} = {a1 } + {a3}.
6. HOLOMORPHIC MAPPINGS
166
Note that T1 is given by Al = (12 °) E SL(2, R.), and T2 by A2 = (0 12) E SL(2, R). Hence, we get Al, A2 E r(2). (6.6.4) LEMMA. The group t(2) is generated by ±A1 and ±A2; that is, every
A E 1'(2) is expressed as A = ±A1"'A2'Ai'A;2...A"A"', m,,n, E Z, 1 <
j5l
PROOF. We have that Ail=(21),Az1=(01 ), and
_ Al=(-2n 1
0 1
Az=
'
2n
1
0
1
Z.
,
Take an arbitrary A = (a a) E r(2). We have AA2 =
By the condition, b' = b - 2na
a b-2na _
a b'
d - 2nc
c d'
c
±a, n E Z. One may take n = n1 E Z so that Ib'I < jai.
(6.6.5) Next, we get
AA2' Am =
1 (a b' 0 _ a - 2mb' b'_ (a' b' (c d') -2m 1) - (c - 2md' d') - (c' d')
If b'#O, one may takem=m1 EZsothat Ia'I < Ib'I.
(6.6.6)
Set AM = A, A(') = AAz' .4m' A(k)
_
Inductively, one obtains
(a(k)
C,td(k)) =
1",
A(k-1)q2
ja(k)I < Ib(k)I < Ia(k-')l < Ib(k'1)I,
k = 2,3,...
(cf. (6.6.5) and (6.6.6)). The sequence (10)11o is strictly decreasing, and one 96 may continue this process so long as b(k) 0. Hence, there is a number I E Z with bl = 0, so that A(1) =
a(1)
(d1)
0
d(l)) =
fAc`n/2
'
Therefore, one has an expression A = ±Ar' Az' Ai3 AZ' ... 4" AZ' , r, , ss E Z, 1<_j<_1. Summarizing the above, we have the following.
6.7. THE LITTLE PICARD THEOREM
167
(6.6.7) THEOREM. The universal covering of C \ {0, 1} is the upper half plane H (or the disk 0(1)), and the deck transformation group al (C\{0,1}) is isomorphic
to r(2)/ {±Q °)}; C \ {0,1} = H/t(2). In particular, C \ {0, 1} carries a complete hyperbolic metric.
The function w = A(z) given by (6.6.2) is called the lambda function, and is one of those called modular functions. It has a close relationship with elliptic functions which will be dealt with in Chapter 7. EXERCISE 1.
Let D and b be domains of a, and let it : D - D be the
universal covering. Show that if D is not biholomorphic to a, C, or C', then b is biholomorphic to 0(1).
6.7. The Little Picard Theorem If a rational function f (z) is not a constant, for any a E C the equation f (z) = a has a solution z E C, and if f has a pole, then f has the value no there. However, this is not the case in general, if f is a holomorphic or meromorphic function on C. For instance, the entire function f (z) = e= does not take the value 0, nor, of course, the infinity no. Thus, the number of points which f misses is two. In general, if a meromorphic function f, regarded as a holomorphic mapping into C, misses a point of C, that point is called an exceptional value. The next theorem is called the little Picard theorem, and shows that the above "two" is the maximum. (6.7.1) THEOREM. The number of exceptional values of a non-constant meromorphic function f on C is at most two. PROOF. Assume that the number of exceptional values of f is more than two. Then we are going to show that f reduces to a constant. Let three distinct points a, b, and c of C be exceptional values of f . By a linear transformation, we may assume that {a, b, c} = {0, 1, oo}. Thus we have a holomorphic mapping
f :C-+C\{0,1}. It follows from Theorem (6.6.7) that the universal covering of C \ {0, 1} is the unit disk A(1). Let 7r : A(1) --+ C \ {0, 1} be the universal covering mapping. Since C is simply connected, it follows from Theorem (5.3.18) and (5.3.21) that f has a holomorphic lifting
f :C
o(1),
aof(z)=f(z)
Liouville's Theorem (3.5.22) implies that f is constant, so that f is constant. D
6. HOLOMORPHIC MAPPINGS
168
By the same idea as above, Montel's Theorem (6.4.2) can be greatly improved, as Montel himself showed. We first generalize the notion of normal families.
Let D C C be a domain, and let fn : D - C, n = 1, 2, ... , be a sequence of holomorphic mappings. We say that {fn I converges uniformly on every compact subset to f: D C if for any point P E D there are a neighborhood U C D of
P and a number no such that (6.7.2)
fn(U) C
fool,
n > no,
or (6.7.3)
fn(U) c C \ {0},
n > no,
and that in the case of (6.7.2) (reap., (6.7.3)) the sequence { ffIU} (reap., fib o fn 1111 = { 11f. I U }) of holomorphic functions converges uniformly on every com-
pact subset of U to f (reap., w o f). In this case, f is, of course, a holomorphic mapping. A family F of holomorphic mappings from D into C is said to be normal if each sequence in F has a subsequence which converges uniformly on every compact subset. (6.7.4) REMARK. Lemma (2.2.7) holds for a domain D C C. In fact, if D = C, D is itself compact; if D C, by a linear transformation, D is homeomorphic to a domain of C for which Lemma (2.2.7) is valid.
(6.7.5) LEMMA. Let F be a family of holomorphic mappings from D into For F to be normal, it is necessary and sufficient that for an arbitrary point P E D there exists a neighborhood U c D of P such that the restriction.FlU = if IU; f E .F} of .F to U is normal.
PROOF. The necessity is clear. By Remark (6.7.4) we take Un C D, n = 1, 2,.- - , as in Lemma (2.2.7). Take a sequence {f,,,) in F. For a point P E Un there is a neighborhood Vp C D of P such that { f IVp } contains a subsequence which converges uniformly on Vp. Since Un is compact, there are a finite number
of P, EU and V, = Vp,, 1
j
I, such that I
Un C U Vj. j=1
Let (fl,) be a subsequence of If,) which converges uniformly on V1. Let { f2,} be a subsequence of If,., } which converges uniformly on V2. Inductively, we get a subsequence (fl,, }°__1 of { f,} which converges uniformly on U. For n = 1, the subsequence of If,) obtained as above. For n = 2 we we denote by { similarly take a subsequence {f(2),,) of {fill,}. Inductively, we get U (n),1100=17
n = 1,2,.... Then, the subsequence f f(,).,) of (/,, } converges uniformly on every compact subset of D. 0 The next theorem is due to MonteL
6.7. THE LITTLE PICARD THEOREM
169
(6.7.6) THEOREM. Let F be a family of holomorphic mappings from D into C. If there are three distinct points a, b, c E C which are exceptional values for all
f E F, then F is normal. PROOF. We take an arbitrary point P E D. By Lemma (6.7.5) it suffices to show that there is a neighborhood U C D of P such that FLU is normal. Let { fn }R t be a sequence in F. Since C is compact, we may assume that (fn (P)) converges to a point Q E C. By a linear transformation we may also assume that {a, b, c} = {0, 1, oo}. It follows from Theorems (6.6.7) and (6.3.3) that C \ {0, 1, oo} = C \ {0, 1} carries a complete hyperbolic metric do\{o.i} We assume that P E C; if P = oc we make the same arguments using the complex coordinate i. For the sake of simplicity, we may assume after a translation that
P = 0. Taker > O so that 0(r) C D. For a point z E i(r/2) we take a line segment C,,(0: t E [0, 1] -. tz). Then by (6.2.13) we have do(r)(0, z) = La(r)(Cz)
=logy+1z' r - jzj
<<
It follows from Theorem (6.3.10) that (6.7.7)
dC\{o. 1}(f..(0),fn(z))
(6.7.8) The case of Q = limn.,- fn (0) E {0, 1, oo}:
For instance, we suppose that Q = 0. Since do\{o,l} is complete, for an arbitrary e > 0 we have
dc\{o,l}(fn(0),C(0;e)) = min{dc\{o,I}(fn(0),w);w E C(0;e))
-y+oo
(n-oo).
It follows from (6.7.7) that there is a number no such that
fn(A(r/2)) C 0(0;E),
n > no.
Thus {f,, IO(r/2)} converges uniformly to 0. The other cases of Q = 1,oo are discussed similarly.
FIGURE 78
6. HOLOMORPHIC MAPPINGS
170
(6.7.9) The case ofQEC\{0,1}: There is a number no such that dC\{o.l}(fn(0),Q) < 1,
n > no.
Thus, by (6.7.7)
z E A(r/2).
dc\{0,1}(Q,fn(z)) < 1 +log3,
n >_ no.
Since dC\{o.l } is complete, the set {w E C \ (0,1); do\{o.l } (Q, a') < 1 + log 3}
is compact; in particular, it is a bounded set of C. Therefore, {fnl0(r/2)} is uniformly bounded, and Theorem (6.4.2) implies that {fnl0(r/2)} contains a subsequence which converges uniformly on compact subsets. D EXERCISE 1. Let F be a family of holomorphic functions f on a domain D c C such that for some fixed number a E R, Im f (z) > a. Show that F is
normal. EXERCISE 2. Prove the little Picard Theorem (6.7.1) by making use of Montel's Theorem (6.7.6). EXERCISE 3. Show by an example that the number three of exceptional values
in Montel's Theorem (6.7.6) cannot be smaller.
6.8. The Big Picard Theorem In this section we write
A'(r) = 0(r) \ {0} (r > 0),
0' = A'(1).
The universal covering of 0' is given by w:H3z-4e2"'sE0'.
(6.8.1)
FIGURE 79
A fundamental domain is
{z=x+iy;0<_x<1,y>0}.
6.8. THE BIG PICARD THEOREM
171
As described in §3, A' carries the complete hyperbolic metric ha., which is, in view of (6.2.18) and (6.8.1), expressed by (6.8.2)
ho. (z) = (xI2(I g IzI2)2Idz12,
z E '.
The hyperbolic length La. (C(0; r)) of the circle C(0; r) (0 < r < 1) with respect to ho. (cf. (5.3.24)) is given by 2n
(6.8.3)
La. (C(0; r)) = f
2 rJ 1
0
r2l rd8
= logrl .
The area A&.(A'(r)) of 0'(r) with respect to h.o. (cf. (5.3.25)) is given by
27rIr
(lot t)2
dlogt = 27r
logt]r0
2wr
logri. We generalize the notion of isolated essential singularities of holomorphic functions for meromorphic functions. Let P E C be a point, and let U be a neighborhood of P. Let f be a meromorphic function in U \ {P}. If f cannot be extended meromorphically to U, P is called an isolated essential singularity of f. As f is regarded as a holomorphic mapping f : U \ {P} C, this is equivalent to saying that f cannot be extended to a holomorphic mapping from U into C. The next theorem is called the big Picard theorem.
(6.8.5) THEOREM. If a meromorphic function f on U\ {P} has an isolated essential singularity at P, then f assumes all points of C infinitely many times in an arbitrary neighborhood of P with P deleted, except for at most two points of C.
PROOF. We may assume that P = 0, and f is defined on A* (R) (R > 0). Suppose that there are three distinct points a, b, c E C such that f ({a, b, c}) is a finite set. Then there is a 0 < Ro < R such that
f'i({a,b,c}) n0'(Ro) = 0. We show that in this case f will be meromorphically extended to 0' (R0). By the change of variable z/Ro, f may be assumed to be defined on 0', and f (0')n {a, b, c} = 0 as well. We regard f as a holomorphic mapping (6.8.6)
f :0' -aC\{0,1}.
172
6. HOLOMORPHIC MAPPINGS
By Theorem (6.6.7) C \ {0, 1} carries the complete hyperbolic metric hc\(o,1} Take a sequence {zn}°O=1 of 0' so that (6.8.7)
lim zn = 0,
Iznl > Izn+1I,
n-.oo
an = d (zn),
n-.oo
lim an = a E C.
Take a disk neighborhood U(a) of a. If for a positive number r < 1 (6.8.8)
f(A*(r)) C U(a),
by Riemann's extension Theorem (5.1.1) f extends to a holomorphic mapping from 0(r) into C. Assume that (6.8.8) does not hold. Then, there is a sequence {wn}10°_1 of 0' such that
lim wn noo
= 0,
On = f(wn) E C \ U(a).
Since a \ U(a) is compact, by taking a subsequence one may assume that
Q = lim f (wn) E C \ U(a). Moreover, taking subsequences of {zn} and {wn} and renumbering them, one may assume that Iznl>Iwnl>Izn+1I,
n=1,2,....
FIGURE 80 Set
Rn={zEC,IwnI
6.8. THE BIG PICARD THEOREM
173
Then, ORn = I'n U An. For an arbitrary point z E r we have by the contraction principle, Theorem (6.3.10), and (6.8.3)
dc\{o,1}(f(z),an) < Lc\{D,1}(f(rn))
< Lo (re) =
21r
--.0
11o81zn11
(n , oo).
an = a, f (I'n) a (n oo) if a E C \ {0,I }. If or E (0, 1, oo}, similarly to (6.7.8) one sees that f (1'n) -, a (n -+ oo). Therefore one always has Since
(6.8.9)
a
(n -' oo).
f (An) - Of
(n -+ oo).
f (I'n)
In the same way, one sees that (6.8.10)
Since one of a and j3 is in C, we assume that a E C.
FIGURE 81 Set
ro =
f1,
Q=oo,
la-01, 0000,
and
E=0 (a;3ro 4 )
\a;-4
)\O
It follows from (6.8.9) and (6.8.10) that there is a number no such that (6.8.11)
En(f(I'n)Uf(An))=0,
n>-no.
Take a curve Cn in An from zn to wn. Then f (Cn) intersects C(a; ro/2). By the Prinzip von der Gebietstreue, Corollary (4.3.14), f (Rn) is a domain, and its boundary is a subset of f (1'n) U f (An). Thus, it follows from (6.8.11) that (6.8.12)
f (Rn) D E.
6. HOLOMORPHIC MAPPINGS
174
We consider the area of E with respect to hC\{o,l} It is clear that Ac\{o.l) (E) > 0.
(6.8.13)
On the other hand, by the contraction principle, Theorem (6.3.10), and (6.8.4)
Acs(o.l)(E) < f f`hcN{o.l}
R
< AA. (A*(IzmD)) =
in.
hA .
I log IznI I
.0
(n - co).
This contradicts (6.8.13).
If a meromorphic function f on C has an isolated essential singularity at oo, f is called a transcendental meromorphic function; that is, if f is not transcendental, f is rational. The following is a direct consequence of Theorem (6.8.5). (6.8.14) COROLLARY. A transcendental meromorphic function on C takes infinitely many times all values of C except for at most two values.
For example, f (z) = e' misses 0 and oo, and in fact takes all other values infinitely many times. Hence the above "two" cannot be made smaller. The proof of Theorem (6.8.5) essentially depended on the fact that the universal covering of C \ {0, 1, oo} is the upper half plane. This fact was proved in §6 by making use of the Riemann mapping Theorem (6.6.7). Hence it may be said that the Riemann mapping Theorem (6.6.7) played a key role in the proof of the big Picard Theorem (6.8.14). The Riemann mapping theorem is a deep result, but is a speciality of the one variable case. It is, however, sufficient for the proof to construct a complete hermitian metric h on C \ {0, 1, oo} for which the contraction principle holds. There is a quantity K(h), the so-called Gaussian curvature of h. and the contraction principle for h holds if there is a constant Co > 0 such that
K(h) < -Co. From this viewpoint, the big Picard theorem can be generalized to the higher dimensional case. For this, cf. [181 and X91.
Problems (2,-2), and dotty (2, 2. Show in Theorem (6.3.10) that if f' h p, (a) = hp, (a) holds at a point a E D1, then the equality holds at all points of D1. 1. Calculate
z).
3. Let D be a hyperbolic domain. Show that for arbitrary points z; E D, i = 1.2, there exists a continuously differentiable curve C from zl to z2 such that dD(zl, 22) = LD(C).
PROBLEMS
175
4. Let f be an entire function, and let N be a natural number. Prove that if the cardinality of f -i (w) is not more than N for all w E C, then f is a polynomial of degree at most n.
5. Obtain a biholomorphic mapping from D = {z E A(1);Imz > 0} to the upper half plane H. 6. Obtain a biholomorphic mapping from D = {z = re'5; 0 < r < 1, 0 < 0 < a}
(0 < a < 2ir) to 0(1). 7. Show that by w = z+1/z, 0(1) is univalently mapped onto D = C\ [-2,21. Where is C \ 0(1) mapped to? 8. Show that. by w = z(1 + ez)2 with e = -e-'°, a E R (this is called Koebe's function), 0(1) is univalently mapped onto (w E C; w 54 re, r > 1/4). 9. For a meromorphic function f on a domain D C C, set f.,,(z) 3 z) s
f
This is called the Schwarzian derivative of f. Show the following.
(a) if; z} = 0 if and only if f (z) = (az + b)/(cz + d). (b) For (c e) E SL(2,C), { -+b ;z} = {f;z}. (c) Set w = f (z), and assume that f'(z) iA 0. Then, {w; z}
(dw\J {z; w}. dz
(d) Set w = f (z), and let g(w) be a meromorphic function in w. Then, {go
f;z}={g;w}
(d)2 +{w;z}.
10. (Vitali) Let E be a subset of a domain D which has an accumulation point in D. Let (f,,) be a sequence of holomorphic functions which is uniformly bounded. Prove that if {fn } converges on E, then { f } converges uniformly on compact subsets in D. 11. Let f f } be a sequence of holomorphic functions on a domain D that converges at a point zo E D. Prove that if {Refs(z)} converges uniformly on compact subsets in D, then so does {f,) itself.
12. Let D be a hyperbolic domain, and let f : D - D be a holomorphic mapping. Show that if there is a point zo E D such that f(zo) = zo and f'hD(za) = hD(zo), then f E Aut(D). 13. Let D i = 1, 2, be rectangle domains of C, and let f : Dt D2 be a biholomorphic mapping. Show that f extends continuously up to the boundary of DI. Moreover, show that if each vertex of D1 is mapped by f to a vertex of D2, then D1 is similar to D2. 14. Let D be the interior of an n-gon in C, and let f : H
D be a biholomorphic
mapping. Suppose that for -oo < at < .. < a < +oo, b; = f (a,) are
6. HOLOMORPHIC MAPPINGS
176
vertices of D. Let ajx (0 < aj < 2) be the inner angles of the vertices b,. Prove that f can be expressed as follows: (z-aj)a1-1dz+C2
f(Z) = C1 J 11
(an
+oo),
j=1 Z n-1
1 (z) = C1
JJ (z - aj)°'-1dz + C2 J j=1
(an = +oo),
where C1 (# 0) and C2 are constants. (The above formula is called SchwarzChristoffel's formula) Hint 1. Suppose that an 0 +oo. Then, by Schwarz' reflection principle the function f (z) is analytically continued to a multi-valued holomorphic function on C \ {a1};=1. Hint 2. By making use of a branch oft = (f (z) - bj )1 /01j, one sees that t is analytically continued to the lower half plane in a neighborhood of aj, and has a zero of order 1. Thus, one gets an expansion f(z) - bj = co(z - aj)°3(1 + c1(z - aj) + )°3,
and sees that the principal part of f"(z)/f'(z) is (aj - 1)/(z - aj). Hint 3. Note that the multi-valuedness of f (z) is caused by the even number of repetitions of reflections with respect to line segments. Hence, letting f (z) denote another local branch of f (z), one gets
f' (z) = Af (z) + B
(A (0- 0) and B are constants),
and f " (z) / f(z) is invariant, so one-valued, and at oo holomorphic with value 0. It follows that f" (z)
_-
f'(z)
aj
4z _1 - aj
.
Now, integrate this. Hint 4. In the case when an = +oo, take a linear transformation 0 E Aut(H) such that 0(an) # +oo. Transform the formula obtained above by ¢-1.
15. Replace H by 0(1) in problem 14. Then show that f(z) can be expressed as r2 n
fl(z - a1)°'-1dz + C2.
f(z) = C1
J j=1
Here, Ia.. I = 1 and bj = f (aj).
16. Set
f(z)=
j
1
v'z(l _-Z 2 )
dz.
Prove that f (z) maps H biholomorphically to the interior of a square in the first quadrant with one edge a real interval from 0 to t(1/4)2/1 2x.
PROBLEMS
177
17. Let P(X) be a polynomial of degree > 2 without double root, and let f (z) be a holomorphic function on A'(1) (resp., C). Show that if P(f (z)) has only finitely many zeros (reap., no zero), f (z) has at most a pole at 0 (resp., f is a constant). 18. Show that the number of exceptional values of a non-constant rational or meromorphic function on C' is at most 2. 19. Prove the big Picard Theorem (6.8.5) by making use of Montel's Theorem (6.7.6).
CHAPTER 7
Meromorphic Functions
In this chapter we investigate more deeply the properties of meromorphic functions. Beginning with an approximation theorem, we
prove two kinds of existence theorems, the introductory part of value distribution of meromorphic functions on C, the infinite product expression of meromorphic functions, and the basic part of elliptic functions. These lead to function theory of several complex variables, the value distribution theory of one and several variables, the theory of Riemann surfaces, and the theory of elliptic curves and automorphic forms.
7.1. Approximation Theorem Let U C C_ be an open set. In the present chapter we mainly deal with the case of U # C. By a linear transformation we may assume U C C without loss of generality.
(7.1.1) LEMMA. Assume that U C C is a bounded open subset. Let K C U be a compact subset, and let f be a holomorphic function on U. Then, for an arbitrary e > 0 there is a rational function F such that all poles are contained in 9U and If (z) - F(z)I < e,
z E K.
That is, f can be approximated uniformly on K by such F.
PROOF. Let n be a natural number, and let k and j be integers. Set (7.1.2) k)=z=x+iy,2n<x'2n1,2
En (7>
En =
I
<
=y
U En (j, k). 179
180
7. MEROMORPHIC FUNCTIONS
Let U. denote the interior of En. Then 00
Un C Un+1,
U = U Un. n=1
F-1-
I
Figure 82
Since K C U is compact, there is a number no such that (7.1.3)
K c Un,
n > no.
Take z E K and n >_ no arbitrarily. If z is not a boundary point of En (j, k)'s, then there is a unique (jo,ko) with z E En(jo,ko). By Cauchy's integral formula, Theorem (3.5.16), and Cauchy's integral Theorem (3.4.14) we see that for E,, (.j, k) C U 1 f (z) d( 27ri JOE, (3,k) (- Z
_ f (z), (j, k) = (jo, ko),
otherwise. 10, Here, the orientation of 8En(j,k) is anti-clockwise. If En(j,k) and E,(j',k')
share an edge, then the sum of the curvilinear integrals over the edge with respect to 8En (j, k) and 8E,, (j', k') is zero. Therefore we obtain (7.1.4)
f(z) = 2'ri
I
S
(zz
Both sides of the above equation are continuous in z E K, and so it holds for all z E K. Since K is compact, d(K; 8U,,) = inf {d(z; 8Un); z E K}
= min{d(z; (9Un); z E K} > 0, d(K; 8U,,) > d(K; 8Un_ 1).
Set 60 = d(K; BU,,). Choose and fix n
no so that
(7.1.5)
10
1
2n
2
Let ry1,... , ryi denote the line segments of length 2-n that form the boundary 8Un. The rectangle with an edge -y which lies on the opposite side of En (j, k) C
7.1. APPROXIMATION THEOREM
181
U with the edge y intersects 8U. Let . be a point of the intersection. It follows from (7.1.5) that for z E K and ( E y
IS - I 2 < vF 20 < 1 Iz - .J.
(7.1.6)
Figure 83
For these z and C we have 1
(-z
(7.1.7)
_
1
1
z__
r
1
(-G
,
=o
By (7.1.6) this power series converges uniformly and absolutely in z E K and (E -y,,. Therefore, for an arbitrary E > 0 there is a number N. such that
+
I'll
1
z
j_o
(S - G)' (z - G)J+t
G E.
We set
F (z)
I 27ri
(z(
-)i+ f (()d(
1 F(z)
L F,.(z), =1
M = max{If(()I;( E 8U}.
7. MEROMORPHIC FUNCTIONS
182
Then, we have
If (z) - F(z)I y=1
27ri
7r
< 2"M
f-dC + F1,(z)I C3
(7.1.8) LEMMA. Let K C C be a compact subset, and let a, b E C \ K be two points in the same connected component of C \ K. Let F be a rational function which possibly has a pole only at a. Then F is uniformly approximated on K by national functions which possibly have a pole only at b.
PROOF. By the assumption we have that F(z) = Ek _,v Ck(Z - a)k with 0 S N < oo and ck E C. It is sufficient to prove the assertion for every 1/(z-a)k with k > 0. Take a curve from a to b which does not intersect K.
Figure 84
Set bo = min(Iz - wI; z E K, w E C) > 0, and take points of C:
a=ao,ai,...,a, =b,
jai+1-ail <
2 Lo
It suffices to prove that 1/(z - a,)k (k > 0) is uniformly approximated on K by rational functions with a pole only at ai+1. It follows that for z E K
ai - a,+1
z-a,+1
<
1
2
As in (7.1.7), we get 1
z - a,
_ j-_0
(a, - a+1)3 (z - ai+1)j+1
which converges uniformly in a neighborhood of K. By Theorem (3.5.19) the (k - 1)-th differentiation of both sides of the above equation yields 1
(z - a,)k
_C. j=o
(k - 1)! (z - a,+1)j+k+1
which converges uniformly on K. Thus, we obtain our assertion.
7.1. APPROXIMATION THEOREM
183
The next is called Runge's theorem. (7.1.9) THEOREM. Let D C C be a domain and let K C D be a compact subset. Let f be a holomorphic function in a neighborhood of K. If any connected component of D\K is not relatively compact in D, then f is uniformly approximated on K by holomorphic functions on D.
PROOF. Take a bounded open set U so that K C U c D and f is holomorphic
in U. It follows from Lemma (7.1.1) that f is uniformly approximated on K by rational functions with poles at finitely many points a E 8U. Take one a E 8U C D\K, and denote by V the connected component of D\K containing a,,. If V is bounded, then V n8D # 0. Take a point b E V noD. Then a and b belong to the same connected component of C\K. It follows from Lemma (7.1.8) that a rational function with a pole only at a,, is uniformly approximated on K by rational functions with a pole at b,,. If V is not bounded, we take b E V so that A rational function with a pole at b is expanded to a Taylor series KC about 0, and hence approximated uniformly on K by polynomials. Therefore,
we deduce that f is uniformly approximated on K by rational functions with poles at most on 8D. Applying the above theorem to the case of D = C, we have (7.1.10) COROLLARY. Let K C C be a compact subset such that D \ K is connected. Then, a holomorphic function in a neighborhood of K is uniformly approximated on K by polynomials.
(7.1.11) LEMMA. Let U C C be an open subset, and let K C U be a compact subset. Let {V}0 EA be the set of connected components of U \ K which are relatively compact in U. Then, A is at most countable, and the set
k=KUUV. aEA
is compact.
PROOF. Since U \ K is open, every Va contains rational points. The set of rational points is countable. Hence, A is at most countable. We first show that k is bounded. Otherwise, there is a point as E Va such
that K C 0(jao1/2). Note that V. has no intersection with 8U. Since C \ A(Iaa1/2) (C C \ K) is connected and contains a., Va ? C \ A(Ja.1/2). Thus, Va is unbounded. This is a contradiction, and hence k is bounded. Take an arbitrary sequence of points z,, E K, n = 1, 2, ... , which converges to a point zo E
C. Suppose that zo 0 K. If zo E U, then there is a connected component V of U\K which contains zo and is not relatively compact in U. For a sufficiently large n, z,, E V. There is some V. containing z,,,. Then, V. = V. This contradicts the choice of Va. Therefore, zo E 8U. Set
bo = min{d(z; 8U); z E K} > 0.
7. MEROMORPHIC FUNCTIONS
184
Figure 85
For a sufficiently large n, d(z,,; 8U) < 60 /2, and
A(z,,; d(zn; 8U)) C U \ K. Hence, for V,, such that V,, 3 z,,,
a
A (zn; d(zn; 8U)) C V,,.
Since 0(z,,; d(z,,; 8U)) is not relatively compact in U, neither is Vn. This is again
absurd. Thus, k is closed and bounded, and so compact.
Let D C C be a domain and let D = say that
U be an open covering. We
is a Runge increasing covering if the following conditions are
satisfied: (7.1.12) i) U C Un+1, n = 1, 2, ... .
ii) Every holomorphic function in a neighborhood of Un is uniformly approximated on Un by holomorphic functions in D.
By Lemma (2.2.7) there exists an increasing sequence of open subsets Vn, n = 1, 2, ... , such that Vn C= Vn} 1 i and D = u
1 V.. Applying Lemma (7.1.11)
to Vn, we obtain Vn. Let Wn denote the set of interior points of Vn. It follows
that (7.1.13)
Wn = Vn,
Vn C Wn,
n = 1, 2, ...
.
Set U1 = W1 and n1 = I. Since U1 C D, there is a Vn2, n2 > n1 i such that U1 C Vn,. Set U2 = Wn,. Repeating this inductively, we obtain {Un}n°_1, which is by Theorem (7.1.9) a Runge increasing covering. This shows the following.
(7.1.14) THEOREM. Every domain D C C has a Runge increasing covering.
7.2. EXISTENCE THEOREMS
185
7.2. Existence Theorems Let D C C be a domain and let f be a meromorphic function on D. Let a E D be a pole of f . Then, f is expanded to a Laurent series about a:
_ f
C-k a)k
(z
C-k+1
C-1
(z - a)k-I
z - a
x
+ E
a)n,
k > 0.
n=0
Here, we call (7.2.1)
Q. (Z)
(z
c- a)k + (z c- a)k-I + ... + z -la
the principal part off at a. If a = oo, f is expanded to 00
n=0 Ck
Cl
zk
z
00
+>C_nZ n=0
The principal part of f at oo is given by Qoo(Z)=CkZk+...+clz=zk+...+ Cl
(7.2.2)
It is a fundamental question whether there exists a meromorphic function which
has a given principal part at every point of a given discrete subset of D. The following result, Mittag-Leffler's theorem, answers this question.
(7.2.3) THEOREM. Let {an}l I (N < +oo) be a series of distinct points of D without accumulation point in D. Suppose that at every a,, a rational function such as (7.2.1) or (7.2.2) is given. Then, there exists a meromorphic function on D which has the principal part Qa at every point an. PROOF. If D = C, then N < +oo. Thus, we set N (7.2.4)
f(z) = EQa.,(z) n=l
which clearly has the prescribed property. Assume that D jA C. By a linear transformation, we may assume that D C C. It follows from Theorem (7.1.14) that D has a Runge increasing covering 1. By the assumption each U contains at most finitely many an's. Set
Q., (z) = E Q.. (z), aEU..
v= 1,2,... .
7. MEROMORPHIC FUNCTIONS
186
For p such that U. fl
0, we set Q = 0. First we set f1 = Q1. Then,
Q2 - fl is holomorphic in a neighborhood of U1. Take a holomorphic function 92 on D such that
192+Q2-fl<1Set f2 = 92 + Q2. Suppose that f (v >_ 2) was taken so that f - Q is holomorphic in a neighborhood of U, and (7.2.5)
If-(Z) - f"-1(z)1 < 2° 2,
zE
f is holomorphic in a neighborhood of U,,, there is a holomorphic function 9i'+1 on D such that Since
zEU,,.
191-+1(z)+Q,,+1(z)-f,(z)I <2- ,, Set
Inductively, we take f,,, p = 1,2..... It follows from
9v+1 +
(7.2.5) that cc
f(z) = fi(z) +
E(f,,+1(z) - fv(z))
V=1
x E D, = fa(x) + E(f,, +1(z) -z f,,(z)), v=a
defines a meromorphic function on D. By the construction, f clearly has the prescribed property. 0 EXAMPLE. Let R be a meromorphic function on C, that is, a rational function. Let 1 (N < +oo) be the set of poles of R, and let Q., be the principal
part of R at an. As in (7.2.4), R(z) - En 1 QaR(z) is bolomorphic on a, and hence a constant by Theorem (3.7.8). Thus 00
R(z) = co +
(7.2.6)
Q,,,, (z),
co E C.
n=1
The above right side is called the expansion by partial fractions. EXAMPLE. Let D = C, and set f (z) = 1/ sin z. Then the poles off are mr, n E Z, the orders are 1, and the residues are (-1)". Set
(-1)" I
9(x) _ n=1
\
l+n7r nEz
(z + nor
ir
nor
Since 1
1
z+nmr
n7r
_
IzI
<
1
IzI
nirlx+nal = rr2 (1 - 1 I) n2'
l
n'r l/
7.2. EXISTENCE THEOREMS
187
g(z) converges absolutely and uniformly on every compact subset K of C except for a finite number of terms. Thus g(z) is meromorphic function on C. Set h(z}
1
+9(z)
z
11
- 11 + z 1
1
+ (-1)
x +7r
7r
nr
11
z-7r
11
1
z-n7r
7r
n7r
Noting the absolute convergence, we have
h(z - 7r) = -h(z),
h(z - 27r) = h(z).
Thus, 27r is a period of h(z), and hence
F(z) =
sin z -
h(z),
z E C,
is an entire function with period 21r. For z = x + iy
_
1
(7.2.7)
I sin zI
2
2
Iei:-y - e-a:+yl
a-y - ey
- 0,
IvI -, +oo .
Here the convergence is uniform. On the other hand, h(z) is written as
h(z) =
1
z
00 (-1)n
+ n=1
2z z2 _ nzWz
The above right hand side converges absolutely and uniformly on every compact subset except for a finite number of terms. We take its partial sum:
()
2m
sm x = E(-1) n n=1
_ m I -2z 2z z2 - n2s 2 n=1 ` z2 - (2n - 1)2ir2 + z2 - 4n2 2z
m
_ 27r2z(4n - 1) = F, {z2 - (2n - 1)27r2}(z2 - 4n29r2)
Forz=x+iywith 0<=z<_2,r and IyI >41r,wehave m
IBm(x)j
c2iL n +cln2)2,
(y2
where c; > 0, i = 1, 2, are constants. For t > 0, the function
0(t) =
C21vIt (y2 + C, t2)2
takes the maximum value O(to) = 9c2/16
c,y2 at to = IyI//. Therefore,
we have m
E
c21bin
= n 1 (?l2 +clt2)2
_<_
C2 1Y1
J
t 1
dt +
9c2
y2
7. MEROMORPHIC FUNCTIONS
188
0(t)
(tdt.It.I+I
I
Figure 86
We compute the integral of the right side:
dt = C21yI t 2(y2 + c1) (y2 + Ctt2)2
c2IyI f
Thus, there is a constant c3 > 0 such that Ism(z)I < C3
lyl,
S.(z)I 5 c3/IyI, we get
Since lg(z)I = I (7.2.8)
Ih(z)I 5
I
z I+
lg(z)I < '+1C3
-
1Y
z=x+iy, 05x527r,
Iyj>41r.
It follows from (7.2.7) and (7.2.8) that F(z) is bounded on {z = x + iy; 0 5 x 5 21r}. By the periodicity, F(z) is a bounded holomorphic function on C, and so by Liouville's Theorem (3.5.22) F(z) is constant. It follows from (7.2.7) and (7.2.8) that liml,,_.«, F(iy) = 0, and hence F = 0. Thus, one gets the following identity:
sinz
z
_
+ n=-oo (-1) ao
1
z
+
G +n7r
na
2z
E(-I)n
z2
- n272'
where >' stands for the sum with the term for n = 0 omitted. EXERCISE 1. Show that 7r cot 7rz = 1
EXERCISE 2. Show that
>r sing 7rz
+ L/
r
z
I
n=-oo (-_----
n=00-oo (z - n)2
+ 1) . n
7.2. EXISTENCE THEOREMS
189
We are next going to show the existence of a holomorphic function possessing prescribed orders of zeros at prescribed discrete points in a domain D. This is called Weierstrass' theorem:
(7.2.9) THEOREM. Let D C C be a domain, and let {an}n 1 (N S oc) be a discrete set of points of D. Let vn, 1 < n < N, be integers. Then there exists a holomorphic function f (z) which has zeros of order vn at an, n = 1, ... , N, and no zeros elsewhere.
PROOF. By Theorem (7.1.14) there is a Runge increasing covering {Un}n 1 of D. Because of the construction of Un (cf. (7.1.13)), the following holds: (7.2.10) Any connected component of D \ U,, is not relatively compact in D. For every n we set
Pn(z) = fJ (z - aj)"'. ajEU
For a, E Un+1 \U,, we take a point bj E D\Un+1 which belongs to the connected
component V of D\Un containing a,. Then, the function log(z-a,)/(z-bj) has a one-valued branch in a neighborhood of Un. To show this, we take a piecewise linear curve C from a, to b, in V. The curve C consists of line segments L. from zk to zk+1, where
ai = zo,z1,... ,z1 = b,.
The function (z-zk)/(z-zk+1) is holomorphic and non-vanishing in C\Lk. Since C \ Lk is simply connected, 1og(z - zk)/(z - zk+1) has a branch in C \ Lk D U,,. Therefore, the function
- a, EOlogz-zk+1 z - zk log z -b; 1-1
z
(7.2.11)
k=
has a branch in a neighborhood of U,, . We take n = 1 and set f1(z) = P1(z). For a, E U2 \ U1, we take b, E D \ U2 as above. Set 92(z) =
P2(z)
1
-
n(z - b,)"' fl (z)
a,EUIul
z - a, z - bi /
Then log g2(z) has a branch in a neighborhood of U1. For an arbitrary E2 > 0, there is a holomorphic function h2(z) in D such that (7.2.12)
1 log92(z) + h2(z)i < e2,
z E U1.
Set
P2(z) f2(z) =
II(z - bj)"j
e112(2).
7. MEROMORPHIC FUNCTIONS
190
Then, in a neighborhood of U1 the function log
L2 (-z)
f1 (z)
= logg2(z) + h2(z)
has a branch, and by (7.2.12) it satisfies (7.2.13)
log
f2 (z)
I < C2-
h(z)
1
Here, we take E2 = 1/2. Repeating this process, we construct holomorphic functions f,(z) in neighborhoods of U such that (7.2.14) i)
fn(Z) Pn(4)
ii)
log f `+ (}) has a branch in a neighborhood of TI., and
Z E U,,, n = 1,2,...
0, 00.
z
,
zEU,.
logf fn(Z))1<2-, Set
(7.2.15)
f(z) = f1 (z)
f-+1(z) n=1 fn (z)
= fm (Z)n_fi
fn+t(z) fn (z)
On every U,n we have
x fn+1(z) = exp
n=m log n=m
(nirm
fn (z)
fn+1(z) < fn (z)
E1
n=m
2n
log
fn+1(z) fn (z)
1
2-
Therefore, the infinite product (7.2.15) converges absolutely and uniformly on compact. subsets of D, and so f (z) is a holomorphic function in D. It follows from (7.2.14) and (7.2.15) that f (z) satisfies the required conditions. 0 In the proofs of Mittag-Leffier's and Weierstrass' Theorems (7.2.3) and (7.2.9) the approximation theorem due to Runge (Theorem (7.1.9)) played an important role. In the case of several variables, an analogue of Mittag-Leffler's theorem holds on a domain where the approximation theorem holds (this is called the first Cousin problem, and was solved by K. Oka in 1936-1937). The analogue of Weierstrass' theorem (this is called the second Cousin problem), however, does not follow. When we applied the approximation theorem in (7.2.12), we took logg2(z). In the case of several variables, this argument does not work, and K. Oka proved that some topological condition on D is necessary. This fact is referred as Oka's principle.
7.2. EXISTENCE THEOREMS
191
If a holomorphic function f on a domain D cannot be analytically continued to a neighborhood of any point of OD, 8D is called a natural boundary of f, and D is called the domain of existence of f . This makes sense only when D 54 C, and so we assume in the following that D C C. (7.2.16) THEOREM. There is a holomorphic function f on D such that D is the domain of existence of f. PROOF. For positive integers n, we set
D = { x E D; d(z; 8D) > 1 } f 0(z).
nJ
111
r = 8D .
As in (7.2.12), we cover C by squares En(j, k) with sides of length 1/2".
Figure 87
Take a point z" (j, k) from every E" (j, k) n r,, with En (j, k) n r,, 36 0. Let x"j. j = 1, ... , k", be distinct points of zn(j, k)'s, and order them as Z11,... ,Z1k,,Z21,....Z2kz,Z31,... .
is discrete in D and every point of 8D Then Denote this sequence by It follows from Theorem (7.2.9) that there is is an accumulation point of a holomorphic function f (z) in D which has precisely zeros of order one at the
z,'s. By the identity Theorem (2.4.14), 8D is the natural boundary of f. 0
EXERCISE 3. Show that f (z) _ n a z is a holomorphic function whose domain of existence is 0(1). EXERCISE 4. Show that the real axis is the natural boundary of .(z) in (6.6.2) The next theorem is most fundamental in the interpolation problem:
(7.2.17) THEOREM. Let {a}1 be a discrete set of points of D, and let P ,,(z)
=Cno+c"1(z-an)+...+Cnd(Z-an)d
7. MEROMORPHIC FUNCTIONS
192
be arbitrary polynomials. Then, there is a holomorphic function f (z) in D such that about every an, f (z) is expanded to
f (z) = Pn(z) + higher order term. PROOF. By Weierstrass' Theorem (7.2.9) there is a holomorphic function F(z) such that locally about every an
F(z) _ (z - an)d^t1Fn(z), where Fn(z) is a holomorphic function in a neighborhood of an. By MittagLeffler's Theorem (7.2.3), there is a meromorphic function G(z) in D such that locally about an
G(z) - Pn(z) = H. (z) F(z) is a holomorphic function. Set f (z) = F(z)G(z).
Then, about an we have
f (z) = Pn(z) + F(z)Hn(z) and hence f (z) satisfies the required property. 0
7.3. Riemann-Stieltjes' Integral For use in the next section we here describe necessary facts on RiemannStieltjes' integral. Let i(t) be a real valued function on a bounded closed interval [a, bj C R. For a partition (d) : a = to < t1 < ... < tj = b, we set
I0(ti) - 0(ti-1)i.
V (0: (d)) _ i=1
V(O) = sup{V(O; (d)); (d) is a partition of [a,bj} 5 00.
We call V(a) the total variation of 0; if V(0) < oc, 0 is said to be of bounded variation. For c E [a, b] we have (7.3.1)
I210(c)I - I0(a)I - I0(b)II < 10(a) - 0(c)I + I0(c) - 0(b)I < V(OI[a,c))+V(01[c,d]) = V(0).
Specially, if V(46) < oo, 0 is bounded. For x E R, we set
x+ = max{x, 0},
x- = max{-x, 0}.
It follows that
Ixl=x++x-,
z=x+-x-.
7.3. RIEMANN-STIELTJES' INTEGRAL
193
Set I
V:1 (09 (d))
= E(O(ti) i-1
- 0(ti-1))}
V(O; (d)),
Vt(O) = sup{Vt(¢; (d))} 5 V(O)Then we see that (7.3.2)
V+(0) + V (¢) = V(O),
Vt(0I[a,c])+Vt(01(c,bl) = Vt(c), 0(b) -.,(a) = V+(0) - V-(0).
a:5 c!5 b,
We call V+(0) (reap., V-(O)) the positive (reap. negative) variation of 0. EXERCISE 1. Prove (7.3.2).
(7.3.3) THEOREM. i) A function 0 on [a, b] is of bounded variation if and only if ¢ is written as a difference of monotone increasing functions. ii) If V (O) < oo, there exist limits limt-4±o 0(t) fore E [a, b]. Here, at the end points t = a, b, the one-sided limits exist. iii) If 0 is continuously differentiable in a neighborhood of [a, b], then
V(0) = f 1-0'(t)Idt. n
PROOF. i) It is clear that a monotone increasing function is of bounded variation. Since V (O±iO) < V (O) + V (L,), the difference of two monotone increasing functions is of bounded variation. Conversely, assume that V(0) < oo. We define vt(t) = V±(OI[a,t]) for t E [a,b]. Clearly, vf(t) are monotone increasing, and by (7.3.2)
0(t) = 0(a) + v+(t) - v-(t). ii) This is clear.
iii) The mean value theorem implies this. 0 Let ip : [a, b] -+ R be a function. Take a partition (d) of (a, b] as above, and
take points t,-1 < t:5 = ti, 1 5 i 5 1. If the limit lim
IE''(W(0(t1) - 0(ti-1)) i=1
exists, we denote it by Ja
b
iP(t)dO(t),
which is called Riemann-Stieltjes' integral of +' with respect to 0. This is just the curvilinear integral of v' when 0 is considered as a curve in C. EXERCISE 2. Let 0(t) = 01(t)+i02 (t), a < t < b, be a curve in C. Show that the curve 0 has finite length if and only if 0i, i =1, 2, are of bounded variation.
7. MERONJORPHIC FUNCTIONS
144
(7.3.4) THEOREM. i) If V(o) < oc and ?k is continuous, then fo iv(t)do(t) exists.
ii) Under the same assumption as above, fa f(t)di (t) exists, and b
0(t)d+i(t) = [0(t)w(t)]b -
J
jb
(t)dO(t)
PRooF. i) This is a special case of (3.2.8) applied to 0 : [a, b] - R C C. ii) Take a partition (d) and points , as _
!5
©(C)(
_5 :
(tc) - Cti-1)) 1-1
As I(d)I -- 0, the required equality follows.
(7.3.5) EXAMPLE. Set 0(t) = 0 for t < 0, and 0(t) = 1 for t ? 0. Let iP(t) be a continuous function on [-bo, bo) (bo > 0). Then
f
I
b(t)do(t) _ 0(0),
0 < 6 < bo.
a
To see this, we take a partition, -6 = to < t1 < do such that
< tI = b. Then, there is a
t" -I <0
7.4. Meromorphic Functions on C (resp., {bµ}µ 1) be the Let f be a meromorphic function on C. Let zeros (resp., poles) of f counting multiplicities (e.g., if al is a zero of order k, then ai = a2 = . . . = ak a), j > k). The following is called Jensen's formula. 1
7.4. MMEROAIORPHIC FUNCTIONS ON C
195
(7.4.1) LEMMA. Assume that f (z) = z° (cr. + %+j z + ) with ck° 34 0 and ko E Z. Then we have (7.4.2) 1
2"
J loglf(reie)Id0 _ logr - - E aY
log
0<16,1
r
+ ko log r + log Ick° I .
bµ
PROOF. If the case of k0 = 0 is proved, we apply it to f (z)z-k Then equation (7.4.2) follows. Thus, we may assume that f (z) has no zero nor pole at the origin 0. Assume first that there is no a, nor b,, on C(0; r). For a E 0(r), set
E(z) =
=_-2 r r
-°r
+1
=
r(z-a) -az + r2
For Z E A(r), IE(z)I < 1, and IE(z)I = 1 for z E C(0;r). Set h(z) = f (z)
r(z - b,,)
r,
Ib I
r2
r(z -
Then, h(z) is holomorphic and non-vanishing in a neighborhood of 0(r). Therefore, log Ih(z)I is harmonic there, and If (z)I = Ih(z)I on C(0; r). By the mean value Theorem (3.6.10), iii) 12*
2a
0
z
log
If(Tei0)!d0 =
J
log Ih(re`9)Id0 = log Ih(0)I
=logIf(0)I+ E
logIaYI
logIbµl Ibi,I
Now we consider the case where f has zeros or poles on C(0; r). The right side of (7.4.2) is continuous in r. Therefore, it suffices to show the continuity of the integral of the left side of (7.4.2) in r. Suppose that f (re'B°) = 0, or oo. By the parameter change 9 - 00, one may assume that 00 = 0. For 0 (reap., t) close to 0 (reap., r) one has f (te'B) = f (r + te's - r) = (te'B - r)kg(tec0),
where k E Z and g(te`a)
0.
t
Figure 88
t>r
7. MEROMORPHIC FUNCTIONS
196
It is our aim to estimate f ah log If (teil) I dO, so that it suffices to estimate
fe log Ite'e - rIdO
As in the above figure, it follows that
J
log jre'B - rld8 < f 6 log Iteie - rlde < 0.
6
b
Thus, it is sufficient to estimate f'6,, log Iete -1IdO: 6
J a log J&° - 1IdO =
where lo(6)161
I
log -j
+
+ ... f dO =
(t2! Z
r
6
a log
lOIdO + o(6),
0 as 6 - 0. It follows that
r logIOjdO=2blog6-26. Therefore, for an arbitrary e > 0 there is a 60 > 0 such that
IJ6 logIf(te`B)jd9l <e,
It - rj < be.
s
Hence, there is a finite union I of open intervals of [0, 21rj such that I contains arg a,,, arg b,, with a,,, b,, E C(0; r) and
flog If(te
)Idol < e,
It - rl < bo.
Since
1f
2a
logIf(te'e)ldO
1 f 2a
t -.r,
logIf(re'e)IdO,
(0.2,.)\1
(o.z-m)\1
the following holds: (7.4.3)
lim 2a t-r
f
zx
0
log f (te)IdO =
1
j2W
0
log I f (TeiB)IdO.
1og+ x = log max{x,1 } (>_ 0),
x > 0.
It follows that (7.4.4)
logx = log+ x - log' x, n
log+
fl =1
I log X1 = log{ x + log+
n
x; < E log+ x i=1
n
x; < F log+ x; + logn.
log+ i=1
i=1
!,
7.4. NIEROMORPHIC FUNCTIONS ON C
197
EXERCISE 1. Prove (7.4.4). We set 2w
+n(r,f) =
f log' if(re'e)IdO, 1
J0
which is called the proximity function of f. As in the proof of (7.4.3), we see that m(r, f) is continuous in r > 0. It follows from (7.4.4) that (7.4.5)
27r
z' 0
log lf(re'B)1d8 = m(r,f) -
m
(r, fll/
Set
(7.4.6)
n(r, f) = the number of b,,'s in 0(r), n(0, f) = the number of by's such that bµ = 0,
N(r,f) = jr n(t,f) -n(O,f)dt+n(0,f)logr. 0
We call n(r, f) and N(r, f) the counting functions (of poles) of f . (7.4.7) LEMMA. Let the notation be as above. Then we have
o
tog
r _ Jrfl(tf)....fl(03f)d t
Ibl
PROOF. First, note that n(r, f) is a monotone increasing function in r. By Theorem (7.3.4) and Example (7.3.5) one has log Ibµl
=1r log t d{n(t, f) - n(O, f )} rr n(t, f) - n(O, f)dt. o
t
0
We define Nevanlinna's order function or characteristic function of f by
T(r, f) = N(r, f) + m(r, f). Let g be a meromorphic function on C, and let a E C. Then, it follows from (7.4.4) that (7.4.8)
T(r, f - a) < T(r, f) + log' lal + log 2,
T(r, f +g)
7. MEROMORPHIC FUNCTIONS
198
(7.4.9) THEOREM. Let f (z) be a meromorphic function on C such that about 0
0andkEZ. Then
f(z)=ckzk+ck+lzk+1+... with ck
T(r,f)=T(r, f)+logIckI.
/
\\
(7.4.10) COROLLARY. For a E C
N (r,
f
1
a)
where 0(1) stands for a bounded term as r - oc. PROOF. It follows from (7-4.8) and Theorem (7.4.9) that
N(r, f l a) SN(r,f
l
a)+m(r,f
I a)
=T (r,T1 ) =T(r, f -a)+ 0(l) =T(r, f)+O(1).
0
The inequality of Corollary (7.4.10) is called Nevanlinna's inequality. As seen
in the above proof, if the number of a-points of f is larger, i.e., N(r. 1/(f - a)) has a larger growth as r oo, then that of the proximity function m(r,1/(f -a)) is smaller. In the Nevanhnna theory the order function T(r, f) is considered as the standard, and the quantitative properties are investigated by comparisons of other quantities such as N(r,1/(f - a)) with T(r, f ). EXAMPLE. For coprime polynomials P(z) and Q(z) we have (7.4.11)
T (r,Q ) =degQ logr+O(1),
where deg P (resp., deg Q) denotes the degree of P (resp., Q), and deg Q = max{deg P, degQ}.
For we may assume without loss of generality by the first main Theorem (7.4.9) that deg P < deg Q. In this case, we get
P)
(771
r,Q = O(1),
N(r,Q) Thus, (7.4.11) follows. For an entire function f we define the maximum function of f by
M(r, f) = max{I f (z)j; z E A(r)} = max{I f (z)I; z E C(0; r)}.
7.4. MEROMIORPHIC FUNCTIONS ON C
199
(7.4.12) THEOREM. Let f be an entire function. Then
m(r,f)<_log M(r,f)S R+rm(R,f),
R-r
0
PROOF. By definition the first inequality is trivial. We show the second. By the change of variable Rz, we may assume that R = 1. Take zo E C(0; r) so that
logIf(zo)I = M(r,f) Apply Lemma (7.4.1) to f o ff'=o , where 020(z) = (z - zo)/(-zoz+ 1). Then one gets
f
log If(zo)I = log if o O:.,1(0)1 <
zx
log If o z
2n 0 Applying the computation used in (3.6.2), one obtains
logIf(e'e)I
logIf(.zo)I = 2a I 0
,
lea-
I2 dO 2
<_
(1-T2)do
IOgIf(eiB)I
2s
1+rM(1.f)
`=
13
We use the notation
T(r, f) = O(log r) for the property that T(r, f )/ log r is bounded as r - oc. (7.4.13) LEMMA. An entire function f is a polynomial if and only if T(r, f) _ O(log r).
PROOF. By (7.4.11) the "only if" part is clear. Suppose that there are positive
numbers C1 and ro such that T(r, f) < C1 log r for r > ro. Since T(r, f) _ m(r, f ), it follows from Theorem (7.4.12) that
logM(r, f) S 3m(2r, f) 5 3C1(logr+log2),
r > ro/2.
Therefore, M(r, f) <_ 231 r3', for r > ro/2. The coefficient estimate of Theorem (3.5.20), i) implies that f is a polynomial. (7.4.14) LEMMA. Let f be a meromorphic function on C. Then, the number of
poles off is finite if and only if N(r, f) = O(logr). PROOF. The "only if" part is trivial. Suppose as above that
N(r,f)SC1logr,
r>ro.
Since n(r, f) is a monotone increasing function in r, for r > ro we have
n(r,f) <
1
)dt < 8r J r n(tf
< 1 C1 log r2 S 2C1. log r
-
lo$rN(r2,f)
7. MEROMORPHIC FUNCTIONS
200
(7.4.15) THEOREM. A meromorphic function f on C is a rational function if and only if T(r, f) = O(logr).
PROOF. The "only if" part follows from (7.4.11). Suppose that T(r, f) = O(log r). Then, N(r, f) = O(log r), so that the number of poles off is finite by Lemma (7.4.14). Denote them by {b,}y=1, counting orders, and set t
g(z) = f(z) [1(z - b,.)
µ=1
Then, g(z) is an entire function, and by (7.4.8) it satisfies
T(r, g) < T(r, f) + T (rift(z - bµ) µ=1
= O(log r).
Then, Lemma (7.4.13) implies that g is a polynomial, and so f is a rational
function. 0 We define the order of a meromorphic function f on C by (7.4.16)
logT(r,f)
P1 = r-oo lira
log r
When f is entire, Theorem (7.4.12) implies (7.4.17)
Pf
- r-=
log log M(r, f )
logr
(7.4.18) EXAMPLE. Let P(z) be a polynomial of degree n, and set f (z) = eP(=) It follows from (7.4.17) that Pt = n.
Historically, until the time of Hadamard the order of entire functions f was defined by (7.4.17), and the value distribution theory, i.e., the quantitative theory of the solutions of f (z) = a had been developed. This treatment faced a difficulty in dealing with meromorphic functions. The definition of (7.4.16) is due to R. Nevanlinna. For more details, cf. [15], [16], and [17]. For the generalization to higher dimensions, cf. [19], [20], and [9). We 1 be a discrete sequence in C such that 0 < Let define the exponent A of convergence of {a,,}, 1 by 00
(7.4.19)
A=inf(µ>0;Ej- .p
.
Fbr example, if a = v, v = 1, 2, ... , then A = 1. In the same way as (7.4.6), we 1. define the counting functions, n(r) and N(r) of
7.4. MEROMORPHIC FUNCTIONS ON C
(7.4.20) LEMMA. Let
201
A, n(r), and N(r) be as above. Then
17
A =rlim
rkm log N(r) log r
log g
PROOF. It follows from Example (7.3.5) and Theorem (7.3.4) that for p > 0 (7.4.21)
rlim
rr
tM 1
do(t)
(n(r) r
rµ +
oc
rr n(t) 0
tµ+1 dtl
oc, it follows from (7.4.21) that for an arbitrary e > 0 there is If an ro > 0, satisfying (7.4.22)
e > IA f
)
r
pn(r) f
dt
dl =
n(r)
r > r0.
Setting p = limr(log n(r))/logy, one obtains p <_ p. Therefore, p < A. Conversely, we arbitrarily take p > p and p < p' < p. Then, there is an r1 > 0 such that n(t) < tµ' for t > r1. Hence, we have n(r)
1
0
rµ <
f tµ+i+i dt <
Jr
r -- oo, dt < oo.
By (7.4.21), A < p, and so A < p. Thus, the first equality has been proved. The second equality follows from the following:
f2' n (r) < iog2
ntt)dt
_<_
log
2N(2r),
log' N(r)=log' fntt)dt=log+(Jrntt)dt+0(1)) < log+(n(r) log r) + 0(1) < log+ n(r) + log+ log r + 0(1).
(7.4.23) THEOREM. Let f be a meromorphic function of order p on C. Let a E C and let A be the exponent of convergence of the solutions of f (z) = a with counting multiplicities, that is, a-points of f . Then, A < p. PROOF. This follows from Corollary (7.4.10) and Lemma (7.4.20).
By Picard's theorem of Chapter 6 (more directly, by Corollary (6.8.14)), there are infinitely many a-points of a transcendental meromorphic function f except for at most two values a E C. If the number of a-points of f is finite, a is called a Picard exceptional value. E. Bore! proved that there are at most two a E C such that the exponent of convergence of a-points of f is less than the order p of
7. MEROMORPHIC FUNCTIONS
202
f ; such an a is called a Borel exceptional value. Of course, a Picard exceptional value is Borel. R. Nevanlinna defined the defect 6(a) of a by
6a= 1 - lim N(r,1/(f - a))
()
r--'x
7'(r, f)
Corollary (7.4.10) implies that
i) 0 < 6(a) < 1; ii) if a is a Picard exceptional value, then 6(a) = 1.
In fact, Nevanlinna established an even deeper theorem:
1: b(a) 5 2. aEC
This formula means that there are at most countably many a E C with 6(a) > 0,
and that their sum is not greater than two. The above inequality is called Nevanlinna's defect relation, and immediately implies that the number of Picard exceptional values is not greater than two. EXERCISE 2. Let f3, j = 1, 2, ... , n, be meromorphic functions on C. Show the following:
i) T r, En f! '5 E" E" I T (r, f,) + log n,
ii) T (r,fl ..i f,) :5 E T(r,f,) EXERCISE 3. Show the following for f (z) = eZ.
i) N(r, f) = 0 and m(r, f) = r/r. (Thus, T(r, f) = r/ir.) ii) For a E C', we have
N(r, fla)=a+O(logr),
m(r, f1a)=O(logr).
EXERCISE 4. What is the exponent of convergence of an
/=
n°, n = 1, 2, ... ,
with a>0? EXERCISE S. i) For f (z) = sin z, cos z, show that 2rr
+O(logr) 5T(r, f) 5 72 rr +0(1).
ii) For f(z) = tanz, show that T(r, f) =
2r
+O(1).
7.5. Weierstrass' Product Let f be a meromorphic function on C. By Theorem (7.4.23) the exponent of convergence of zeros of f does not exceed the order of f. Conversely, for a discrete sequence of C, whose exponent of convergence is p, we may ask whether there exists a meromorphic function of order p whose zeros are exactly (a,). In the present section we describe Weierstrass' product, which answers this question.
7.5. WEIERSTRASS' PRODUCT
203
Let p E Z+ and define Weierstrass' irreducible factor E(z; p) by
E(z;O) = 1 - z,
p>1.
E(z;P)_(1-z)eZ+sF+...+°,
(7.5.1) LEMMA. i) For IzI < 1/2, I log E(z; p)I < ii) For an arbitrnry z E C we have
2IzIp+'.
log IE(2; 0)I < 1og(1 + IzI),
log JE(z; p) I < A(p) min{IzI". Izlp+i }.
p ? 1,
where A(p) = 2(2 + logp).
PROOF. i) For IzI < 1 - 1/2(p+ 1)(> 1/2) we have by (2.5.3) p
2
(7.5.2)
log(1-x)+z+ 2 + +zP
I logE(z;p)I =
zp+2
zp4-
ff
p+1
Ip+
1
-1p+1p+2+...I
(1+IzI+IzI2+...) zIp}1
< 2IzIp+
(p+ 1)(1 - IzI) =
ii) The case of p = 0 is clear. Assume that p > 0. For IzI > 1 we have 12. (7.5.3)
log IE(z;p)I <- log(1 + IzI) + IzI +
< 2Izl + zI + I z2 2 2
+ ... +
+.... +
I pp
IzIp
p
< 2+ 2 +-..+ pl IzIp (2 + logp)IzI".
By (7.5.2) and (7.5.3) it remains to show the given estimate for z in the annulus
{1 - 1/2(p+ 1) < IzI < 1}. For IzI = 1, (7.5.3) implies that logIE(z;p)I < 2+ logp. Since log IE(z;p)I is harmonic in A(1), the maximum principle (Theorem (3.6.10), iv)) implies that log IE(z; p)I < (2 + logp),
IzI < 1.
If 1 - 1/2(p+ 1) < IzI < 1, then IzIp+i > (1 - 1/2(p+ 1))p+' >_ 1/2, and hence log I E(z; p)I 5 (2 + logp) < 2(2 + logp)IzIp+1
7. MEROMORPHIC FUNCTIONS
204
(7.5.4) LEMMA. Let {aY}4 1 be a discrete sequence of C such that 0 < Ia,4 < IaY+1I
Let p E Z+ be the minimum such that E'1 1/Ia,IP+i < oo. Then, the
infinite product
17(z)E(QY;p)
(7.5.5)
V
converges absolutely and uniformly on compact subsets, and hence defines an entire function satisfying 1=1 n(t) + IzIP+1 r=x
(7.5.6) log I11'(z)I = (p + 1)A(p)
S IZIP
n(t)
I.
J0
Here, n(r) is the counting function of {a,}, A(O) = 1, and A(p) = 2(2 + logp)
forp>0. PROOF. Let R > 0 and IzI <_ R/2 <_ Ia,,9I/2. It follows from Lemma (7.5.1)
that
IlogE( z-p)I<_ aY
Y sVp
2 V=Yo
P+1
z
ao
RP+1
< oo.
V=Vp 2PIa Yj P+
4Y
Thus, the infinite product converges absolutely and uniformly on compact subsets.
Now, we prove the estimate (7.5.6). Assume that p = 0. Then,
log 117(z)15 E log (1 + l Y=1
z
f0 log (1 + I tI } dn(t) //
11
f tIzn(t IzI) + f'° Izln(t)
I
))dt
o
r'° -
= Urn n(r) log
\C1 + r l
t(t + IzI)
o
dt
It follows from (7.4.22) that
0
r-'oo.
Therefore, one gets
n(t)
1og lH(z)I 5 IzI fo'm
t(t+IzI)
{:I
5 IzI fo
t(1 + IzI) IzI
I;I
<_
o
n(t)
t
dt r"0
dt + IzI
J1:1
JI;I
n(t) dt. t2
n(t)
t(t +IzI)
di
7.5. WEIERSTRASS' PRODUCT
205
By (7.4.21) the integral of the last term is finite. Assume that p > 0. Then, from Lemma (7.5.1) it follows that z
--Iy
E IZa +IaE I?I=I
logl17(z)I :5A(p)
la 1
r
=A(p))
J
I
{lzV
=A(p)
+
I
tp
[ZIP+1
all
I t! dn(t)+ J-" Itp+lldn(t) I=I
n(t)
+PIzIp
JI=I n (t) 1+1
+ (p+
n(t)J
Note that limt-.
J
dt
0
0
N
I=I
n(t)t-p-'
I-lp+l
JJJ
= 0 by (7.4.22). Hence,
logIf(z)I<(p+1)A(p)
r
tp+1dt+IzI"l J{zl
dtI.
By (7.4.21) the integral of the last term is finite. 0 The above 17(z) is called Weierstnnss' product or the canonical product assoLet A be the exponent of convergence of ciated with the sequence
and let p be the order of 17(z). Then, p < A 5 p + 1. We take A' so that if A < p+ 1, A < A' < p+ 1, and ifA=p+1, A'=p+1. In the case of A'
ta'-p-'dt + log IW(z)I < 0 (ZIP <_ O (IzI-1 + IzIA') < O (Iz`A').
IzIp+'
f:l 00 tA'-p-2dt 1
In the case of A' = p+ 1, it follows from (7.4.22) and (7.5.6) that log I17(z)I <_ 0. o(Izf"), where o(IzV') stands for a term such as Therefore, by (7.4.17) p 5 A'. Letting A' \ A, one gets p 5 A. Combining this
with Theorem (7.4.23), we see that
pSA=pSp+1.
(7.5.7)
Note that 17(0) = 1, and if p >_ 1, then about 0 zp+I
z
log E
(a,
al,
From this and the estimate of log I11(z)I obtained above, it follows that k
(7.5.8)
dzk
log 17(0) = 0,
0:5 k:5 p,
T(r,17) = o(rp+1).
7. MEROMORPHIC FUNCTIONS
206
=1 be a discrete sequence in C such that 0 < Ia.,j :5 and let A be its exponent of convergence. Let p E Z+ be the minimum oo. Then, the order of Weierstrass' product 11(z) such that associated with is A. Moreover, if an entire function f(z) of order A has exactly the zeros and satisfies (7.5.8), then f(z) = 11(z). (7.5.9) THEOREM. Let
PROOF. It remains to prove the uniqueness. Set
ex:) = f (z) n(z)
f (z) g(z) = log 17(z)'
It follows from Exercise 2 in the previous section, the first main Theorem (7.4.9),
and (7.5.8) that T(r,e9(=)) <-
T(r,f) +T(r,11) = o(rp+').
Thus, by Theorem (7.4.12) max{Reg(z); Izi << r} = o(rp+1).
In the same way, for -g(z) we have that max{-Reg(z); IzI < r} = o(rP+'). Therefore,
max(IReg(z)I; tzl <_ r} = o(rP+1).
It follows that
rer
27r
J
I
(re'9) + 9(
I2
do
=
o(r2(v+l) )
Setting g(z) = E' o c,,z", one obtains Iznl2r2n = o(r2(p+1)). n=0
Thus, for n > p, c = 0, and so g(z) = 0
Then, condition (7.5.8)
REMARK. The existence of an entire function with given zeros was already established in Theorem (7.2.9). It is an important point of Weierstrass' product 17(z) that it is obtained with estimate (7.5.6). W. Stoll generalized this to the case of several complex variables (1953). EXAMPLE. We prove the infinite product expression: w
zl sinz=zri/ (1\ n7r I
ez 100
z2
n1
n27r2
7.5. WEIERSTRASS' PRODUCT
207
Here, if stands for the product excluding n = 0. The zeros of (sin z)/z are nir, n E Z \ {0}. Since E n-1 = oo and E n-2 < oo, we set p = 1, and 00
17(z) = rj (1 + x) a-O" . (1 - z) es/n" nir
n=1
nir
00
n9r
)ez/n" '
which converges absolutely and uniformly on compact subsets in C. Since the exponent of convergence of the zeros is 1, the order of 17(z) is 1. It follows from Exercise 5 in the previous section that the order of sin z is 1. Since sin z z
=1-+... z2
d
3!
dz
L.
sin z - U, z
we have sin z _ Z
00
?) ez/n"
- -00' (I - n7r
(7.5.10) EXAMPLE. The gamma function r(z) is defined in Re z > 0 by
F(z) = fe_tt71dt.
(7.5.11)
The integral is uniformly convergent on compact subsets in Re z > 0. Integrating by parts, we have
r(z + 1) = zF(z).
(7.5.12)
Since F(1) = 1, r(n) = n!, n E N.
Forx>0weset /'n On (x) =
J0
(I -
t )'ttx_1dt,
n = 1, 2, ...
.
We show that
cn+1(x) - r(x),
(7.5.13)
n -+ co.
For that purpose we show that the following convergence is monotone and uniform on compact subsets of (0, oo): (7.5.14)
/C1-n)\ (1-n+l) n+1 <
t_>_0.
It follows from (2.5.3) that for n > t > 0 n
loge-' - log
Cl -
to
t1 = n
vn,-1
V=2
7. MEROMORPHIC FUNCTIONS
208
The right side is clearly monotone decreasing in n. Let k > 0 be an arbitrary
number. Take no so that 0
x
(010
oo
tv
<
yn1 E E v=2 v=2
710Y
+
noY
GY
(1)# V1
+
('°
For an arbitrary e > 0, we take no > 2k such that 2-"0 < E. Moreover, take n1 > no so that no
ny
Ls yY-1
(1)P
n>n1.
2
Y=2
Then, for n > n1 and t E [0, k]
0 < log a-t - log l l -
tn)" < 2e.
Thus, the uniform convergence on compact subsets is proved. Now, we prove the convergence of (7.5.13). Let I C (0, oo) be an arbitrary bounded and closed interval. Take an arbitrary e > 0 and no E N so that
J * e-ctz-ldt < e,
x E I.
a
This and (7.5.14) imply that for n > no and x E I
0
rnp
t} n
t1-
o
n
is-1dt
/
= 1n I 1 - i J to-1dt < J na \
n
e-ttt-ldt < E.
no
There is an n1 > no such that for all nf> nl to
0
e-tt=-ldt - f
0
0
(1- t )n tx_ldt < E. \
n
Therefore,
n > nl,
0 < 1`(x) - 'n(z) < 3e, By repeating the integration by parts for
we get
nzn! On (-T) =
z E t.
x(x + 1)...(x + n).
7.5. WEIERSTRASS' PRODUCT
209
By making use of the expression of the right side, we extend the defining domain of qn (x) to Re z > 0, and define
0n(z) - z(z +
nX n!
Rez>0.
n)'
It follows from the above definition of 0n(z) that 1
6, (W l
TT
=
v= II
(1+ L) e
logn converges to a positive constant C, called It is well known that Euler's number. It follows from Lemma (7.5.4) that the convergence 00
(7.5.15)
1
On(z)
- zec' [I (1 + t/z) e'=1
is absolute and uniform on compact subsets of C; thus the limit is an entire function. It follows from (7.5.13) and the identity theorem (sharing the same
=-
values on the real positive line) that
00
zec,
(7.5.16) r(z)
z) e_tL
fi (1 + V
B=t
Therefore, we deduce that r(z) is analytically continued to a meromorphic function on C which has poles n E Z, n < 0, of order one, and has no pole nor zeros elsewhere. It also follows from (7.5.15) that (7.5.17)
r(z) = hm
n--0o z(z +
nZn! n)*
Equation (7.5.16) is called Weierstrass' infinite product expression, and (7.5.17) is called Gauss' infinite product expression.
The gamma function r(z) plays an important role in studying a number of other special and important functions, e.g., Riemann's zeta function ((z). Here, ((z) is defined by
((z)=Es °O
Rez>1.
n=1
It is known that {(z) can be analytically continued to a meromorphic function on C which has an pole of order one only at z = 1, and has zeros of order one at the negative even integers (called trivial zeros). It is the famous olden Riemann this has been hypothesis that all the other zeros of ((z) lie on the line Re z = z; an open problem since 1859. EXERCISE 1. Obtain the infinite product expression of cos z. EXERCISE 2. What are the exponents of convergence of the following sequences, and their Weierstrass' products?
i) an = n(log n)2,
n = 2,3,... .
ii) a = n!, n = 1, 2, ... .
7. MEROMORPHIC FUNCTIONS
210
EXERCISE 3. Show that for an arbitrary p = 0 there is an entire function of order p. EXERCISE 4 (H. Bohr and J. Mollerup). Show that if a function ,(x) in x > 0 satisfies the following three conditions, then o(x) = r(x): i) 0(1) = 1,
ii) O(x + 1) = x¢(x),
iii) log O(x) is a convex function.
7.6. Elliptic Functions Let f be a meromorphic function on C. If a number w E C satisfies f (z + w) = f (z),
z E C,
w is called a period of f . For example, f (z) = e= has a period 2iri, and f (z) _ sin z, cos z have a period 27r. Let 11 denote the set of all periods of f . Then 11 forms an additive group; that is, 0 E 11, and if wi, w2 E fl, then wl - wz E fl. We call fl the period group of f. (7.6.1) LEMMA. Let f and 12 be as above, and assume that f is not constant. Then ft is one of the following three cases: i) f2 = {0}. ii) There is an wi E C \ {0} such that t1= {nwi; n E Z}. iii) There are R-linearly independent wi,w2 E 0 such that fl = {nlwi + n2(02; nl, n2 E Z}.
PROOF. Assume that n 96 {0}. Since f is not constant, by the identity Theorem (2.4.14) fl is discrete. Take wi E fl \ {0} such that (7.6.2)
Iw1I = min{iwl;w E R).
Take an arbitrary element w E iZ \ {0}, and set r = w/w1. Assume that r E R. By making use of Gauss' symbol [r], we have that w = [r]wi + (r - (r])wi. Hence, (r - [r])w1 E 0, and J (r - [r])wi I < Iwi J. It follows from (7.6.2) that r = [r] E Z. Therefore, if there is no element of fl which is linearly independent of w1 over R, fl is of case ii).
Figure 89
7.6. ELLIPTIC FUNCTIONS
211
Assume that IT = {w E 11;w/w1 ¢ R} 0 0. Let w E Sl', and denote by , w] the parallelogram with vertices, 0, wl, w, w1 + w. Let IQ[wl, w] I be the area of Q[wl, i). Take w2 E St' such that
Q[W l
1Q[W1, W2]I = min IQ[wi,w]l
Let w E fl be an arbitrary element. Then there are unique r1, r2 E R such that W = r1Wi +r2W2
= [r1] i1 + [r21W2 + (rrl - [ri1)wl + (r2 - [r2])W2-
We want to show that r1, r2 E Z. By the above equation, it suffices to show that if 0 S rj < 1, j = 1, 2, then rj = 0. Since IQ[Wi, w]I = r21Q[wl, w2)I, the definition of w2 implies that r2 = 0. Thus, w = rlWl. The choice of w1 implies rl = 0. In this case, iii) holds. 0 The above Sl is a discrete subgroup of C, and acts on C by wES2.
As described in Chapter 5, §3, the set C/U of all !2-orbits [z] = {z + w; w E U} gives rise to a Riemann surface. If Sl = {0}, then C/tl = C, and this is the most likely case for f . For instance, if f (z) is a non-constant polynomial, or rational
function, then fl = {0}. For, if there exists an w E 12 \ {0}, then for zo E C f (zo + nw) = f (zo), n E Z, and hence f (z) = f (zo). Next we consider the case of ii). Take the universal covering (cf. Figure 90)
a: C9 z-+e2"fz/4'3 EC'.
Thus, C/il = C'. Then, f(z) is expressed by a meromorphic function g(w) in w = e2w'x/"' . We take a ring domain R(rl, r2) _ {rl < IzI < r2} so that it does not contain any pole of g(w). Then, g(w) is expanded there to a Laurent series 00
cnwn
g(w) n=-oo
Therefore, we have the expression, 00 cne2wiz/W,
f(z) _
n=-oo
212
7. MEROMORPHIC FUNCTIONS
If f (z) is entire, the above expansion converges uniform on compact subsets in C.
FIGURE 90
We next deal with the remaining case of iii), which is the the main theme of the present section. In what follows, we assume that 0 is of case iii). Then, f is called a doubly periodic function or an elliptic function; w1 and W2 are called the fundamental periods. (The origin of the name, "elliptic function" will be explained at the end of the section.) The set Q[wi,W2] is called the period parallelogram of f , or of Q. In the sequel, we assume that Q[wl, w2] contains the edges twI and tw2i 0 <_ t < 1.
O
FIGURE 92
FIGURE 91
Let
7r:C-+ C/S2
be the universal covering. Then, 7r(Q[wl,w2]) = 7r(Q[wi,w2]) = C/S2, and so C/O is compact. Topologically, C/St is homeomorphic to the surface of a doughnut as in Figure 92. We call C/St a 1-dimensional complex torus, or an elliptic curve. The periods wI, W2 are called generators of S2, and we write SZ = Il[w1,w2]
The subgroup 52 is also called a lattice of C, and every element w E ft is called a lattice point. Using -W2 if necessary, we may assume that W2
7.6. ELLIP'T'IC FUNCTIONS
213
For two elements [a] and [b] of C/11, we define the addition by [a] + [b] = [a+ b].
For an arbitrary [a] E C/f2, the transition C/f1 D [z] - [z] + [a] E C/O
(7.6.3)
is biholomorphic. Let wi,w2 E 11 be other generators. Then,
wi = awl + bw2, w'2
=
= a'wi + b'w2,
w,
w2 =
cxol + dw2,
c'w'1 + d'1J2,
where a, b, c, d, a', b', c', d E Z. Set A = (a d ) and A' = (a: b' ). Then, AN
_
(a ° ). Let GL(2, Z) denote the group of all 2x2 invertible matrices with elements in Z. Then we have (7.6.4)
A
(a d)\ =
E GL(2, Z),
/ 1
/
w1) = I
l c d) (WI)
2
Conversely, if wi and w2 are given by (7.6.4), then wi and w2 are the generators of 12. If Im w1 /w2 > 0 and Im w'/w'2 > 0, then A E SL(2, Z). Summarizing the above, we have (7.6.5) LEMMA. Let f1[wl,w2] and fl(wi,w2] be lattices of C. Then, fl[w1,w2] _ 12[wi,w2 1 if and only if there is a matrix (a bd) E GL(2, Z) such that (w 2)
- (c d)
k
In particular, if Im wl /w2 > 0 and Im wi /w2 > 0, then (a d) E SL(2, Z).
Setting r = wl/w2, one gets a holomorphic automorphism F : C 9 z -+ zw E C, and so a biholomorphic mapping (7.6.6)
f : C/fl[wi,w2] ' C/12[1,x].
Let it : C -+ C/11[wl,w2] and Al : C -+ C/fl[l,r] be the universal covering mappings. Then,
--
C (7.6.7)
fox=atoF,
F
7
C W1
C/12[wl, wz]
-+ C/11[1, r].
(7.6.8) THEOREM. Two complex Lori, 12[1, rl], j = 1, 2, where Im rl > 0, are biholomorphic to each other if and only if
rl
cr2
+ d'
(c d)
E
SL(2, Z).
7. MEROMORPHIC FUNCTIONS
214
PROOF. The "if" part follows from Lemma (7.6.5). Conversely, suppose that there is a biholomorphic mapping
f : C/f1[1,r1] - C/fl[1,r2]By (7.6.3) we may assume without loss of generality that f ([0]) = [0]. Let xj : C C/n[1, rj] be the universal covering mappings. As in (7.6.7) there
is a lifting F : C -. C of f o a, such that F is biholomorphic, F(0) = 0, and f o z, = n2 o F. It follows from Theorem (6.1.1) that F(z) = az with a E Co. Then, F(f1[l,r,]) = fl[1,r2J; in particular, F(1) = a E (1[1,T2], and F(r1) = ar, E f2[l, r2]. Thus, a and art generate 11(1, r2]. By Lemma (7.6.5) there is a matrix (a e ) E SL(2, Z) such that
(a1
(a
I
d)(1)'
Therefore, r, = (are + b)/(cr2 + d). 0 We investigate holomorphic and meromorpbic functions on 1-dimensional complex tori. (7.6.9) THEOREM. A doubly periodic holomorphic function f is necessarily constant.
PROOF. Let il[wi,w2] be the period group of f. Then, if(z)I < max{If(w)I;w E Qw1,4,-2 },
z E C.
Therefore, f is bounded on C, and by Liouville's Theorem (3.5.22) f is a con-
stant. D By this theorem we see that for a given lattice f2[w1, w2] there is no nonconstant holomorphic function with period group f2[w1,w2]. One may next naturally ask about the existence of non-constant doubly periodic meromorphic functions. Later, we will construct such functions by means of Weierstrass, but for now we deal with the general properties of doubly periodic meromorphic functions. Let f) = il[w1, w2] be a lattice, and let f be a meromorphic function on C such that every w E ft is a period of f. In this case, f is said to be 11-invariant. We define the degree deg f off as the number of poke off in the period parallelogram Q[w1, w2], counting multiplicities.
(7.6.10) THEOREM. Let f and fZ = Q[w1iw2] be as above. Then we have the following.
i) EaEQ(,,,,21 Res(a; f) = 0. ii) If f is not constant, then deg f > 2. iii) For an arbitrary a E C, the number of a-points of f, counting multiplicities, is deg f .
7.6. ELLIPTIC FUNCTIONS
215
PROOF. i) Let a;, i = 1, 2.... n, be poles of f contained in Q[w1, w2] (n deg f ).
FIGURE 93
Take a small c E C so that the interior of the shifted parallelogram c+Q[w1,w2]
contains all a;. Let Cj,1 < j < 4, be its perimeter, so that C2 = w1 + C4 and C3 = w2 + C1, and the orientation of Ej4=1 Cj is anti-clockwise. Then it follows from the residue Theorem (4.2.7) (cf. (4.2.14)) that Res(a;; f) =
. f (z)dz
tai J tai
E j=1
IfE c'
tai{f&z ,
21ri
EJ 7=fr' 1
f (z)dz
f(z) dz +
j
CM+ s
- f(z+w2))dz+
c.
(f(z)-f(z+w1))dz} 11
=0. In the last equality, the periodicity of f was used. ii) By Theorem (7.6.9) it suffices to show that there is no f with deg f = 1. If so, f has only one pole a1 E Q]w1, w21 with order 1. Then,
E Res(a; f) = Res(a1; f) 54 0. aEQ[w,,W2J
This contradicts i). iii) Let b1, 1 <_ i <_ m, be points of f -1(a), counting multiplicities. As in the proof of i), we may suppose that not only the poles of f, but also all bj are contained in the interior of Q[w1,w2]. It follows from the argument principle,
Theorem (4.3.4) and the periodicity of f and f that
m - deg f =
1
tai ,fE,., C
f'(Z) adz = 0.
fO-
EXERCISE 1. Let f (z) be an fl-invariant meromorphic function on C. If f (z) is an even function, then f'(z) has a pole or zero at w such that 2w E ft
7. MEROMORPHIC FUNCTIONS
216
In what follows, the notation
E it \ {0}.
FIGURE 94
(7.6.11) LEMMA. For a > 2, the sum EWEn IwI-Q is convergent. PROOF. Set 1 k = {Awl + mw2; Inl -<- k, Imp <_ k}. Then the number of lattice
points of O k is (2k + 1)2, and that of ilk \ 1k-I is 8k. Let d denote the shortest distance from the origin 0 to the perimeter of the parallelogram with vertices ±w1 and ±w2. Then one has 1 °r-°! 1 < °r° 1
ru
Wen
Iwlp -
ru G WESl,,\Stk_1
k=, WERE\Stk_I
(kd)a
00
8
1
0
<00.
do A.0-1
(7.6.12) THEOREM. The infinite sum (7.6.13)
Iw)2
P(z) = z2 +wen E/ { (=
- 11,
z E C,
converges absolutely and uniformly on every compact subset except for a finite number of terms, and defines a doubly periodic meromorphic function whose period group is Q.
PROOF. Take an arbitrary R > 0. For (z) < R and (wI > 2R we have 1
(z - w)2
_
1
w2
l
_ 2zw-z2 = (z - w)2w2 2+ W I
1
I&I3 =
\
IOR
1
.
IwI3
/1
Therefore, by Lemma (7.6.11) we get the convergence stated above. We show the periodicity. Termwise differentiation implies (7.6.14)
p(z) = -2 F wen
(z lw)3'
7.6. ELLIPTIC FUNCTIONS
217
Thus, the period group of p(z) is clearly fl. For an arbitrary wo E n p(z + wo) - p(z) = C,
(7.6.15)
where C is a constant. By definition (7.6.13) p(z) is an even function; i.e.,
p(-z) = p(z),
(7.6.16)
z E C.
Setting z = -wt/2 in (7.6.15), one obtains
pl2/-pi
2)=P
T)-p121-0=C.
Hence, the periodicity, p(z + wo) = p(z) follows. Since the poles of p(z) are exactly at the lattice points of 12, 0 is the period group of p(z). D It follows from (7.6.13) and (7.6.14) that p(z) is even, p'(z) is odd, and
degp = 2,
(7.6.17)
deg p = 3.
The above obtained doubly periodic meromorphic function p(z) associated with ft is called Weierstruss' pe function, and it plays a fundamental role in the study of the 1-dimensional complex torus (elliptic curve) C/12. In the rest of this section we describe a very introductory part of the theory. (7.6.18) THEOREM. i) For two points z, z' E C, p(z) = p(z') if and only if
[Z]=(fzl
ii) The poles of p'(z) are the points of 12, and the zeros of p(z) are the it-orbits of w1/2, w2/2, and (w1 +w2)/2. iii) Set e1 = p(w1/2), e2 = p(w2/2), and e3 = p((wi + W2)/2). Then these are distinct, and p(z)2 = 4(p(z) - e1)(p(z) - e2)(p(z) - e3)
iv) Set g2 = 60 EWEn w-4 and 93 = 140 El ,En W-6. Then we have
p'(z)2 = 4p(z)3 - 92p(z) - g3.
PROOF. i) By (7.6.13) p(z) is even, and its period group is 12. Therefore, if
(z] = [±z'), then p(z) = p(z'). On the other hand, assume that p(z) = p(z'). Since deg p = 2 and p(z) = p(-z), it follows from Theorem (7.6.10) that (z'] = (±z]. ii) Let wo be w1/2, w2/2, or (w1 +w2)/2. By Exercise I of the present section,
p'(wo) = 0. Since deg p = 2, it follows from Theorem (7.6.10), iii) that p(z) has p(wo) as double point, and so p(z) does not take the value p(wo) on Q(wi,w2J except for wo. Hence, el, e2, and e3 are distinct. iii) Set
f (z) = (p(z) - el)(p(z) - e2)(p(z) - e3)
7. MEROMORPHIC FUNCTIONS
218
The set of poles of p'(z)2 and f (z) is f2, and the orders of those poles are 6. Hence, G (z)2/ f (z) is entire, and by Theorem (7.6.9) there is a constant c E C such that p'(z)2 = cf (z) We look at the Laurent series of both sides about 0:
P(z)2
4 _6-T ...,
c f(z) = z6 .}...
Therefore, c = 4. iv) By making use of (2.4.9) we have the Laurent series of p(z) about z = 0: (7.6.19)
p(z) = z2
F(n + 1) `wl2
+ EI
W2
'+ES1
w2
n=o
}
Z + E(2n + 1)G2n+7Z2n n=1
Here, Gk = EWEn w-k, k ? 3. If k is odd, Gk = 0. It follows from (7.6.19) that (7.6.20)
V (Z) =
p (Z)2 = 4p(z)3 =
-2 Z3
z8 4
+ 6G4z + 20G62 +
+
4
36G4 z2
,
- 8066 ..F ... ,
+ 6066 +
.
Noting that 92 = 60G4 and 93 = 14066, one has V(z)2 - 4p(z)3 + 92p(z) _ -93 + ... ,
Thus, the above right side is a doubly periodic entire function, which must be
the constant -93. D EXERCISE 2. Show that p(z) satisfies the differential equation
p"(z) = 6p(z)2 - 2 . We are going to show that an fl-invariant meromorphic function can be expressed in terms of p(z) and p'(z).
(7.6.21) LEMMA. Let f (z) be an fl-invariant memmorphic function on C. If f (z) is even, then deg f is even. PROOF. We may assume that f (z) is not constant. There are only finitely many zeros of f'(z) contained in the period parallelogram Q[w1,w2] of fl. Take
a value a E C' which f (z) does not take at zeros of f' and on the boundary of Q[w1,w2]. Set f-'(a) n Q[w1,w2] = {a;} a1 with d = deg f. Since f is even,
f (-a;) = f (z,) = a. If [-ail = Jail, then 2aj E f2. By Exercise 1, f'(a;) = 0,
7.6. ELLIPTIC FUNCTIONS
219
which contradicts the choice of a. Therefore, the ft-orbits of a; and -aj are distinct, and hence, there exist j 96 i and w E 11 such that
aj = -a, +w. It follows that d is even. (7.6.22) THEOREM. For an fl-invariant meromorphic function f on C, there are polynomials R;(T), i = 1, 2, in T such that
f (z) = Ri (p(z)) + p'(z)R2(p(z)) PROOF. Note that f (z) can be written as the sum of even and odd functions:
f(z) = (f(z) + f (-z))/2 + (f (z) - f (-z))/2. If f (z) is odd, f (z)/p'(z) is even. Therefore, it suffices to prove that if f (z) is a non-constant even function, f (z)
can be written as a rational function of p(z). In that case, deg f is even. We set deg f = 2k with k E N, and take a as in the proof of Lemma (7.6.21); in the same, we take Q E C,13$ a. Let al, ... , ak E Q[wl, w2] n f -' (a) be such that k
f -'(a) = U (aj + f2) U (-aj + f2) j=1
in the same way as above, let b1, ... , bk E Q[wl, w2J n f -' (,) be such that k
f -'(0) = U (bj + 12) u (-bj + n). j=1
The zeros of j1k1(p(z) - p(aj)) are of order 1, and coincide with those of f(z) - a. Set
f (z) - a
f(z) -
_ - cj=1(p(z)
p(ai))
ni=1(p(z) - p(bj))
Then, c is a non-zero constant. Thus, f (z) is expressed by a rational function of p(z). REMARK. The set Mer(fl) of all fl-invariant meromorphic functions forms a field, and coincides with the meromorphic function field of the Riemann surface
C/fl. One sees by Theorems (7.6.22) and (7.6.18), iv) that Mer(l) is a field of an algebraic extension of C(p) by adding p' which is of order 2, where C(p) denote the transcendental extension of C by adding p. The functions X = p(z) and Y = p(z) give a parameter expression of the solutions of the following algebraic equation: 3
(7.6.23)
Y2=4X3-g2X-g3 =4fl(X-e4) j=1
220
7. MEROMORPHIC FUNCTIONS
EXERCISE 3. Show that an arbitrary polynomial P(X) of degree 3 can be reduced, by a change of variable X = Y + a with a E C, to the form P(Y + a) = aoY3 + a2Y + a3. A polynomial equation of degree 3 of the type (7.6.23) is called Weierstrass' canonical form. Now, we consider the Riemann surface S determined by the multi-valued function Y = Y(X) defined by (7.6.23). The function Y(X) has branch points eI, e2 and e3 whose order of the branch is 1.
C,
e, "
FIGURE 95
Let CI be the line segment from eI to e2, and let C2 be a line from e3 to 00 of C, disjoint from C1. Prepare two copies of such a Riemann sphere, and consider
C, as double, C. Then, identify one C, with another C,-. The resulting space S is topologically homeomorphic to a doughnut (precisely, its surface), which is the same as a 1-dimensional complex torus. x 30
FIGURE 96
In fact, let ir(z) denote the point expressed by X = p(z) and Y = g'(z). Then the mapping (7.6.24)
7r:zEC-7r(z)ES
is holomorphic. Since p'(z) is odd, it follows from Theorem (7.6.18), i) that the restriction 7riQ[w1,w2] of 7r to Q[wt,W2] is injective. Therefore, one sees that (7.6.24) gives a universal covering, and (7.6.25)
C/f [wl,w2] = S.
That is, (7.6.23) gives an algebraic representation of the 1-dimensional complex torus C/SZ[whw2]. The study of Weierstrass' pe-function p(z) associated with
7.6. ELLIPTIC FUNCTIONS
221
1, which is an analytic object, is equivalent to the study of the 1-dimensional complex torus C/a, which is a geometric object, and also to that of the algebraic equation given by (7.6.23), which is an algebraic object. They are the same study from three different viewpoints. (7.6.26) LEMMA. Let 92i g3, e 1 < j < 3, be as in Theorem (7.6.18). Then
92-27g3=1611(e;-ej)200. i<3
PROOF. This is a special case of the discriminant of a general algebraic equation, and follows from the relations between the roots and the coefficients. In fact, we have
e1+e2+e3=0,
ele2 + e2e3 + e3ei = -L,
e1e2e3 = 4 .
Thus, it follows from (el - e2)2 = (el + e2)2 - 4ele2 = e3 - 4e1e2, etc. that (el - e2)2(e2 - e3)2(e3 - el)2 = (e3 - 4eje2)(eI - 4e2e3)(e2 - 4ele3)
63eie2e3 - 4(eie2 + e2e3 + e3ei) + 16eie2e3(e3 l + e2 + e3),
ei+e2+e3(el+e2+e3)(ei+e2+e3 - e1e2 - e2e3 - e3ei) + 3e1e2e3 3
=493, eie2 + e2e3 + e3ei = (eie2 + e2e3 + e3ei){(eie2 + e22 e3 + e2ei
- ele2e3(el + e2 + e3)} + 3eie2e2 2 3 92
_ -
2
{(eie2 + e2e3 + e3ei)
4
3 - 2ele2e3(el + e2 + e3)} + 16 93 3
92
3
2
64+1693 Getting together the above equations, we have the required identity.
Assume that T = w2/wl, and Imr > 0. Then we set 92 = 92(w1,w2) =
601: m ,n
1
1
(mwl +nw2)4
93 = 93(W1,w2) = 140''
m,n
1
(mil + nw2)6
= wi92(1,T), =
1
1 93(1,T). 1
7. MEROMORPHIC FUNCTIONS
222
By Lemma (7.6.26), one can define J(T) =
(7.6.27)
92(l,r)3
g2(1, r)3 - 27g3(1, r)2
_
92(J1, W2)3
92(w1,w2)3 - 27g3(wl,w2)2
The function J(r) is holomorphic in H. By (7.6.6) the 1-dimensional complex torus C/1l[wl, w2] is biholomorphic to C/f?[l, T]. By Theorem (7.6.8) C/fI[1, r] is biholomorphic to C/S1[I,r'] if and only if
r
ar + b
a
cr + d'
c
b
dE SL(2, Z).
In this case, we compare J(r) with J(-r'):
g`(1,T')=60 (m + 1riT')4 m, n
4
CT + d J
,,n
_ (cr + d)460E' { (na + mc)r + nb +
and}-4.
m,n
Since (na + me, nb + md) = (n, m) ( d) , (0, d) E SL(2, Z), (7.6.28)
92
(1, c27+ d) = (cr + d)4g2(1, r).
93
(1' cr+d) =
r+
Similarly, we have (7.6.29)
(Cr+d)6g3(I,r).
In general, if a holomorphic function g(r) in H satisfies
(cr + d) = (cr + d)kg(r),
(a d)
E SL(2, Z),
where k E Z, then g is called an automorphic form of weight k (with respect to SL(2, Z)). The ratio G(r) of two automorphic forms of the same weight is an SL(2, Z)-invariant meromorphic function in H; i.e.,
G (°) = G(r),
(c d I E SL(2, Z).
It follows from (7.6.28) and (7.6.29) that g2(r) is an automorphic form of weight 4, and g3(r) is of weight 6. Hence, in (7.6.27) the denominator and numerator of J(r) are of weight 12, and so J(r) is SL(2, Z)-invariant. In fact, we have the following:
7.6. ELLIPTIC FUNCTIONS
223
(7.6.30) THEOREM. The function J(r) maps H/SL(2, Z) bijectively onto C. Moreover, two 1-dimensional complex tori, C/Sl[wl,w2] and C1f1[wi,w2] with r = w2/wi E H and r' = w2/wl E H, are biholomorphic to each other if and only if J(r) = J(71We have given the proof only of a part of the statement. For the complete proof, cf. [1] and [4]. We collect 1-dimensional complex tori which are biholomorphic to each other. We consider them as one point, and form a set M. The above Theorem (7.6.30) shows that the mapping
J:M=H/SL(2,Z)-+C gives rise to a biholomorphic mapping; the fundamental domain of the action of SL(2, Z) on H is described in Figure 97.
FIGURE 97
EXERCISE 4.
(a iI
Show that SL(2, Z) is generated by A = (°I o) and B =
and then determine the above fundamental domain.
Finally, we explain why the algebraic curve S defined by (7.6.23) is called an "elliptic" curve, and doubly periodic meromorphic functions such as p(z) are called "elliptic" functions. The arc length 6 rof the circle of radius 1 is given by dx = 1-x =f' J ;
o
dy
1y2
As the inverse functions of 9 = 9(x) and 6 = 9(y), we have x = cos 9,
FIGURE 98
y = sin 8.
FIGURE 99
7. MEROMORPHIC FUNCTIONS
224
Instead of a circle, we consider the case of an ellipse
=1,
x2+
b2
Let F(x) be the length of the curve from (1, 0) to (x, y). Then (7.6.31)
1 + (b2 =1)x2dx
F(x) = f t
1 + (b2 - 1)x2
- J:
(1 - x)(1 + (
- 1)x2)
Thus, the integrand contains the square root of a polynomial of degree 4. In general, let P(X) be a polynomial of degree 3 or 4, and let R(X, Y) be a rational function of two variables. Then, the integral
G(X) =
J
X R(X,
P(X ))dX
is called the elliptic integral. The inverse function X = C-1(z) of X = G(z) is called an elliptic function. As in the case of trigonometric functions, it is easier to deal with G'' (z) than with G(X) . The idea of extending the defining domain of elliptic integrals to complex domains, and of considering their inverse functions
is due to Abel and Jacobi, and was a landmark of mathematics. It not only gave a transparent viewpoint to the complicated theory dealt with only over real
numbers, but also made it possible to grow to a great theory of mathematics which has had a primary influence on the advancement of mathematics since then. For example, we consider the following integral: G(x) =
(7.6.32)
I=
dx
4x2-92x-93
This is called an elliptic integral of the first kind, whereas (7.6.31) is called an elliptic integral of the second kind; it has rather more complicated properties. Substituting x = p(z) in (7.6.32), we have by Theorem (7.6.18), iv) p-'(i)
G(x) =
dp(z) Jp-t(z)
p'(z)
=p-'(z)+C
J
t,(z)
(CEC).
That is, G is essentially the inverse function of p(z). We know the addition theorems of trigonometric functions; for example, sin(u + v) = sin u cos v + cos u sin v.
7.6. ELLIPTIC FUNCTIONS
225
The elliptic function p(z) satisfies a sort of an addition theorem, too, but it is not so simple. Take v E C such that 2v Q. By Theorem (7.6.18), ii), p(v) 96 0. Set
p'(u)
f(u) = i + p'(v) 2 p(u) - p(v)
The poles off are points of ft and the fl-orbit (v] of v, and their order is 1. It follows from (7.6.19) and (7.6.20) that about u = 0
f(u) = 1 - u + 6G4u + 20Geu3 + 2 E
u+
+ p'(v)
_1(2n +
p(v)
u _ p(v)u + 2 p (v)u2 + .. . Hence, one obtains f (U)2
(7.6.33)
= u2 + 2p(v) - p (v)u +
Next, we find the Laurent series of f (u) about v. The point v is a pole of f of order 1, and Res(v; f) = 1. We have
f(u) -
1
U-v
(u - v)(p (u) + p'(v)) - 2(p(u) - p(v)) 2(p(u) - p(v))(u - v)
Let g(u) be the numerator of the above right side. Then, g(v) = 0, and
g'(u) = -p'(u) + p'(v) + (u - v)p"(u). Thus, g(v) = 0. Since g"(u) = (u - v)p"'(u), g"(v) = 0. Therefore, we get
f(u)= (7.6.34)
f (U)2
=
v+c1(u-v)+...,
uI 1
(u
v)2 + 2c1 +
.
One infers from (7.6.13), (7.6.33) and (7.6.34) that p(u - v) + p(u) - f (u)2 is a doubly periodic holomorphic function on C, and so a constant. By making use of (7.6.33) and (7.6.19), we calculate the constant term of its expansion about
u=0: p(u - v) + p(u) - f (u)z = p(-v) - 2p(v) + .. .
Hence, p(u - v) + p(u) - f (u)2
p(u - v) =_
1C 4
= -p(v) + ... -p(v); i.e., p (u) + V(v)12
` p(u) - p(v)
f)
-
p(u) - p(v)
Replacing v by -v, we have P, (U)
(7.6.35)
p(u + v) = 1 4
(p(u)
- p(v) )
p(u) - p(v)
7. MEROMORPHIC FUNCTIONS
226
This is called the addition theorem for the pe function.
EXERCISE 5. Write p(2z) and V(2z) as rational functions of p(z) and p'(z).
The elliptic integral described in the present section is extended to the socalled Abelian integral on a general Riemann surface (cf. the reference books [1o]N[14]).
Problems 1. Prove the following. 1)
1
oos z
=
I+
ii) tan z=-
(-1)- (z V=-0o °° y=_"c
(
L + (2v
2
2
1)A)
.
2
zl-i + (2v -1)A ). 1
2. Let D C C be a bounded domain whose boundary is a closed Jordan curve. By making use of Jordan's Theorem (3.3.8), show that a holomorphic function defined in a neighborhood of D is uniformly approximated by polynomials on D. 3. Prove that the exponent of convergence of a discrete sequence {a }1 U (a,, # 0) is given by log n
- jog la.l' 4. (Blaschke's theorem) (a) Let f (z) $- 0 be a bounded holomorphic function on A(l), and let {z,};0--1 be the zeros of f, counting multiplicities. Then, prove that 0c
D1-Iz;l)
(b) Let
be a sequence in A(1) such that E,__1(1 - Ia, I) < oo. Set 00
z -aj
9(z) = fl 17j l - asz, i=1
a
(ifaj =0,nb=1).
aj
Then, prove that g(z) converges absolutely and uniformly on compact sub-
sets of 0(l), and defines a holomorphic function with Ig(z)I < 1, whose zeros are {aj}. 5. Let f (z) be an entire function without zeros, whose order is p. Show that p E N and there is a polynomial of order p such that f (z) = ep(Z). 6. Let A E C' and let P(z) -* 0 be a polynomial. Show that eaZ + P(z) has infinitely many zeros.
PROBLEMS
227
7. Let p > 2 run over all prime numbers, and set 1
((z) = f 1 -
Rez > I.
Prove that this infinite product converges absolutely and uniformly on com-
pact subsets, and moreover, show that it coincides with Riemann's zeta function.
8. Let f (z) be a meromorphic function, and let (- bd) E SL(2, Z). Show that
of +b
T (r),cf+d
T(r,f)+O(1).
9. Let 0 = S2[w1 i W2] be a lattice of C, and let f be an fl-invariant nonconstant meromorphic function. Let a1, ... , ad (resp., b1,... , bd) be zeros (reap., poles) of f in Q[w1,w2], counting multiplicities (d = deg f ). Prove
that 10. Let e; E C,1 < i < 4, be distinct, and set P(w) = n; 1(w - e;). Show that by a linear transformation of the variable, the elliptic integral
is reduced to an elliptic integral
11
dz ,
Q(z)
where Q(z) is a polynomial of degree 3. 11. Let p(z) be Weierstrass' pe-function associated with a lattice 9 of C. Show
that if u+v+wEfi, p(u)
p'(u)
p(v) 00
1
1 = 0.
p'(w) 1 12. Observe that the solution of the dynamics of the pendulum shown below is described by an elliptic integral. p(w)
9 (gravity)
FIGURE 100
Hints and Answers
Here we give hints and answers to some of the exercises and prob-
lems presented in the text. The reader is urged to think over the exercises and the problems for at least a week before looking at these hints and answers.
Chapter 1 §2. Ex. 2. Assume that A = [To, Tl J is covered by two open subsets, Uo and U1, such that Uo n UI n A = 0. We may assume that To E Uo. It suffices to show that A C Uo. Set a = sup{t E (To,T1]; [To,t] C Uo}. V U , which leads to a contradiction (why?). Therefore, a E Uo. In the same way, if a < Ti, a contradiction follows; hence, a = Tl , and (To, Tl ] C Uo. §3. Ex. 2. For z E UQ take i(z;r(z)) C UQ,, r(z) > 0. Moving z
and or,, we obtain an open covering {A(z; r(z)/2)} of A. There are finitely many {A(z,,;r(za)/2)} which cover A. For UQ,, let V be the union of all {0(z,%;r(za)/2)} such that A(za;r(za)) C UQ,.
Problems. 1. E.g., set = x + iy. Then, x2 - y2 = 0, and 2xy = 1. The first implies x = ±y, but by the second, x =may. Hence, x = y = ±1/v"2-.
2. u=f
x2+y2+x/V`,v=± y/
x2+y2+x. (z-1)(z-Pj)zn-1=(z-1)(z"-I+
4. Use the identity 5. Use the absolute convergence.
+1).
6. lim z = (z1 - azo)/(1 - a).
7. Take an arbitrary ao E r. If n E, = 0, EQO C UQEr(C \ E,). Thus, there are finitely many EQ, , 1 < i <_ 1, such that EQO C LL, (C \ E-,) ), and so fl;=, EQ, = 0.
§1. Ex. 3. b= 2- 4-e
Chapter 2 .
§2. Ex. 1. We consider them at z = 1. For any b > 0, there is an irrational number 9 > 0 such that `e2"O - 11 < b. Since {e2n"'9} is dense in C(0; I), there is a subsequence e2n°"'8 -' -1, L - 00. Thus, I(e2"O)n - lI -' 2, and so 229
HINTS AND ANSWERS
230
j f" } is not equicontinuous.
§3. Ex. 1. IF,-=. fk(z)I :5 E'=,,, Ifk(z)I < Ex=m Mk. Then, use Theorem (2.3.2).
§4. Ex. 1. Set rn = (nlogn)I/n 10grn =
n(logn)2 , 0. Hence, r --+ 1, and
the radius of convergence is 1.
Ex. 2. Use Theorem (2.4.4). n!n-"/(n + 1)!(n + 1)-(n+1) = (1 + 1/n)n -+ e. The radius of convergence is e.
§5. Ex. 2. z' = 1/z. §6. Ex. 1. It follows from (2.6.3) and Theorem (2.6.4). Ex. 2. J:°O_1 I -n I = IzI E"O 1 n < oo. Then, use Theorem (2.6.6). §7. Ex. 2. Using (2.7.3), we have d(P, P') = 21z - ZII/V/(1 ++ Iz12)(1 + Iz'I2). Letting Iz'I -+ oo, we get d(P, N) = 2/ 1 + IzI2. Ex. 3. Setting z = x + iy, by (2.7.3) we have T1 = 2x/(1 + IzI2), T2 = 2y/(l + IzI2), 1 + T3 = 2IzI2/(1 + IzI2), and 1 - T3 = 2/(1 + IzI2). A circle in the complex plane is written as aIzI2 + bx + cy + d = 0. Multiplying this by 2/(1 + IzI2), one gets bT1 + cT2 + (a - d)T3 + a + d = 0. This is the equation of a hyperplane in R3, and its intersection with a sphere is a circle.
§8. Ex. 1. Let e' be a unit element, too. Then, e' = e e' = e. Let b' be the inverse of a. Then, bab' = eb' = b', and ab' = e, so that b = b'.
Problems. 1. Use Taylor series expansions about z = 0. All are 1.
2. suplfl=n,a.ndinflfI=0. 3. As far as the author knows, the proof requires Lebesgue integration. If the reader has not learned it, he may just read this proof as a story, keeping this fact in mind. Assume that there is a subsequence {f, }OV°_ 1 which converges at every point of 10, 27r]. Set f (x) = lim f" (x). Lebesgue's bounded convergence theorem implies that for f (t)dt = lim fo f" (t)dt. A simple computation yields fo f" (t)dt 0. Thus, fox f (t)dt - 0. By Lebesgue's theorem on differentiation and integration, fo f (t)dt is differentiable almost everywhere, and equals f (x) almost everywhere. Therefore, f (x) = 0 almost everywhere, and so fo " If(t) I dt = 0. It follows again from Lebesgue's bounded convergence theorem that lim fo ff" I f(t)Idt = 0. On the other hand, I
f0 " I sin ntl dt = n fo n" I sin t l dt = fo n I sin tl dt = 2 fo sin tdt = 4. This is a contradiction. 4.
i) nP/(n+ 1)P , 1. ii) 1/ " IfI" = 1/IqI" - oc. iii) log(n!)'/"2 = n =
log j < n f, +' log xdx = n (n + 1) (log(n + 1) - 1) -+ 0. The radius of n convergence is 1. iv) j. 1
5. By the change of variable z -' (z, we may assume that ( = 1. Set sn = En=0 a and s = limn-00 sn = Ln 0 an. For z E i(1), f (z)/(1 - z) _ En0 z" n 0 anz" = E o snz", and then f (z) - s =(I - Z) En 0(3n - s)2".
For z E D (1; cos r) there is a positive constant K (= 2 sec r = 2/ cos r) such that 11-zl < K(1-Izl). For an arbitrary e > 0, there is an n9 E N such that Isn-sl <
HINTS AND ANSWERS
231
E, n > no. Then, If (z) - sl 5 11 - zI (En° 0 Is,, - sl - IzI" + En=na+1 EIZIn) < (Ell 0 Isn - si + E En 0 I z i") = 11- ZI Ell, Isn - 81 + fl 1- z I/(1- Izl) < -z Z I 1- zI En°_0 Is,, - sl + EK. Taking z so that 11 - zI E"°_o Isn - sI < E, we have
f(z)-sI<(K+1)E. 6. Let f (z) = eie _a:+1 lal < 1, jzI 5 r < 1, and Iz'I r. If (z) - f (z')l < (1 - Ial2)Iz - z'I/I1- r12 < Iz - z'I/I1 - r12. cos z, tan(z + ?r) = tan z. 7. Since sin(z + r) = - sin z and cos(z + r) Suppose that there is 0 < r' < r such that tan(z +'7r') = tan z. Then, sin r' = 0. It follows that sin(z + 2r') = sin z. This contradicts the fact that 2r is the fundamental period of sin z. Z2,,) 10. By induction, = F2"=+'-O -1 z". Let n -+ 00. 11. Set f (z) = (Z - Z3)(z2 - z4)/(z - z4)(z2 - z3). This is a linear transformation, and f (z2) = 1, P Z3) = 0, f (Z4) = oo. By Theorem (2.8.4) such j is unique, and so P Z) = (z, z2, 23, Z4)12. Set g(z) = (f (z), f (z2), f (z3), f (z4)). Then, g(z) is a linear trans-
formation, and g(z2) = 1, g(z3) = 0, g(z4) = oo. By the above problem, 9(21) = (zl, z2, 23, 24)
17. g(z) = R2e'B(z - a)/(-az + R2), a E 4(R), 8 E R. 18. i) f (z) = ((2 + i)z + i)/(z + 1). ii) f (z) = (6iz + 2)/(z + 3i).
Chapter 3 §1. Ex. 1. For n ? 0, ((z + h)" - z")/h =
(")zn-'hj-1 - nz (h -' 0).
Forn<0, use (1/f)'=-f'/f2. Ex. 2. Set h = h1 +ih2. By the Cauchy-Riemann equations, f (z+h)- f (z) _
u(x+hl,y+h2)-u(x,y)+i{v(x+h1,y+h2)-v(x,y)) = geiu-h1+ 7ii h2+i-h1+
it-h2 + o(Ihl) + i `) h + o(IhI). Ex. 3. Let A be the Jacobian matrix of f at (x, y). For two arbitrary unit vectors X , Y, I I AX I I = I I AY I I Because the angle formed by X and Z = (X + Y)/2 is equal to that formed by Z and Y, by the conformality, the angle formed by AX and AZ is equal to that by AZ and AY. Therefore, IIAXII = IIAYII
There are r > 0 and 9 E R such that (U: U;) = r (s;ne e }. Thus, u= =
and u,=-v,,. Ex. 4. Use the Taylor expansions. Ex. S. Use the mean value theorem in two variables. §2. Ex. 3. Assume that 10(t)-¢(t')I < KIt-t'I. For (d) : To = to < t1 < ... < & = T1, L(C; (d)) = E1j=l 1/(t,) - O(ti-1)I < K Et=1 t, - t,_ I= K(T1 - To). Ex. 7. fc xdz = 4i.
§4. Ex. 2. Use (2.4.9). The primitive function is En= 2n Z2,,-1 11
§5. Ex. 2. Use the maximum principle for If,, (z) - f n(z)I. Ex. 4. Set f (z) = E a,,,zm. By Theorem (3.5.20), ii), Em=O lam12r2m 2,r fo I f (reie)12d8 < M2r2". Let r -» oo. an = O, m > n. Ex. 5. We may assume that f (z) = 0 on ry. Take finitely many suitable
0 < a; < 2,r, 1 < i <_ 1, so that C(0;1) = Ue'°j7. Set g(z) = H f(et°'z).
HINTS AND ANSWERS
232
Then, g is holomorphic in A(l) and continuous on.(1). Since IgI - 0 on 8A(1), g(z) = 0. By Theorem (2.4.16) some f (esot z) _- 0, and so f (z) - 0. Ex. 6. Apply the identity theorem. §6. Ex. 1.
1(x2 - y2) = 0. x2 - y2 = 2 (z2 + z2). Thus, 2 Fz- (x2 - y2) = 2z.
The primitive function is z2 = x2 - y2 + i(2xy). Thus, the harmonic adjoint function is 2xy. Ex. 2. Use directly the rule of partial differentiations of a composite function,
and the Cauchy-Riemann equations. Or, use locally the facts that a harmonic function is a real part of a holomorphic function, and that a composite of holomorphic functions is holomorphic.
Ex. 3. Let Iz1 < r < R. Applying the Poisson integral, and using u >_ 0, we have ((r - IzI)/(r + Iz1))u(0) < u(z) < ((r + IzI)/(r - IzI))u(0). Now, let r / R. Ex. 4. In Ex. 3, let R -+ oo. Then, u(z) = u(0). Ex. 5. u(z) = Zx fo re -z - Ire +z dB.
Problems.
1. Check the Cauchy-Riemann equations. 2. Check by definition: ga;T(Tz-) = g f (z) = 0. 4. For an arbitrary e > 0, there is a partition (do) : a = to < t1 <
< ti, = b such that L(C) - e < L(C; (do)) <- L(C). Take a partition (d) : a = so < Si < . . < s, = b with b = I(d)I < min{t3 - t;-t} = 61. For each t, there is only one s,,) containing t3 . Let (d) be the partition formed by all tj and sj. Then, [s, ,j
0:5 L(C;(d))-L(C;(d)) _
k-0(ti)1'f'1001009-0-
(s,,, _
)I}. By the uniform continuity of 0, there is a b2 E (0, b1) such that for It - t'I < 62, 10(t) - 0(t')I < e. For 0 < 6 < 62, 0 < L(C; (d)) - L(; (d)) <_ 2ke. Thus, L(C) - (2k + 1)e 5 L(C; (d)) < L(C), 0 < 6 < 625. 9(1 - i)r3.
6. i) f o i = (1 - az)/(1 - f3z), which is holomorphic about z = 0. 7. Use (3.5.4). 8. Consider f and 11f. The assertion does not hold if f takes the value zero: for example, let D = A(1) and f (z) = z. 9. By Ex. 4 in §6, Ref is constant, and hence Theorem (3.1.8) implies that
f is constant. 10. Let p, q E N, and set g(z) = zD f (z)4. For r1 < IzI = r2 < r3, the maximum principle implies that Ig(z)I < max{riM(r1)4,r3M(r3)Q}. r2l4M(r2) < max{rPj"9M(r1),r3/QM(r3)}. Take a > 0 so that rl M(r1) = r3M(r3). Then, let p/q -' or. rZ M(r2) < ri M(r1). Take the logarithms of both sides, and substitute a = (log M(r1) - log M(r3))/(logr3 - log r1).
= Z i Jo' e"nie''de = 2xs fo ' `ae-1 t & _ d(. If Ial < 1, f (()/(1 - a() is holomorphic in a neighborhood of Ishri fc(o;1) A(l), and so the integral equals f (o). If IaI > 1, use 1/(1-a() = 1/(-1/((-1/a). 11. We have -L fc(o;1) z- o dz
12. Use the mean value theorem.
1
233
HINTS AND ANSWERS
13. y-x+2xy. 14. f(z) = zez.
15. Use the Poisson integral.
16. Use t,(z) = (z - i)/(z + i), and (3.6.7). 17. Note that Re -1F2,-117 =y/fit-z2.
Chapter 4 §2.
Ex. 4. Let C be the union of circles around every point of E. Then, fcw=0.
Ex. 5. Let a (reap., bµ) be the zeros (resp., poles) of f of order m (resp., zm. fl(z-b',)"". Here, set nµ). Then, g(z) = f(z) if a, = oo, and (z - b )"- = z-"- if b, = oo. Then, g(z) is a non-zero constant. Ex. 6. Let g(z) be the right side of the last equation. Then, h(z) = f (z) -g(z)
is holomorphic on C. Thus, h(z) __ c E C. Comparing the Laurent series expansion of f o z about i = 0 with that of g o z, one sees that h(oo) = 0. Ex. 7. Res(±i; f) = 1/2. Ex. 8. Res(tai; f) _ +i/4a3. Ex. 9. Res(0; f) = E 01/n!(n + 1)!, Res(oo; f) Eri o 1/n!(n + 1)!. §3. Ex. 2. Set w = (f'(z)/ f (z))dz. Let M (resp., P) be the number of zeros (reap., poles) of f, counting multiplicities. Then, M - P = Ea c Res(a; w) = 0. Ex. 3. z =2.,/w- +w+En°-_1(-1)"-1(2n-1)!!2-2n+1 21r{(2a2\-
1)/ a - 1 - 2a}.
§4. Ex. 1. i) 21r/(1 - a2). ii) 1r/2 a +a. iii)
Ex. 2. i) 0. ii) a/2'. iii)
iv) 7r(2n - 2)!/22n-'((n - 1)!)2. Ex. 3. i) 2irie_ / '2+'/2/3 + 1rie-' /3. ii) (ir cos a)/2. iii) a/2e. iv) or/4. Ex. 4. i) a/2sin(a7r/2). ii) a(a + 1) ... (a + n - 2)a/(n - 1)!sinair. iii) it/2a1+0 cos(air/2). Ex. 5. i) (1 + log a)/2a. ii) (log2 a - log2 b)/2(a - b). iii) -wr/4.
Problems
00 1. i) En=-00
(-1)' sin l
-
n
1
VL.k=maxio.n}
-1 °oos1
t
T
.
ii) sin &
cos 1 00 n
(
1
sin 1
z-1 - 2
1
z-1
-
+
nz-2n-1
111) En--O( + z-1 + z-1 +1 When n > 0, the pole is at co, and the residue is 0; when n = -1, the pole is at 0, and the residues are 1 at 0 and -1 at oo; when n < -1, the pole is at 0, and the residues are 0 at 0 and at oo. ii) The poles are 7r/2 + nzr, n E Z, and the residues are all I. iii) No pole. The residues are 1 at 0, and -1 at oo. iv) The poles are -2 with residue -2, and -3 with residue 3. The residue at oo is 1. v) The poles are -1 with residue 6, and oo with residue -6. 3. i) 7r. ii) 21ri/(n + 1)!. iii) When n >_ 0 and m > 0, 0; when n < 0 and
n. 2. i)
m
0, 0 for m < -n - 2, and 27ri(_ 1)(-1)m+n+l for in >_ -n - 1; when
nm<0,0forn < -m- 2, and27ri(_m_1) f o r _ -m-1;when n<0 ;='O and m < 0, 0. iv) 21ri(-1)"-1 (m + n - 2)!/(n - 1)!(m - 1)!(a v) 0. 4. i) On C(0;1), l7z4j - 10 - 2z5 + z3 - z + 11 >_ 7 - 6 = 1. Apply Rouche's theorem. ii) If Izi > 2, 1z41 - 61zI - 3 > 1, and so there is no solution there. b)m+n-1
Thus, there are four solutions in !zI < 2. On jzI = 1, 16zl - Iz° + 312! 2, so that
HINTS AND ANSWERS
234
there is one solution there. Thus, there are three solutions in R(1, 2).
5. On IzI = R', IzI - If (z) I > IzI - If(z)I >_ R' - OR. Hence, for R' < R sufficiently close to R, the number of zeros of z and z - f (z) in A(M) are the same. Thus, there is one z E A (R) such that z - f (z) = 0. 6. Use Rouche's theorem on I z I = 1. 7.
Let f (zo) = 0. Apply Rouche's
theorem for f (z) and for f (z) + (f. (z) - f(z)) = f. (z) on a small circle Iz - zol = b. 8. i) ( (a + 1)/a - 1)7r/2.
ii)
iii) 21rie ' /3 - 7ri/3. iv) 27r(1 - a/ Q - 1). v) 7r/2ea. Vi) 7r/aea.
7r (cos '1 7r + b sin 2 7r) / sin air.
vii)
Use the curve in Figure 101. -7r tan 2 ir. viii) Use the equality log3 z - (log z + 27ri )3
= -127rlogz+
6ai loge z - 8iri. 3 7r + Z47r3. ix) Use the curve in Figure 101. -7r2/4.
FIGURE 101
Chapter 5 §1. Ex. 1. Use Theorem (2.8.9). Ex. 2. Apply the reflection principle to extend f meromorphically to all C. Hence, f is rational and holomorphic in C, so that f is polynomial.
Ex. 3. Let f (z) = E a,, z". The first condition implies a" E R. The second condition implies that a" = 0 for even n. Thus, f (z) is odd. §3. Ex. 3. Use Theorem (3.5.14).
Problems 1. Glue two C along [0,1]. 5. Let If (P) l = max{ I f 11 = M. Show that {Q E X ; if (Q) j = M} is open and closed.
7. The length is log R_r. The area is 47rr2/(R2 - r2).
Chapter 6 §2. Ex. 1. Use 7j'(z) = (z - i)/(z + i), which maps H biholomorphically onto A(1), and maps geodesics to geodesics. Ex. 3. dH(z7, z2) = 1og{(1 + Izi - z2I/Izi - z2I)/(1 - Iz1 - z2I/I1I - Z21))§3- Ex. 2. For arbitrary zz E A(1), j = 1,2, such that ir(z,) = wj, by Theorem (6.3.10). For an arbitrary e > 0, there is doitl(zl,z2) >_ dD(wt, a piecewise C' curve C C D from wi to w2 such that LD(C) < dD(wl, w2) + C. Fix zi as above. Let e be the lifting of C with the initial point z1. Let z2
be the terminal point of C. Then, 7r(z2) = w2, and LD(C) = L,1(,)(C) (cf. (6.3.2)). Therefore, DA(l)(zl, z2) < L,&(1 (C) = LD(C) < dD(wi, w2)+e. Hence,
HINTS AND ANSWERS
235
do(l)(zl, z2) < dD(wi, w2) §4. Ex. 1. Using Lemma (2.2.7) or (7.1.12), find countable open coverings, U,, C V,, of D such that {fn I V,, }' is normal. Then, apply the diagonal argument used in the proof of Theorem (2.2.6). Ex. 2. Apply Schwarz' lemma to g o f -I : 0(1) A(1). Ex. 3. Use the uniqueness of Theorem (6.4.4). Ex. 4. Parametrize the parabola y2 = 4c2(x + c2) by x = t2 - c2, y = 2d. Set
z = x + iy = (t + ic)2. Replace t by a complex variable w, and take fz such that = i. Set w = f - ic. As w = t E R runs from -oo to oo, y(t) with z = z(t) = x(t) + iy(t) runs from -oc to oo. Thus, w = f - is is the required mapping. §5. Ex. 1. f(z) _ (z - 1)/(z + 1).
Ex. 2. f (z) = (iz + 1)/(-iz + 1). Ex. 3. It follows from Theorem (6.5.7) and Theorem (3.6.11). §6. Ex. 1. If D 0 C, C, C', then C \ D contains at least three points, which may be assumed to be 0, 1, oc. Set Do = C \ (0, 1, co}. The inclusion mapping
t: D 3 z - z E Do is holomorphic, and has a lifting c o a: D -' 0(1), where 0(l) is the universal covering of Do (Theorem (6.6.7)). Thus, there is a bounded non-constant holomorphic function on D. It follows from Theorem (6.4.6) that D is biholomorphic to A(1). §7. Ex. 1. Considering f - ia, we may assume a = 0. Using (2.8.12), we have lip o f (z) I < 1 for f E F, z E D. Thus, {,I o f; f E F} is normal, and so .F is normal. Ex. 2. Let f (z) be a non-constant entire function. Assume that f (z) does not take three values 0,1, oo. Set f,, (z) = f (vz), v = 1, 2,.... Then, f does not take 0, 1, oo, and hence If, } is a normal family. Since f (0), (f,, } contains a subsequence, converging uniformly on compact subsets. On the other hand, E IanI2lvl2n - oo (v -' oc), where f (z) a,, z". This zx ff * I implies that any subsequence of does not converges uniformly on C(0; 1), and so a contradiction. Ex. 3. Set f,,(z) = e1z, v = 1, 2, ... , which do not take the values 0, oc. Since ff,(0) = 1 and f,,(0) = v, any subsequence of does not converge uniformly on compact subsets.
Problems 1. 2 log 3. log 25 9
.
: A(1) -+ Dj be the universal coverings, and let F : i (1) -+ A (l) be a lifting of f oirl. Take zo E 0(1) such that lr,(zo) = a. The assumption implies f `gt (zo) = gI (zo ), and by Theorem (6.2.3) F'g1 = gI at all points. Hence, 2. Let i r k
f`hD, = hD, 3. Use Exercise 2 of §3, and take a geodesic in the universal covering 0(1).
Take W E C so that the cardinality of f -'(w) is the maximum. Set fI(w) = {zJ}1_I. Then, f'(z,) 34 0. There are neighborhoods A(w;co) of w 4.
and U; of zJ such that f I Ui : U; - A (w; eo) are biholomorphic. We may assume
HINTS AND ANSWERS
236
that U, are bounded. Take R > 0 such that A(R) D U,UJ. By the choice of w, If (z) - wI > co on C \ A(R). Hence, in a neighborhood of oo, 1/(f (z) - w) is bounded and holomorphic. By Riemann's extension Theorem (5.1.1), f (z) has at most a pole at oo. Thus, f is rational. 5. Define a biholomorphic mapping f : A(1) - (z E C; Re z > 0) by f (z) _ (1 +z)/(1- z). The real axis in 0(1) is mapped by f onto the real positive axis. Thus, f (D) = {z; Re z > 0, Im z > 0}, and w = f(z)2 has the required property.
6. Composeg: DE) z-+z" =09=)'Q E {wEA(1);1mw>0} (log1=0) with the one obtained in 5, and then with ty(w) = (w - i)/(w + i). 7. Set z = e'B. Then w = 2 cos 6. We see that C(0;1) is mapped onto 1-2,2].
Using the branch f = 1, we have z = (w - w - 4)/2, which expands to Thus, z = 0 is corresponding to oo. Thus, A(1) is mapped z = w+Y + .
univalently onto D. One sees also that i5\ d(1) is mapped to D. 10. Suppose that (f,, } does not converge uniformly on a compact subset K C D. Then, there is an Eo > 0 such that for every N E N there are vN, µN > N and zN E K with I f,,,, (zN) - fMN (zN) I > co. Run N = 1,2,..., and choose h, and zN - zo E K. Thus, we have subsequences to get f,,N - 9, fµN jg(zo) - h(zo)j > co. On the other hand, g = h on E, and hence g = h. This is a contradiction. 11. Apply (3.6.7). 12. Let a : A(1) D be the universal covering, and let f : 0(1) A(1) be a lifting of f . We may assume that a(0) = zo and j(0) = 0. By the assumption and
Lemma (6.3.11), f E Aut(O(1)), and f(z) = e2ztOz,0 < 6 < 1. If 0 E Q, then for some m E N f"' = f o . o f (m times) = ido(l). Thus, f"' =idD, and f -' = E N, v = 1, 2, ... , so f"'' 1. Let 6 rxf Q. Take an increasing sequence m,, < that elm x'e -+ 1. Applying Theorem (6.7.6) to choose a subsequence, we may
assume that f'- - idD and f" -' - g. Then, f o g = idD. 13. The first half follows immediately from Theorem (6.5.7). Then, using Schwarz' reflection principle, extend f to f E Aut(C). One sees that f (z) _ az + b (a 96 0). Thus, Dl is similar to D2. 17. Use the little and big Picard theorems.
19. We may assume that U = A(1) and P = 0. We also may assume that f omits 0, 1, oo. Let z,, E A`(1) be such that Iz.-I < 1/2, limz = 0, and a. If a = oo, we consider 1/ If (z), so we may assume that a E C. lim f
Set !v(() = f (ZvC), v = 1, 2, .... Then, fy : W(2) - C \ {0, 1}, V = 1, 2, ... , form a normal family by Theorem (6.7.6). Hence, we may assume that U-11 converges uniformly on C(0;1). Set f (z) = F a,z" (Laurent series). Then
E"
a,, z" z", v = 1, 2, ... , converge uniformly on C(O;1). There is an M > 0 I2" < M. Thus, for n < 0, an = 0. such that 2* fo" I f,.(e'e)I2d9 = E l o
HINTS AND ANSWERS
237
Chapter 7 §2. Ex. 1. Let g(z) denote the right side. It follows that g(z + 1) = g(z). Set
f (z) = it cot az - g(z). Looking at the principal part at z = n E Z, one sees that f (z) is entire. We want to show the boundedness of f (z). By making use of the periodicity, f (-z) = - f (z), and of f (z) = f (z), it suffices to show that f (z) is bounded on the set of z = x + iy, 0 :5 x <_ 1/2, y >_ 1. Since I cot lrzj <
(e"' + e-"y)/(e"y - e-"y), cot az is bounded for y >_ 1. Note that lg(z)I < 1+f(1+ +ioo1 z2n < y2) + / (1 + 2y) fi C1 + f(1 + 2y) fo y3dt = 2y)/ (; + 1-1 2 +y dt 1+2fF°O_1(2+y)/{(n-1)2+y2}
e-
C1 + v/ 2-(l +2y)2ir/y < C2 < oo. Therefore, f (z) is constant. The constant term of the expansion of f (z) about z = 0 is 0, and hence f (z) = 0. Ex. 2. Differentiate both sides of the equation obtained in Ex. 1. Check that termwise differentiation is possible.
Ex. 3. Let z = e2"'8 with 0 E Q. For 0 < r < 1, f (re2A`s) is, up to finite terms, the sum of r"' + r("+1)} + - . Thus, lim,._1 f (re2"'B) = +oo. Since such points are dense in the boundary of d(l), f (z) cannot be analytically continued over any boundary point of 0(1).
Ex. 4. There is a sequence of points zn E H such that it accumulates to an arbitrary point of R, and )(z,) = A(z1). §4. Ex. 1. We have log+ E,"_1 xj < log+ n max{x,} < ma(log+ x,} + log n log+ xi + log n.
Ex. 3 i) N(r, f) = 0. m(r, f) = A fp/2rcos0d0 = . ii) n(r,1/(f -a)) _ + 0(1), and hence N(r, l/(f - a)) = x + O(logr). It follows from i) that m(r,1I (f - a)) = O(log r).
Ex. 4. I/a. Ex. 5. Use Ex. 3.
§5. Ex. 1. cost = lln _oa (1 - na+x 2) ez/(na+n/2) Ex. 2. i) E 1/Jan I" is convergent if and only if f W 1/xµ (logx)2µdx converges. Thus, it converges for 0 < 14 < 1, and diverges for p > 1. Weierstrass' product 11 rim 1(1 - z/n!). is = 0. 2 1 - n 1ogn ) e`/"(lOg") . ii) Ex. 3. Take Weierstrass' product associated with an = nl/P, n = 1, 2, ... .
Ex. 4. It follows from i) and ii) that O(n) = (n - 1)! for n E N. Condition iii) implies that for 0 < x1 /< x2 < X3 < X4, (1og0(x2)- log¢(x1))/(x2 -TI) -:5 (log 0(X3) - log 0(x2))/(x3 - X2) < (logO(x4) - log¢(x3))I(x4 - x3). For
0 < x < 1, set x1 = n - 1 < X2 = n < x3 = n + x < x4 = n + 1, and substitute them in the above expression: log(n - 1) S (log ¢(n + x) - log(n -1)!)/x log n.
Combine this with O(n+x) = (n+x-1) .. xO(x) to get (n-1)t(n-1)!/x(x+ 1) < O(x) < n=(n - 1)!/x(x + n - 1). Therefore, (n/(n + x))4(x) < n=n!/x(x + 1)...(x + n) <_ ,(x). Let n -. oo. Then, by (7.5.17) o(x) = I'(x).
HINTS AND ANSWERS
238
§6. Ex. 1. Since f (z) is even, f'(z) is odd. Suppose that w is not a pole of f',
nor f. Since f'(w) = f'(w - 2w) = f'(-w) = -f'(w), f'(w) = 0. Ex. 2. Differentiate the equation of Theorem (7.6.18), iv). Ex. 4. A4 = (o°),A3 = A-' = (i o'). Bn = (o In)
n E N. Take
S = (C d ) E SL(2, Z). SBn = (-b _nc) _ (a d, ). a) The case of a = 0. Then c = ±1. Taken such that d - nc = 0. Then, SBn = (° o) = A or A-'. b) The case of a 6 0. If b is a multiple of a, taken such that b' = 0. Then SBn = (a d) Then, ( d, ° ) A = ( d, ' ). This is reduced to case a). If b is not a multiple of a,
we may taken so that Ib'I < IaI/2. Set (° b') A = (a;; d ). Then, Ia"I < jaI/2. Repeat this. It is reduced to case a). Ex. 5. Use (7.6.35), Ex. 2, Theorem (7.6.18), iv).
Problems 1. i) Let f (z) be the right side. For v such that RV - 2) 7rl > 2Izl, l x + I(v 2)'rl-k-1 < 2 1 , , l(2v21 n Hence, f (z) converges abFk 1 Izlk solutely and uniformly on compact subsets except for finitely many terms. It follows that f (-z) = f (z), f (z + 27r) = f (z). Set F(z) = C0.9 Z - f (z). As in I
7. .
---
the computation of 1/ sin z, one sees that F(z) is constant. Since F(0) = 0, F(z) = 0. ii) is similar to i). 2. Take K = D in Corollary (7.1.10). 3. Let po be the exponent of convergence of {an }. Set pl = lim(log n)/ log Ian I.
Let n(t) be the counting function of {an}. If
>Ianl-" < oo, (7.4.21) implies
'dt that f °° n(t)t-"-'dt < oo. For all larger > 0, we have 1 > fZr (1 - 2-°)n(r)/ar". Setting r = Ianl, one obtains n/lan1° < C (a' constant). Hence, pl < po. If p, = oo, then po = oo. Suppose that p, < oo. For an n(t)t_'
arbitrary f3 > pl, take p, <,3' < 0. For all large n, log n/log I an I < 0' < 0. Take
e > 0 small enough so that (1 + e)(logn)/Iog Ianl < 3. Thus, jani-a < n-1-`, and E Ian10 < 00. We have p, >_ po. 4. i) Use (7.4.2). ii) Use (7.4.2), >loglz3l = Elog{1 - (1 - Iz,l)}, and
Theorem (2.6.4). ii) Usenj(z-a,)/aj(ajz-1) = la3I{l+z(1-Ia,I2)/aj(ajz-1)}. 5. Let g(z) be a branch of log f (z) on C. Then, f (z) = e9(z>. By the assumption, IReg(z)I = loglf(z)l < O(IzIP+'). Using (3.6.7), one sees that Ig(z)I = O(IzIP+'). Thus, T(r,g) = O(logr). Lemma (7.4.13) implies that g(z) is a polynomial. Let p be the degree of g(z). Then, log M(r, f) = corP(1 + 0(1)) with co54 0, and so p = p. 6. The order of eaz+P(z) is 1. Suppose it has only finitely many zeros. Then, eaz + P(z) = Q(z)e"`z, where Q(z) is a polynomial and µ E C'. Differentiate both sides deg P + 1 times. Then, ea= = R(z)e"', where R(z) is a polynomial. Hence, A = µ, and P(z) = (Q(z) - 1)e'Z, and so Q = 1. Thus, P =- 0, which is a contradiction. >p_' converges absolutely, and so is ]l(1 +p-Z). The 7. Since Rez > 0, uniform convergence on compact subsets is proved in the same way. Let 2 =
P1 < P2 < ... be primes. Then, (1 - pi s)-'(1 -
p2:)-'
... (1 - Pnz)-' _
HINTS AND ANSWERS z+P12z+...)(1+p2 2
(1+P1
'
Z
+P2 +P1 n=1
22
+P2 -Z
+ (pp) +
2s+...) ... (1+p;z +p,-2z .
239
1+plZ+
Letting n -, oo, we get n(1 -pnz) _
n- = ((z)-
8. This follows from Theorem (7.4.9), (7.4.8), and (2.8.5).
9. Let Q[w1,w2] be the period parallelogram of Q. Take a E C so that the boundary of E = a+Q[w1,w2] contains no point of aj +fl nor bj +52. Thus, we
may assume that aj and bj are contained in the interior of E. Eaj - E bj = = dz. Using the periodicity, we have sai fee z 'L z dz = mw1 + nw2 E sai fOE z 75T
n.
10. Set v = 1/(w - e4) and e = ej - e4, j = 1,2,3. Then, f dw/ P-(w) _ f-1
Cl
v?...(1 _esv)
dv.
11. The meromorphic function f (z) = p(z) - ap(z) -,Q has the period group 1, and has degree 3. Let aj, j = 1, 2, 3, be zeros of f . Since the poles of f are on Il, a1 + a2 + a3 E fl by problem 9. For arbitrary u, v, there are suitable a,,0 such that f (u) = f (v) = 0. Then, necessarily f (-u - v) = 0. Therefore, if u+v+w E !Q, there are a, Q such that f (u) = f (v) = f (w) = 0. The determinant of a system of linear equations with non-trivial solutions must be 0. 12. The dynamical equation is d29/dt2 = -(g/R) sin 0. Multiplying both sides
by dO/dt and integrating, we have (dO/dt)2 = (49/R)((Lb//)2 - sine 0/2), where vo = R0'(0), 0(0) = 0 (the initial condition). Set 0 < k = vo/ f4R < 1 (to avoid the rotation). Letting x = k sin 0/2, we get (dx/dt)2 = (g/R)(1 x2)(1- k2x2). Thus, t = V19-/R fo dx/ (1 -x2)(1 - k2x2), which is an elliptic integral.
References
In writing this book the author has referred to a number of books already published. Some of them are listed below:
[l] A. Hurwitz and R. Courant, Funktionentheorie, Springer Verlag, Berlin, 1929. [21
[31
L. Ahlfors, Complex Analysis, McGraw-Hill, Auckland et al., 1979; Japanese Translation by K. Kasahara, Gendaisugakusha, Tokyo, 1982. H. Cartan, Theorie Elementaire de Functions Analytiques d'Une ou Plusieures Variables Complexes, Hermann, Paris, 1961; Japanese Translation by R. Takahashi, Iwanami Shoten, Tokyo, 1965. K. Kasahara, Complex Analysis-Analytic Functions of One Variable (in Japanese), Jikkyoshuppansha, Tokyo, 1978. Y. Komatsu, Introduction to Analysis [I] (in Japanese), Hirokawa Shoten, Tokyo, 1962.
Y. Komatsu, Function Theory (in Japanese), Asakura Shoten, Tokyo, 1960.
Y. Komatsu, Exercises in Function Theory (in Japanese), Asakura Shoten, Tokyo, 1960. R. Takahashi, Complex Analysis (in Japanese), University of Tokyo Press, Tokyo, 1990. T. Ochiai and J. Noguchi, Geometric Function Theory in Several Complex Variables (in Japanese), Iwanami Shoten, Tokyo, 1984; English
Translation by Noguchi and Ochiai, Amer. Math. Soc., Providence, Rhode Island, 1990.
For the readers who want to advance to a further study on complex analysis, the author would Wce to give a list of books, which is not intended to be complete. For Riemann surfaces, I recommend the following.
[10] H. Weyl, Die Idee der Riemannsche Fliiche, B.G. Teubner, Stuttgart, 1913; Japanese Translation by J. Tamura, Iwanami Shoten, 1974. (This is a famous classical book, settling the concept of Riemann surfaces.)
241
REFERENCES
242
(11] K. Iwasawa, Algebraic Functions (in Japanese), Iwanami Shoten, Tokyo, 1952; 2nd ed., 1972; English Translation by G. Kato, Amer. Math. Soc.,
Providence, Rhode Island, 1993. (This is a famous introductory book from algebra.)
[12] R.C. Gunning, Lectures on Riemann Surfaces, Princeton University Press, Princeton, 1966. (Comprehensive lecture notes, using sheaf theory.)
[13] Y. Kusunoki, Function Theory (in Japanese), Asakura Shoten, Tokyo, 1973. (This is a nice introductory book of closed and open Riemann surfaces to the research level.) [14] Y. Imayoshi and M. Taniguchi, An Introduction to Teichmiiller Spaces (in Japanese), Nipponhyoronsha, Tokyo, 1989; English Translation by Y. Imayoshi and M. Taniguchi, Springer-Verlag, Tokyo et al., 1992. (This deals with the deformation theory of the complex structure of Riemann surfaces, which is called Teichmiiller theory.) For value distribution theory, I recommend [15] R. Nevanlinna, Le Theoreme de Picard-Borel et Ia Thdorie de Fonctions Meromorphes, Gauthier-Villars, Paris, 1939. (This is a monograph by the creator of Nevanlinna theory, and it clearly unrolls the development of the mathematical theory in front of your eyes.) (16] W.K. Hayman, Meromorphic Functions, Oxford University Press, Oxford, 1964. (This is an introduction to Nevanlinna theory in one variable, up to the research level.) (17] M. Ozawa, Modern Function Theory I - Theory of Value Distribution (in Japanese), Morikitashuppansha, Tokyo, 1976. (This is a monograph at the research level.) [18] S. Kobayashi, Hyperbolic Manifolds and Holomorphic Mappings, Marcel Dekker, New York, 1970. (This is an introduction to the theory of Kobayashi hyperbolic manifolds by Kobayashi himself, and has had a deep influence on complex analysis since then.) (19] W. Stoll, Value Distribution Theory for Meromorphic Maps, Aspects of Math. E7, Vieweg, Braunschweig, 1985. (20] P.A. Griffiths, Entire Holomorphic Mappings in One and Several Com-
plex Variables, Ann. Math. Studies 85, Princeton University Press, Princeton, 1976. For complex analysis in several variables and the theory of complex manifolds, I recommend
(21] L. Hormander, Introduction to Complex Analysis in Several Variables, van Nostrand, New York, 1966. (This is an introduction to the theory from the viewpoint of elliptic differential equations and functional analysis. This book established one research direction of the theory.)
REFERENCES
243
[22] A. Well, Introduction h. 1'Etude des Varietks kahleriennes, Hermann, Paris, 1958. (This is a comprehensive course in the theory of Ki filer manifolds by one of the greatest mathematicians of this century. Mathematically it is very nourishing. (23] S. Nakano, Complex Function Theory in Several Variables - Differential Geometric Approach (in Japanese), Asakura Shoten, Tokyo, 1982. (This is a good book on function theory on Kiihler manifolds by the author, who is famous for the Kodaira-Nakano vanishing theorem.) (24] T. Nishino, Function Theory of Several Complex Variables (in Japanese), Tokyo University Press, Tokyo, 1997. (The author, who was a student of K. Oka, the founder of the theory, gives a treatise based on Oka's idea and work. An English translation is in preparation.)
Index
Abel's continuity theorem, 46 absolute convergence, 13, 213.5 absolute value, 2 accumulation point, 4 act (action), 132 addition theorem for the pe-function, 226 adjoint harmonic function, 85 analytic, 28 analytic continuation, 117, 124 analytic curve, 121 annulus, 91 arc, 6 arcwise connected, 8 area, 138 argument, 3 argument principle, 100 Ascoli-Arzel theorem, 20 at most a pole, 93 automorphic form, 222 big Picard theorem, 171 biholomorphic mapping, 137 Borel exceptional value, 202 boundary, a boundary correspondence, 152 boundary distance, 22 boundary point, S bounded, 4 11 branch, 76, 126 branch point, 128 canonical product, 245 Casorati-Weierstrass' theorem, 118 Cauchy condition, l3 Cauchy product, 13 Cauchy sequence, 10 Cauchy's integral formula, 73, 77 Cauchy's integral theorem, 70 Cauchy-Riemann equations, 54 characteristic function, 192 circle, 5, 39 circle of convergence, 25 closed, 4 closed curve, 6 closed set, 5 closure, 4 compact, 14 complete, 147, 148 complex coordinate, 2 complex derivative, 49 complex differentiable, 49, 87 complex function, 17
complex number, 1 complex plane, 2 complex Poisson integral, 82 complex torus, 212 conformal mapping, 54 conformal metric, 137 conformal pseudo-metric, 131 conformality, 54 conjugate, 2 connected, 5 connected component, 9 constant curve, 7 continuous, 18 contraction principle, 149, 151 converge, %) 19
converge absolutely, 13, 23 converge uniformly, 19 convergent power series, 25 correspondence of circle to circle, 42 counting function, 197 covering, 5 130 covering mapping, 130 covering transformation group, 133 cross ratio, 47 curve, 6 curvilinear integral, 58 70. 88 deck transformation group, 133 defect, 202 degree, 214 derivative, 49 derived function, 49 differential, 87 88 Dirichlet problem, 8.5 discrete, 5 disk, 3 disk neighborhood, 4,.3-9 135 disk of definition, 124 distance, 3 domain, 5, 87 domain of existence, 191 doubly periodic function, 212 elliptic, 48 elliptic curve, 212
elliptic function, 211212, 224 elliptic integral, 224 end, 7 entire function, 118 equicontinuous, 20 exceptional value, 167 exponent of convergence, 200 245
246
exponential function, 3Q exterior point, 5 finite length, 55 fixed point, 4Z Fubini-Study metric. 138 function element, 124 fundamental domain, 164 fundamental group, 130 fundamental periods, 212 fundamental theorem of algebra, 80 Gaussian curvature, 139 Gaussian plane, 2 general curve, 6 generators. 212 group, 44 Hadamard's three circles theorem, 20harmonic function, 82 Harnack inequality, M hermitian metric, 187 hermitian pseudo-metric, 131 holomorphic, 49 87 13(i holomorphic differential, 88 holomorphic function, 136 holomorphic local coordinate, 135 bolomorphic transformation, 137 holomorphic transformation group, 131
homeomorphism, n 131 homotopic, 63 homotopic to a point, 63
bomotopy, fa homotopy class. 6,3.
Hurwitz's theorem, 141 hyperbolic, 48, 147, 142 hyperbolic distance, 145 hyperbolic geodesic, 1.4fi hyperbolic length, 145, 142
hyperbolic metric, 148 identity theorem, 29 imaginary unit. 1 indefinite integral, Z2 infinite product, 13, 34 infinite product of functions. 36 infinity, 38 initial point, Z injective, U interior point, 3
inverse, 1
INDEX
Laplace equation, 82 Laplacian, 82 lattice, 212 lattice point, 212 Laurent series, 83 length, 54, 1.38 length parameter. 56 length parametrization, 56 lifting, 133 limit. 9 1.1 limit function, 19 line segment, 65 linear transformation, 4_0 141 Liouville'a theorem, 72 Lipschitz' condition, 35
little Picard theorem, 167 logarithmic branch point, 128 logarithmic differential, 20 kowdromic, 48 majorant, 24 majorant test, 24 maximum function, 198 maximum principle, 79 mean value theorem, 81. 84 meromorphic differential, 2 meromorphic function. 94 Mittag-LefHer's theorem, 185 Mobius transformation. 44 modular group, 46 monodromy theorem, 1M Montel, 168 Morera's theorem, ZZ multi-valued, 75
mutual reflections, 42 43 natural boundary, 181 neighborhood, 4 negative variation. 193 Nevanlinna's defect relation. 242 Nevanlinna's first main theorem, 191 Nevanlinna's inequality, 198 non-Euclidean, 141 non-singular, Z. 121 normal. 168 normal family, 153 north pole, 31 number of zeros off surrounded by C, 101 fl-invariant, 21A
inverse function theorem, 1
1-valued, 73.
isolated essential singularity, 93. 171 isolated point, 3
open, 4 open covering, 5 open mapping theorem, 103 open set in, 5
Jensen's formula, 194 Jordan curve, Z Jordan's theorem, 66 Koebe's function, 175 lambda function, 167
orbit. 132 order, 93. 200 order function, 191
INDEX
247
order of urpoint. 1112 order of the branch, 128
Riemann surface of the inverse function,
orientation, 8 parabolic, 48 parameter. 6 parameter change, fi partial sum, 12. 23 partition, 54 partition point, .54 period, 210 period group, 214
Riemann's extension theorem, LIZ Riemann's zeta function, 209
period parallelogram, 212 Picard exceptional value, 241
piecewise continuously differentiable, Z piecewise linear curve (Streckenzug), 9, 85 PoincarE metric. 139. 144, 148
point at infinity, 38 Poiason integral, 82 Poisson kernel, 82 polar coordinate, 3 pole, 93 positive orientation, 127 1112 positive variation, 133
121
Riemann-Stielt jea' integral, 1113 ring domain, 91
rotation number. 7G 3S Rouch6's theorem, 141 Runge increasing covering. 184 Runge's theorem, 1113 Schwarz' reflection principle, 113 Schwarz-Christoffel's formula, 176 Schwarz-Pick's lemma, 151 Scbwarzian derivative, 175 sequence, 2 sequence of complex functions, 13 series, 12
series of functions, 23
series of order change, 13 simple curve, Z simply connected, 63 south pole, 3Z special linear group, 44
power series, 25 power series expansion, 28 preserve the orientation, 157
star-shaped, 114
primitive function, a 88
subgroup, 43
principal congruence subgroup, 46 principal part, 185. principal value, 10-8
subsequence, 111
principle of the permanence of the func-
Taylor series, 28
tional relation, 130 Prinzip von der Gebietstreue (open map-
terminal point, 1
ping theorem), 1113
projective special linear group, 41 proximity function, 132 Puiseux series, 1.115
Puiseux series expansion, lS pull-back, 132
purely imaginary, 2 radius of convergence, 25 ratio of the circumference of a circle to its
diameter. 35 rational function, 94 refinement, 55 reflection, 42 reflection points, 122 reflection principle, 43 relative topology, 5 relatively compact, 12 removable singularity, 118 residue, 95 Riemann mapping theorem, L54 Riemann sphere, 337 52 Riemann surface, 127. 135
stereographic projection, 38 Stolz' domain. 411
sum, a surjective, 13 topology, 4
total variation, 132 transcendental, L12. 174 transitively, 44 trigonometric functions, 31 uniformization of Riemum surfaces, 1311 uniformization theorem, 158 uniformly bounded, 211 uniformly continuous, 18 uniqueness of analytic continuation, LLZ
unit circle, 5 unit disk, 3 univalent, 1112
universal covering, 132 universal covering mapping, 132 upper half plane, 45 variation, 192. 193 w-point, 1.42 Weierstrass' canonical form, 224 Weierstrass' irreducible factor, 213 Weierstrass' M-test, 24 Weierstrass' pe-function, 211 Weierstrass' product, 202. 2115
248
Weierstrass' theorem, 189 zero, 93
INDEX
Symbols
E, a element U union
d(z; OD) boundary distance, 22 SL 2, C), 40 PSL(2, C), 41 Aut(A(1)), 43 Aut(C), 43
intersection
N natural numbers (positive integers)
H = {z E C; Im z > 0} upper half
Z integers
Z+ non-negative integers Q rational numbers R real numbers C complex numbers, 1
plane, 45
Aut(H), 45
SL2;R,45 SL 2, Z), 46
C'=C\{0}
r(n), 46
Aut(- ) automorphism group max maximum min minimum mod congruence
8z = 8/8z, 50
[x) = max{n E Z. n < x} Gauss'
I(d)!, 54 L(C) length of curve, 555
8==8/01,50 (d), 54 L(C; (d)), 54
symbol (n) = V
.01J, 55
nl
v.n -vY
Ck-class k-times continuously dif-
ClJ, 55
ferentiable i imaginary unit, 1 z the (complex) conjugate, 2
R(a;ri,r2) annulus, 91
a,82 R(rl, r2) = R(0; rl, r2), 91 Res(a; f) residue, 95 nb(a; C) rotation number with respect to b 96 rrl (D)p fundamental group, 130 Lh(C), 138 Ah(E), 138
Re real part, 2 Im imaginary part, 2 A(a; r) disk, 3 A(r) disk, 3 0 empty set, 4 intersection, 4 A° complement, 4
geodesic, 146 L(zl, z2) geodesic, 146 hD hyperbolic metric on D, 149 LD(C) = LhD(C), 149 dD hyperbolic distance, L49 C(
A\B={aEA;a 0B},4 A the closure, 4 A interior, 5
OA =A\Aboundary, 5
A'(r) = 0(r) \ {0}, 110
C(a; r) circle, 5 I, [a, b] interval of R, 6
0'=0'(1),170
C(4). C(O: 1-' C) curves, 7
V(O; (d)), 192
lim limit, 10 tg relatively compact, 12
at = max{±x, 0}, 192
AA. (0' (r)), 171 V(O) total variation, 192
249
SYMBOLS
250
V±(O), 1.93
log+ x = log max{x, 11, 196 m(r, f ), 192
n(r,f), 192 N(r, f ), 197 T (r, f ), 197
M(r, f) maximum function, 193 p1 order, 200 E(z; p), 243
I', 207 r(z) Gamma function, 207 ((z) Riemann's zeta function, 209 Q[w1,w21 period parallelogram, 211, 212
11[w1,w2] lattice, 212
GL 2, Z 213 deg f , 214
E', 216 p(z) Weierstrass' p (pe) function, 212
Correction List Introduction to Complex Analysis (Version 1998)
p. 6, 1 8: continuous function = increasing continuous function
p. 17, 15: <0=* <E lb - al
p. 28, in FicuRE 12: IbI
p. 29, 1 8 - 9: p. 29, 1 11:D
Jill
lb - al (2 places) a domain D :
p. 30, 1 11: e. _- base e. 1-z ===> p. 32, 1 10: F2.-'--2) -z>
a
1
p. 32, 1 6: mean : intermediate p. 36, 1 14: series
sequence
p. 49, 1 12: f(a+z)-f(z)
f(a+h)- f(a)
p. 55, 1 20: (¢(tl) - O(ti_1))2 (01(t3) - 01(171))2 p. 57, 1 5 ti 9: ti), - I . till _1 (3 places) : a change of parameter of p. 66, 1 1: homotopic to la - zol p. 69, T 8: lal p. 72, 13, 15: zl z (2places) la - ail. Let f be a holomorphic functionon on D. The p. 73, 1 3: la -- WI. The p. 76, 1 2: (t2), : (t2) < 0,(
I[ti-1 +E,t, -E)
1). 77, 1 1: V)Ilt.,- 1, t,j 2az p. 83,15: a: p. 85, 1 7: Delete " lt(c'°)
1). 95, 1 7: f ==> f 1 p. 105, 1 13: a,,, a,,, p. 110, In FIGURE 44: y
2ai p. 114, 1 3: 2a p. 124, 1 4: polynomial
1
rational function
p. 129, 1 10: function ; functions
p. 1'30.13: f:D-'D= f:D--.D' p. 139, 1 6: ,_) an(x) p. 159. 10 ti 9: an injective = a bijective
0
p. 173, 1 9: o 1). 175, 1 18: (`tZ) p. 176, 1 1: 1 2zr p. 180, 1 9, 13:1 p. 183, 119:
DEC
/
z
( ,z,,
2 2a 1(<) (2 places)
p. 186, 1 4: f = f j
p. 207,T10: n!==> (n - 1)! p. 217, 1 13: V(z)2 p. 218, 1 1: z, a,
(z)2
K)
6: polynomials : rational functions p. 229, 1 12: \/-2y/ y/\/2p. 219,
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