lecture
Notes on
COMPLEX ANALYSIS
Lecture
Notes
on
CAOMPLEX ANALYSIS
Ivan Francis Wilde
Published by Imperial College Press 57 Shelton Street Covent Garden London WC2H 9HE
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LECTURE NOTES ON COMPLEX ANALYSIS Copyright © 2006 by Imperial College Press
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ISBN
1-86094-642-9
ISBN
1-86094-643-7 (pbk)
Printed in Singapore by World Scientific Printers (S) Pte ltd
To Erica
Preface
This text forms what is often referred to as "a first course in complex analysis" . It is a slight enhancement of lecture notes first presented to un dergraduate students in the Mathematics Department of Bedford College, University of London, as part of the Mathematics BSc. degree, and then given for many years in the Mathematics Department of King's College, London . During this time they have been continually revised, reorganized and rewritten. The aim was to provide a rigorous and largely self-contained but extremely gentle introduction to the basics of complex analysis. The audience for the course comprised not only single subject mathe matics BSc. and MSci . students but also a number of final year joint honours students as well as postgraduate students who missed out on the subject in their undergraduate programme. There are a number of core topics (such as Cauchy's theorem, the Taylor and Laurent series, singularities and the residue theorem) which simply must be offered to any student of complex analysis. However, quite a bit of preparation is required, so these important results unavoidably tend to appear in rather rapid succession towards the end the course. This leaves very little room for extra topics, especially if they are particularly complicated or involve a lot of additional machinery. The presentation here is for the benefit of the student audience . There has been no quest for ultimate generality nor economy of delivery. Nowadays, it seems that many students do not get to see an account of metric spaces, so this aspect of complex analysis has been presented in quite some detail (in Chapter 3). It is then but a small step for the student wishing to go on to study metric spaces in general . The exponential and trigonometric functions are defined via their power series expansions in Chapter 5 , so a certain amount of manCEuvring is required to extract vii
viii
Lecture Notes on Complex Analysis
those properties familiar from calculus-for example, the appearance of the number 1r is carefully explained . Those for whom this is familiar territory can quickly press on. Most of the core results are contained in Chapters 8-12. The next two chapters, covering the maximum modulus principle and Mobius transfor mations have been moved around a bit over the years . For example, the maximum modulus principle (Chapter 13) could be discussed anytime after having dealt with Cauchy's integral formulae (Chapter 8). The treatment of Mobius transformations (Chapter 14) is essentially a stand-alone topic so could fit in almost anywhere. It might well be read after Chapter 8 so as to provide a little variety before embarking on the study of the Laurent expansion in Chapter 9. It has to be admitted that the final section of Chapter 13 (on Hadamard's Theorem) , possibly Chapter 15 (on harmonic functions) and Chapter 16 (on local properties of analytic functions) could be considered a bit of a luxury. In practice, they were all usually squeezed out because of lack of time . Most of the rest of the material in these notes just about fits into a one semester course. The majority of students embarking on this subject will have studied calculus and will usually have also been exposed to some real analysis. Nevertheless, experience has shown that the odd reminder does not go amiss and so an appendix containing some pertinent facts from real analysis has been included. These are all consequences of the completeness property of lR (so tend not to be very carefully covered in calculus courses--or else are deemed obvious). A number of text books were consulted during the preparation of these notes and these are listed in the bibliography. No claim is made here regarding originality. As an undergraduate student in the Mat hematics Department of Imperial College, it was my privilege to be taught analysis by M . C. Austin and Professor Ch. Pommerenke. Their lectures could only be described as both a joy and an inspiration. It is a pleasure to acknowledge my indebtedness to them both. I.
F. Wilde
Contents
Preface 1.
vii 1
Complex Numbers 1 . 1 Informal Introduction 1 . 2 Complex Plane . . . . 1 . 3 Properties of the Modulus . . . . . . . 1 . 4 The Argument of a Complex Number 1 . 5 Formal Construction of Complex Numbers . 1 . 6 The Riemann Sphere and the Extended Complex Plane
1 7
2. Sequences and Series
1 7 1 7 1 8 21 23 24 25 26
2.1 Complex Sequences . 2. 2 Subsequences . . . 2. 3 Convergence of Sequences 2. 4 Cauchy Sequences 2. 5 Complex Series . . . . 2. 6 Absolute Convergence 2. 7 nth_Root Test . 2. 8 Ratio Test . . . I
o
1 2 4 8 1 2 1 4
I
o
3. Metric Space Properties of the Complex Plane 3.1 Open Discs and Interior Points 3. 2 Closed Sets . . . 3. 3 Limit Points . . . . 3. 4 Closure of a Set . . 3. 5 Boundary of a Set ix
29 29 32 34 36 38
X
Lecture Notes on Complex Analysis
40 41 49 51 5 6
3. 6 Cantor's Theorem 3. 7 Compact Sets . . . 3. 8 Polygons and Paths in C . 3.9 Connectedness 3.10 Domains . . . . 4.
Analytic Functions
59
4 . 1 Complex- Valued Functions 4 . 2 Continuous Functions . . . 4 . 3 Complex Differentiable Functions 4 . 4 Cauchy-Riemann Equations 4. 5 Analytic Functions 4 . 6 Power Series . . . . . . . . . 4 . 7 The Derived Series . . . . 4. 8 Identity Theorem for Power Series 5.
The Complex Exponential and Trigonometric Functions 5 . 1 The Functions exp z, sin z and cos z . 5 . 2 Complex Hyperbolic Functions 5 . 3 Properties of exp z . . . . . 5 . 4 Properties of sin z and cos z 5 . 5 Addition Formulae . . . . . 5 . 6 The Appearance of 1r 5 . 7 Inverse Trigonometric Functions 5 . 8 More on exp z and the Zeros of sin z and cos z 5 . 9 The Argument Revisited . . . . . . . . 5 . 10 Arg z is Continuous in the Cut-Plane . •
6.
.
•
.
The Complex Logarithm 6 . 1 Introduction . . . . 6 . 2 The Complex Logarithm and its Properties 6 . 3 Complex Powers . . . . . . 6 . 4 Branches of the Logarithm .
7.
59 59 6 1 6 6 70 73 7 4 77
Complex Integration 7 . 1 Paths and Contours 7 . 2 The Length of a Contour 7. 3 Integration along a Contour
.
79 79 80 80 8 3 8 4 8 6 89 91 92 94 97 97 98 100 10 3 111 111 11 3 11 5
xi
Contents
7. 4 Basic Estimate . . . . . . . . . . . 7. 5 Fundamental Theorem of Calculus 7. 6 Primitives . . . 8.
Cauchy's Theorem 8 . 1 Cauchy's Theorem for a Triangle 8 . 2 Cauchy's Theorem for Star-Domains 8 . 3 Deformation Lemma . . . . 8 . 4 Cauchy's Integral Formula . . . . . . 8 . 5 Taylor Series Expansion . . . . . . . 8 . 6 Cauchy's Integral Formulae for Derivatives . 8 . 7 Morera's Theorem . . . . . . . . . . . . . 8 . 8 Cauchy's Inequality and Liouville's Theorem 8 . 9 Identity Theorem . . . 8 . 10 Preservation of Angles
9. The Laurent Expansion
10 .
1 27 1 1 1 1 1 1 1 1 1 1
27 33 36 38 39 42 4 5 4 6 49 5 4
1 5 7
9.1 Laurent Expansion 9. 2 Uniqueness of the Laurent Expansion .
1 5 7 1 6 3
Singularities and Meromorphic Functions
1 6 7
10. 1 Isolated Singularities . . . . . . . . . . 10 . 2Behaviour near an Isolated Singularity 1 0 . 3Behaviour as lzl ---> oo . . . . . 10 . 4 Casorati-Weierstrass Theorem . 11.
1 20 1 21 1 23
Theory of Residues 11. 1Residues . . . . . . . . . . . 1 1 . 2Winding Number ( Index ) 1 1 . 3Cauchy's Residue Theorem
1 2. The Argument Principle 1 2.1 Zeros and Poles . . . 12. 2Argument Principle . 1 2. 3Rouche's Theorem 1 2. 4 Open Mapping Theorem . 1 3. Maximum Modulus Principle
1 1 1 1
6 7 69 72 7 4
1 7 5 1_7 5 1 7 7 1 79 1 8 5 1 8 5 18 7 1 89 1 93 1 95
xii
Lecture Notes on Complex Analysis
1 1 1 1 1 1 4.
3.1 Mean Value Property . . . . . 3. 2Maximum Modulus Principle 3. 3Minimum Modulus Principle 3. 4 Functions on the Unit Disc . 3. 5 Hadamard's Theorem and the Three Lines Lemma Mobius Transformations
1 1 1 1 1 5.
4.1 Special Transformations 4. 2Inversion . . . . . . . . . 4. 3Mobius Transformations 4. 4 Mobius Transformations in the Extended Complex Plane Harmonic Functions
1 5 . 1 Harmonic Functions 1 5 . 2Local Existence of a Harmonic Conjugate 1 5 . 3Maximum and Minimum Principle 1 6.
Local Properties of Analytic Functions 1 6 . 1 Local Uniform Convergence 1 6 . 2Hurwitz's Theorem 1 6 . 3Vitali's Theorem . . . . . .
Appendix A
Some Results from Real Analysis
A . l Completeness of IR . . . . . . . . . . . . A. 2 Balzano-Weierstrass Theorem . . . . . . A. 3 Comparison Test for Convergence of Series . A. 4 Dirichlet's Test . . . . . . . . . . . . . . . . A . 5 Alternating Series Test . . . . . . . . . . . . A . 6 Continuous Functions on [a, b] Attain their Bounds A . 7 Intermediate Value Theorem A . 8 Rolle's Theorem . . . A . 9 Mean Value Theorem .
1 95 1 96 200 20 1 20 4 20 7 20 7 209 210 21 5 219 219 220 221 223 223 226 229 231 231 233 235 235 236 236 238 238 239
Bibliography
24 1
Index
24 3
Chapter 1
Complex Numbers
1.1
Informal Introduction
W hat is a complex number? It is any number of the form z = x + iy, where x and y are real numbers and i obeys i 2 = -1. Of course, there is no real number whose square is negative, and so i is not a real number. Accordingly, x is called the real part of z, denoted Re z, and y = Im z is called the imaginary part. (Notice that Imz is y and not iy.) Complex numbers are declared equal if and only if they have the same real and imaginary parts; if z1 x1 + iy1 and z 2 x 2 +iy2 , then z1 z 2 if and only if both x1 = x 2 and Y l = Y2 · We write 0 for 0 + iO. Addition and multiplication are as one would expect; =
=
=
Z1+ Z2 (xl+ X2 ) +i (Yl + Y2 ) z1 z2 = (x1+ i y1 ) (x 2+iy2 ) = x1x 2 +iy1x 2 +iyd y 2+ xd y 2 (x1x 2 - YlY2 )+ i (y1 x 2 + x1 y2 ) . If z x + iy, then -z -x - iy. Suppose that z = x + iy, and z -=/=- 0. Then at least one of x or y non-zero. In fact, z -=/=- 0 if and only if x2 + y 2 > 0. We have 1 X- iy X - iy 1 -=--= z x+iy (x+iy)(x- iy) x 2+ y 2 X y i 2 x2 + y 2 x + y2 ' X 1 1 -y so that Re and Im = Z X 2+y 2 Z X 2+y 2 =
=
=
=
is
=
-
=
-
By definition, complex conjugation changes the sign of the imaginary part, that is, the complex conjugate of z x + iy is defined to be the =
2
Lecture Notes on Complex Analysis
complex number z
=
x- iy. Notice that z z
Re z= x = Proposition 1.1
-2
(iv) Proof.
and
x 2 + y2 and that
lmz=y
=
z-z
-. 2i
For any complex numbers z1, z2, we have
Zl + Z2 = Zl + Z21 Zl Z2 = Z1 Z21 Zl = Zl· ( Zl ) If z2 -=1- 0, then Z2
(i) (ii) (iii)
1.2
z+z
=
Zl Z2
= =
This is just straightforward computation .
0
Complex Plane
There is a natural correspondence between complex numbers and points in the plane, as follows. To any given complex number z = x + iy, we associate the point (x, y) in the plane and, conversely, to any point (x, y) in the plane, we associate the complex number z= x + iy
z=x + iy
+------;
(x,y).
This is evidently a one-one correspondence.
y P
(x,y) <---x-> + iy
X
Fig. 1 . 1
Complex numbers 8.'l points in the Argand diagram.
3
Complex Numbers
The complex plane (also called the Argand diagram or Gauss plane) is just the set of complex numbers thought of as points in the plane in this way. It is very helpful to be able to picture complex numbers like this . The x-axis is called the real axis and the y-axis is called the imaginary axis . If Pis the point (x, y) , corresponding to z = x+iy, then the (Euclidean) distance of Pfrom the origin is equal to Jx2 +y 2 . This value is written lzl, the modulus (or absolute value) of z. Thus, lzl is the length of the two dimensional vector (x, y). If z is real, then y =0 and so lzl = .JX2 =lxl, the usual value of the modulus of a real number. For any complex numbers z1 and z2 , z1- z2 = ( x1 - x 2 ) + i (yl - Y2)
V
so that l z 1- z 2l = (x l - x2 ) 2 + (Yl - y2 ) 2 which is the distance between the points z1 and z2 thought of as points in the plane . It makes perfectly good sense to talk about complex numbers being "close together" -this simply means that the distance between them, namely lz1 - z 2 l, is "small". Examples 1.1
(1) W hat is the set S ={ z: lz- (I r }, where r > 0 and ( is fixed? The complex number z belongs this set if (and only if) its distance from ( =
is equal to r. We conclude that S is the circle in the complex plane with centre ( and radius r. In terms of cartesian coordinates, we see that z = x + iy belongs to S if and only if r 2 =lz- (1 2 (x - �) 2 + (y- TJ) 2 =
where ( = �+iT]. This is the equation of a circle in JR2 with centre at the point (�, TJ) and radius r. By considering values r < R, we see that { z : lz -(I < R } is the disc in the complex plane formed by all those complex numbers whose distance from ( is strictly less than R. Note that { z : l z l 1 } is the circle with radius 1 and centre at the origin . The set { z : lzl < 1 } is the disc with centre at the origin and radius 1 but not including the perimeter { z : lzl = 1 } . ( 2) W hat is the set A = { z : lz- il l z- 31 } ? We see that z E A if and only if its distance from the complex number i is the same as its distance from 3. It follows that A is a straight line-the perpendicular bisector of the line between i and 3 in the complex plane . We can see this in terms of cartesian coordinates . If z = x +iy, then z - i x + i(y - 1) and z- 3 x- 3+iy, s o that z belongs t o A i f and only if x 2 + (y - 1 )2 (x- 3) 2 +y 2 . =
=
=
=
=
4
Lecture Notes on Complex Analysis
Simplifying, this becomes y = 3x - 4, the equation of a straight line. (3) The set { z : Im z > 0 } is the set of those complex numbers z x + iy such that Im z y > 0. This is just the set of all points in the upper half-plane-those points lying strictly above the x-axis. =
=
Proposition 1.2
For any complex number z,
I Rezl :'S lzl llmzl :-=:; l zl � ( l xl +IYI) :-::; lzl :-::; l xl +IYI where x = Rez and y = Imz. Proof. The first two inequalities are direct consequences of the inequality :-=:; a2 +b 2 , valid for any real numbers a and b (take positive square roots) . Furthermore, lzl2 = x 2 + y2 :'S x 2 + y2 + 2 l xi1YI ( l x l + IYI ) 2 . Taking positive square roots gives lzl :'S l xl + IYI· Finally, the inequality (a - b) 2 2 0, for any real numbers a and b, can be rewritten as 2ab :-=:; a 2 + b 2 . Using this, we have
a2
=
Taking the positive square root completes the proof.
1.3
Properties of the Modulus
Further properties of the modulus are Proposition 1.3
(i) (ii) (iii) (iv) (v) (vi) (vii)
0
as
follows .
For any complex numbers z, (, we have
l zl = 0 if and only if z 0; lzl 2 = z z; lzl = l zl; lz ( I = lz l l(l; z z . 0, then ( = lT(Tl ; if(-=/= lz + (I :'S lz l + 1 (1 (the triangle inequality) ; lz1 +· · · + Zml :'S lztl +· · · +lzm l for any Zt, z2, . . . , Zm E C. =
lI
Proof. Parts (i) , (ii) and (iii) are straightforward. To prove (iv) , we observe that lz (1 2 z( z(, by (ii) , = z (z( lzl 2 1 (1 2, again by (ii) . =
=
5
Complex Numbers
Taking positive square roots gives (iv) . Similarly, if ( -=/=- 0, then
and (v) follows . It is possible to prove part (vi) by substituting in the real and imaginary parts and doing a bit of algebra. However,we can give a slick and relatively painless proof as follows:
(z + ( ) (z + ( ) (z + () (z + ( ) zz + (z + z ( + ( ( lz l 2 + 1 ( 1 2 + z ( + z ( lz l 2 + 1(1 2 + 2 Re(z () 2 2 ::S l z l + 1(1 + 2lz(l lzl 2 + I C I 2 + 2 lzl ICI (lzl + 1(1 ) 2.
lz + (1 2
=
=
=
=
=
=
=
Taking positive square roots completes the proof. Part (vii) is the generalized triangle inequality and follows directly from part (vi) by induction . Indeed,for each m E N, let P(m) be the statement that l z1 + · · · + Zml ::S lz1l + · · + lzm l for any z1, z 2, . . . , Zm E C. Clearly, P(1) is true . We suppose that P(n) is true and show that this implies that P(n + 1) is true. Indeed,for any z1, ... , Zn+l in C, let ( Zn + Zn+ 1 · Then,we have ·
=
lz1 + · · · + Zn + Zn+1 1
lz1 + · · + Zn-1 + ( I ::S lz1 l + · · + l zn-1 1 + 1( 1 =
·
·
by the induction hypothesis (namely, that P(n) is true)
by part (vi),and therefore P(n+ 1) is true,as claimed. Hence, by induction, 0 P(n) is true for all n E N.
6
Lecture Notes on Complex Analysis
Remark 1.1
Replacing ( by -(,the triangle inequality becomes
Now, lz l = l z- O l and 1(1 = I (- Ol , and so the above inequality tells us that the distance between the pair of complex numbers z and ( is no greater than the the sum of the distances of each of z and ( from the origin. This is just the statement that if we form the triangle with vertices 0, z and (, then the length of the side joining z and ( is never longer than the sum of the other two sides-hence the name "triangle inequality" . For any complex numbers u, v, w, we see that
l u - wl
=
l (u - v) + (v - w ) l:::; l u - v l + lv - w l ,
so there is nothing special about the origin in the above discussion-it works for any triangle. Evidently, part (vi) is just a special but important case of part (vii), the generalized triangle inequality. Remark 1.2 For any z, ( E C, we have 1 (1 = I(- z + zl :::; I(- zl + lzl, giving 1 (1 - lzl :::; lz - (1 . Interchanging z and (, we obtain the inequality lz l - 1(1 :::; I ( - z l = l z - (1. Thus - l z - (I :::; lzl- 1(1 :::; lz - (1, which can be written as
l l z l - 1(11
lz - (1.
:S
From this, we see that if two complex numbers z and ( are close (i.e., the distance between them, lz - (1, is small) then they have nearly the same modulus. The converse,however,need not be true, for example, i and - i 2. have the same modulus,namely 1, but I i- ( -i) I l 2il =
=
Example 1.2 If the complex numbers u and v are proportional, with positive constant of proportionality, then u + v u + ru (1 + r)u for some r > 0. Evidently, l u + vi = (1 + r) l u i lui + l v l . Geometrically,this is clear. The complex number u + vis got by putting the vector v onto the end of the vector u in the complex plane. If v ru, then u and v "line up" and the triangle with vertices 0, u and u + v collapses to a straight line. Furthermore, if v1,... , Vm are each of the form Vj = riu, for some + Vm is got by placing parallel Tj > 0, then the vector u + V1 + v2 + vectors end to end and so its length will be the sum of the parts, =
=
=
=
·
·
·
We shall see that the converse is also true,as one might expect.
7
Complex Numbers
First, we shall show that the equality lz1 + z 2 l = l z1 l + l z 2 l holds for non-zero complex numbers, z1 and z 2 , only if they are proportional (with positive constant of proportionality), that is, if and only if z 2 = rz1 for some real number r > 0. Indeed, as we have just discussed, if z 2 rz1 , with r > 0, then the claimed equality holds . Conversely, suppose that lz1 + z2 l l z1 l + l z 2 l · Then =
=
and so Re(z1z2 ) = lz1 l l z2 l l z1z2 l · It follows that z1z2 has no imaginary part and so z1z2 = Re(z1z2 ) = lz1z2 l· From this we see that z 2 = r z1 2 where r= lz1z2 l / l z1 l . Now, suppose that z k -=/=- 0, for all k 1, . . . , n , and that =
=
n
n z l i I k L = kL1 Zk I· k=1 = We wish to show that there are positive real numbers r 2 , . . . , Tn such that Zj rjz1 for j = 2, . .. , n . Now, for any partition of the set { 1 , 2, .. . ,n } into two subsets I and J, the equality (* ) implies that =
n
n z i = l I L1 k L1 Zk I= I L Zk + L Zk I kEf kEJ k= k= � I L Zk I + I L Zk I kEJ kEf
kEf
kEJ
n z z � L i k l + L i k l = L iz k l kEf kEJ k=1 and so IL:kEfz k l= l:kEf l z k l · Taking I= {1,j} and applying the first part, the result follows. Let (0 = 0 and (j = (j-1 + Zj, for j= 1, ..., n and let Pbe the polygon [(o, (t] U · · U [(n-1 , (nJ, where [w, z] denotes the straight line segment from w to z . Then the equality in question is the statement that the distance between the initial and final points of Pis equal to the sum of the lengths of its segments. This can only happen if the polygon stretches out into a straight line (and does not turn back on itself) . ·
8
Lecture Notes on Complex Analysis
Example 1.3
For given z1, z2, z3 E
{wE
where J-L= f3 1 ( a + /3) (assuming that the denominator is not zero) .] Now, the set {( : (= ( 1 - J-L)Zl + J-LZ2, 0 � J-L � 1 } is just the line segment from z1 to z2. Let (Jl. (1 - J-L)z1 + J-LZ2 be some point on this line segment. The set LJl.= {w : w = ( 1 - I)(Jl. + 1z3, 0 � 1 � 1 } is the line segment from (Jl. to ZJ. As /-L varies between 0 and 1, so (Jl. varies along the line segment from z1 to z 2 and the LJl.s fill out the triangle. =
1.4
The Argument of a Complex Number
We have agreed that a complex number can be usefully pictured as a point in the plane. Now, we can use polar coordinates rather than cartesian coordinates, giving the correspondences (assuming z =/:. 0)
z= x + iy
(x, y)
f----t
f----t
(r, 0) ,
where r= }x 2 + y2 = lzl, and where (} is given by the pair of equations
x
cos(}= -:;:
x Rez =� = l;r
and
. (} sm
y
=
-:;:
=
y Imz � = l;f .
The value of (} is determined only to within additive multiples of 27T, that is, if(} satisfies both cos(}= xlr and sinO= ylr then so does 0+ 2k1T, for any k E z. Moreover, these are the only possibilities : if 1/J also satisfies cos 1/J xIr and sin 1/7= yIr then 1/7 = (}+ 2 n1T for some suitable n E Z. The angle(} is called the argument of z, denoted arg z. Note that accord ing to the above discussion, arg z is not well-defined. One could call arg z an argument of z, i.e ., a solution to cosO= xlr and sinO= ylr, or one could define arg z to be the set of all such solutions,arg z= { 0, 0±21T, 0±41T, }, =
·
·
·
Complex Numbers
9
y
p
(x, y)
+------+
z= x + iy
X
Fig. 1. 2
Cartesian and polar coordinates.
where(} is any solution . We prefer the first idea, even though it is something of a nuisance . By convention, we can pick on a particular choice . T here is a unique solution (} satisfying - 1r < (} � 1r; this choice of (} is called the principal value of the argument of the complex number z and is denoted by Arg z . Thus, for any z -=/=- 0, Arg z is well-defined and is uniquely determined by the requirement that Arg z= (} E ( - 1r,1r] and cosO= x/r and sinO= y/r. For example, Arg x = 0 for any real number x with x > 0. If x is real and x < 0, then Arg x 1r. Also Arg i= 7r/ 2, Arg( - 1) = 1r, Arg( -i) = -7r/2. =
y -1+ i
X
-1 - i
Fig. 1. 3
Arg( 1 +
i}
=
�,
Arg(- 1 + i)
=
3
;,
Arg(- 1- i)
=
-3
;.
10
Lecture Notes on Complex Analysis
standard trigonometric formulae, we find that
and so fh + fh is a possible choice of argument for the product z1 z2 , for any choices fh and fh of arguments for z1 and z2 , respectively. In general, arg(z1z2 ) = argz1 + argz2 + 2k7r,
for some k
E
Z,
where argz1 , argz2 and arg(z1z2 ) denote any particular choices of the arguments. Of course, different choices will lead to different values for k. In particular, by induction, we obtain De Moivre's formula (cos 0 + i sinO) n Remark 1.3
=
cos nO + i sin nO .
It is not always true that
For example, Arg (- 1 ) = 1r but Arg ( (-1) ( -1 ) ) Arg 1 0 # 1r+1!". The ambiguity of the argument of a complex number reappears when we try to set up the notion of the logarithm of a complex number. =
=
Remark 1.4 Suppose that z has polar coordinates (r, 0) , z f-4 (r, 0). Then zn f-4 (rn , nO). Multiplying complex numbers amounts to multi plying their moduli and adding their arguments. Furthermore, 1 /z = zj(z z) = z/ lz l 2 , so that 1 /z f-4 ( 1 /r, 'f), where 'Pis an argument of z. But z = r cosO - ir sinO = r(cos(-0) + i sin (-0) ) giving z f-4 (r, -0). Hence 1 /z f-4 ( 1 /r, -0). Dividing by a complex number amounts to divid ing by the modulus and subtracting the angle; zjw f-4 (r/ p, 0 - a ) , where z f-4 (r, 0) and w f-4 (p, a ) . Example 1.4
For any w # 1,
1- wn 1 1 +w+w2 +. . ·+wn - = --1-w 1 n n (because (1- w) ( 1 + w+ . . +w - ) = 1- w ) . Setting w =cosO+i sinO (so that wk = cos kO +i sin kO f-4 ( 1, kO) ) , we find that ·
Sn
=
cos 0+cos 20+
·
·
·
+cos nO =
sin ( ( n + � ) 0) - sin ( � ) , ( 29 ) 2sm
11
Complex Numbers
and that Tn
=
sin 0+sin 20+ · · + sin nO
=
·
provided 0 =J. 2k7r for any k E Z. Indeed, if we let ( = cos( � ) - i sin ( � ) Sn +iTn
�
cos(!!.) - cos( (n + l ) 0) 2 2 2sm ( 20) •
( 1 ,- �), then we have
cos 0+i sin 0+cos 20+i sin 20+ · · · + cos nO +i sin nO =w +w2 + ·· ·+wn =
w- wn +l 1- w (wn+l- w)( -'---c,-----:-;--c� (w- 1 )( n w +l (- w( = w( -( cos( ( n + � ) 0) +i sin( ( n + � ) 0) - cos( ! ) - i sin( � ) 2i sin ( ! ) =
=
_
Equating real and imaginary parts gives the required formulae.
Returning to the general discussion, we have declared a complex number to be one of the form z = x + iy, where i2 = -1, and have so far accepted this uncritically. The question is "is this legitimate?" What is this magical number which we denote by i? It is certainly not a real number, so what is it? Now watch closely: (- 1 ) = -1 1 ===? ===? ===?
===?
=
1 __ (- 1 )
� {!: =
JI A JI A RR = JIJI - 1 = 1,
which is some cause for concern . A similar but somewhat less picturesque "observation" is that
v'iO
=
v(-
5) (- 2)
=
v'-5 vC2 = i J5 i Y2 = i2 v'iO
= -
Fa
.
12
Lecture Notes on Complex Analysis
Maybe we should take a little more care in setting up the notion of a complex number. Happily, this can be done, as we will now see ( but the paradoxes above must wait until we discuss complex powers) .
1.5
Formal Construction of Complex Numbers
We define C to be the set of ordered pairs (x,y), with x,y E JR, together with the binary operations of "addition" ( denoted +) and "multiplication" ( denoted · ) given, respectively, by
(a, b) + ( c, d)
=
(a + c, b + d)
and
(a,b) · ( c, d)
=
(ac bd,ad + be) . -
( Secretly, we think of (a,b) and (c, d) as being a + i b and c + id. Then the addition and multiplication laws above are the obvious ones. ) Proposition 1.4 which (0, 0) is the
plic ation.
The set C equipped with these operations is a field in ide ntity for addition and ( 1 , 0) is the identity for multi
It is clear from the definitions above that (a,b) + (0,0) = (a, b) = (0, 0) + (a,b) and that (a,b)· ( 1, 0) = (a,b) = ( 1, 0) · (a,b) . Also, ( - a, -b) is an additive inverse for (a, b) . Furthermore, provided (a,b) -=/=- (0, 0), we see that (c,d) is a multiplicative inverse for (a,b) where c = a/J a2 + b2 and d = -b/J a2 + b2 . ( These are what we would expect if (a, b) is to some how be a rigorous realization of the expression a + ib) . Straightforward computations, using the definitions of + and · , show that, for any (a,b), ( c, d), (e, f) E C, Proof.
(a,b) + (c,d) = (c,d) + (a,b) , ( + is commutative) , (a,b) + (( c, d) + (e, f) ) =( (a,b) + ( c, d) ) + (e, f) , ( + is associative ) , (a,b) · (c,d) (c,d) · (a, b), ( · is commutative) , (a,b) . ( (c,d) . (e,f) ) =( (a,b) · (c,d) ) · (e, f), ( · is associative ) , (a,b) · ( ( c,d) + (e,f) ) = (a,b) · (c,d) + (a,b) · (e, f), ( · is distributive over + ) =
Thus, C is a field, as claimed.
0
13
Complex Numbers
Proposition 1.5 The map ¢ : JR. ____, C given by
set IF = {(a,O ) : a E JR.} is a subfield of C and the ¢ : a f--+ (a, 0) is a field isomorphism of JR. onto IF.
Proof.
From the definitions, we sec that (a,0)+ (b,0) (a+b,0) and that (a, 0) · (b, 0) = (ab,0), for any (a,0),(b,0) E IF. Furthermore, the additive inverse of (a,0) is ( -a,0) E IF and, if a -=/=- 0, the multiplicative inverse is (1/a,0) E IF. It follows that IF is a subfield of C. Next, we note that ¢(a+b) (a+b,0) = (a,0) + (b,0) = ¢(a) +¢(b) and ¢(ab) = (ab,O ) (a,O ) · (b,O) =¢(a) · ¢(b) , ¢(0) = (0,0) and ¢(1) (1,0) and so ¢ is a homomorphism with respect to both operations + and · . Finally, we observe that (a,0) = ¢(a) and so ¢ maps JR. onto IF, and if ¢(a) ¢(b), then (a,0) = (b, 0) and therefore a = b. Hence ¢ maps JR. one-one onto IF and is a field isomorphism. 0 =
=
=
=
=
This means that IF and JR. are "the same", that is, JR. can be embedded in C as IF. This is just the formal proof that the "real line" is still the "real line" when we consider it as the x-axis of the complex plane. This is not an entirely vacuous statement because we are also considering the additive and multiplicative structures involved . ( The plane is more naturally considered as a linear space, so that addition is natural but multiplication is a little special . In fact, it can be shown that IR.n ( with n > 1 ) can be given a multiplication making it into a field only for n = 2, in which case the multiplication is as above. ) Now, any (a , b) E C can be written as
(a, b) = (a, 0) + (0,b) = (a,0) + (0, 1) · (b,0) =¢(a) + (0,1) · ¢(b) =¢(a) + i ¢(b) =a + ib where we have dropped the isomorphism notation ¢ by writing ¢(x) as just x, for any x E R Also, we have set i = (0,1), and we have dropped the ·,denoting multiplication merely by juxtaposition,as usual. Thus,with this new streamlined notation,any complex number has the form a+ib, with 2 (0, 1) · (0, 1) = ( -1, 0) = ¢( - 1) = - 1, a , b E JR., and where i satisfies i 2 i.e., i - 1. We have therefore givell substance to the hopeful out vague idea of "a number of the form x + iy, with x, y E JR., and with i 2 =- 1" , and have recovered our original formulae for addition and multiplication. The complex numbers are well-defined-they form a field and contain the set of real numbers JR. as a subfield. =
=
14
Lecture Notes on Complex A nalysis
What's going on?
We might worry about simply asserting that
e
=
-1
without having said precisely what i was in the first place. However, it turns out
to be all right. We can just go ahead and write any complex number as a + ib, where i2
=
-1 and not worry. It can all be justified. (But we still need to sort
out square roots.)
Remark 1.5 The set of real numbers has a notion of order, defined in terms of positivity. For any real number x, precisely one of the following three statements is true; x 0, x > 0, -x > 0 . A property of positivity is that if x > 0 and y > 0, then xy > 0. It follows that 1 > 0. (To see this, first we note that, clearly, 1 =f. 0 (otherwise,x = 1 x would b e 0 for all x E IR). Hence either 1 > 0 or - 1 > 0, but not both. If - 1 > 0 were true, then we would have (- 1 ) ( - 1 ) > 0 . But ( - 1) ( - 1 ) = 1 2 = 1 so that also 1 > 0. Both - 1 > 0 and 1 > 0 is not allowed, so we conclude that - 1 > 0 is false and therefore 1 > 0 . ) Is there such a notion for complex numbers which extends that for the real numbers? If this were possible then, for example, we would have either i > 0 or -i > 0 (since i =f. 0) . If i > 0 were true, we would have - 1 i2 > 0, which is false. Hence - i > 0 must be true. But then, again, this would ( -i)2 > 0, which is false. We must concede that there is no imply - 1 generalization of positivity extending from the real numbers to the set of complex numbers. For real numbers x and y, the inequality x > y is j ust a way of writing x - y > 0. This latter does not make sense, in general, for ,complex numbers, so it follows that inequalities, such as z > ( , do not make sense for complex numbers. =
=
=
What's going on?
Try as we might, we cannot make (useful) sense
of
inequalities between complex numbers.
1.6
The Riemann Sphere and the Extended Complex Plane
Let 82 denote the sphere { ( x, y, z) E IR3 : x2 + y2 + z 2 1} in IR3 and let N denote the point ( 0, 0, 1 ) , the "north pole" of 82. Think of C as the plane {(x, y, z): z 0 } , containing the equator of 82. Then given any point P in this plane, the straight line through P and N cuts the sphere 82 in a unique point, P', say. As P varies over the plane, the corresponding point P' varies over 82 \ {N} . This sets up a one-one correspondence between C and 82 \ {N}. =
=
15
Complex Numbers
We note that points far from the origin in C are mapped into points near the north pole (and points close to the origin are mapped into points close to the south pole of S 2 , i.e . , the point (0, 0,- 1 ) ) . Notice too that if (Pn ) is a sequence of points in C which converges to some point P in C, then the images P� of Pn converge in 82 to the image P' of P. We also see that if ( zn ) is a sequence in C such that lzn l --+ oo , then the sequence (P�) of their images converges to N in S 2 (and vice versa). The point N is called the "point at infinity" . The extended complex plane, C 00, is defined to be C together with one additional element , that is, C oo = C U {oo }, where {oo} is a singleton set with oo ¢:. C. It does not matter what oo actually is, as long as it is not already a member of the set C. For example, we could take oo to be 0, which is certainly not a complex number. (Note that a and {a} are different mathematical objects, so, in particular , 0 is not the same as { 0 }. Indeed, the objects { 0 }, { 0, { 0}} and { 0, { 0, { 0}}} are different , as different as the numbers 1 , 2, 3.) What's going on?
The issue i s how t o augment a given set to give i t just one
new element. That is, given a set A, how does one construct a new set B such that B
we
\
A is a singleton set? In the case above, A
=
C and B
seek. If it does not matter what the new element is,
the explicit construction above is just
one of many
as
=
Coo is the set
in the case here, then
possibilities.
There is then a one-one correspondence between C 00 and S 2 given by oo f-----4 N together with the correspondence between C and S 2 \ {N}, as introduced above. The extended complex plane, C00, viewed in this way is referred to as the Riemann sphere. This gives a sensible realization of "infinity" . For example, the mapping z f--> 1 / z is not a p riori defined at z = 0 E C. However, if we consider the extended complex plane , or the Riemann sphere, then in addition to the mapping z f--> 1/ z for z E C \ { 0 }, we can define 0 f--> N and N f--> 0, (or, in more suggestive notation, 0 f--> oo ( 1 /0 = oo ) and oo f--> 0 ( 1 /oo = 0) ) . This defines the map z f--> 1 /z as a mapping from Coo --+ C 00. This construction is reasonable in that if Zn --+ z, then 1/ Zn --+ 1/ z even if z or any Zn is equal to 0 or to oo . The point here is to notice that by studying C 00, rather than j ust C, we can sometimes handle singularities j ust as ordinary points-after all, one point on a sphere is much the same as any other. In real analysis, one considers the limits x --+ oo and x --+ -oo . Whilst it must be stressed at the outset that this is just shorthand symbolism, nevertheless , it does invoke a kind of image of two infinities-one positive
16
Lecture Notes on Complex A nalysis
and the other negative. In view of the picture of complex numbers as points in the plane, one might wonder if it might be worth considering some kind of collection of "complex infinities", each being somewhere off in some given direction (perhaps corresponding to some "end of the rainbow" at the "end" of the ray r (cos (}+ i sin 0) as r becomes very large) . The view of C as being wrapped around a sphere, as developed above, suggests that we can bundle all these "infinities" into j ust a single "point at infinity" , namely, the north pole. It should be stressed that whilst C is a field (so one can do arithmetic) , this is no longer true of C 00• There is no attempt to assign any meaning whatsoever to expressions such as oo + oo or 0 x oo. The operations of addition and multiplication are simply not directly applicable when oo is involved.
Chapter 2
Sequences and Series
2.1
Complex Sequences
A sequence of complex numbers is a collection of elements of C labelled by integers; for example, a1,a2, a3, .... To be more precise, a sequence of complex numbers is a map a from the natural numbers N into C. If we write an for the value a ( n) , then we recover the intuitive notion above. The important thing about a sequence is that there is a notion of "further along" , e.g. , the term a710 is "further along" the sequence than, say, al06• Of course, this property is inherited from the ordering within N; an is further along the sequence than am if and only if n > m. We denote the sequence a1,a2, . . . by (an) or (an)nEN· It is often very convenient to allow a sequence to begin with a0 rather than with a1. ( We can still express this in terms of a map from N into C by considering the sequence (bn) where b : N � C is the map b : n f---> an_1, n E N.) The notation (an);:"=o or (an)n�O might be appropriate here.
2.2
Subsequences
A subsequence of a sequence (an) is any sequence got by removing terms from the original sequence (an)· For example , a1, a3,a5,a7,.•. ( where " . . . " means "and all further terms with an odd index" ) is a subsequence of the sequence a1,a2,a3, .... More formally, we define subsequences via map pings on N as follows. Suppose that ; N � C is a given sequence in C. Let '1/J : N � N be any given map such that '1/J(n) > '1/J(m) whenever n > m (i.e . , '1/J preserves the order in N) . Then o '1/J : N � C is a sequence of complex numbers;
o'I/J maps n into ('1/J(n)) E C , for n E N. Any sequence of this form is said to be a subsequence of. 17
18
Lecture Notes on Complex A nalysis
In our example above, suppose that¢( n) =an . Let 1/J : N--+ N be the map 1/J: k f---> 2k 1 for k EN (note that 1/J preserves the order inN). Then the sequence¢ o 1/J is n f---> ¢(1/J( n)) =¢( 2n - 1) = a2n-J = a'I/J(n)> i.e. , the subsequence a1/J(J),a1/J(2),a1/J(J), . ... = a1,a3,.as,. . . . Often the image ,P(k) of k E N under 1/J is denoted by n k , in which case the subsequence¢ o 1/J is denoted by (ank ) keN (or ( ank k� o if we start with k 0 rather than with . k = 1) . -
,
=
2.3
Convergence of Sequences
We have a useful notion of distance between two complex numbers, so we can use this to define convergence in C . Definition 2.1 The sequence (an) of complex numbers converges t o ( in C if for any given e > 0 there is N E N such that · i a n -(I < e whenever n > N. ( is the limit of the sequence. We signify that (an ) converges .to ( by writing an --+ (, as n --+ oo, or limn --+00 an = (. Remarks 2.1
(1) One expects that the smaller the given e, the larger N will need to be. ( 2) This definition looks exactly the same, typographically, as that of the convergence of a seque1.1ce of real numbers. In fact, if ( is real and each an is real, then this reduces to the definition for real numbers. In other words, this is a generalization of the notion for real sequences to complex ones. (3) The value i an- ( I is the distance between an and (, and so ( an ) con verges to ( if and only if the distances between ( and the various terms an become smaller than any preassigned positive value provided we go far enough "along the sequence" . This last part is usefully paraphrased by saying that for any given e > 0, the distance between ( and an is "eventually" smaller than e. O ur first result confirms that any subsequence of a convergent sequence also converges , and to the same limit as the original sequence. Proposition 2.1 Suppose an --+ ( as of ( an ) · Then ank --+ ( ask--+ oo.
n--+
oo
and ( ank ) is a subsequence
19
Sequences and Series
Let c: > 0 be given. Since (an ) converges to (, there is some N E N such that n > N implies that ian-(I < c:. Now , (nk) is a strictly increasing sequence of integers and so there is K such that nK > N . But then nk > nK > N whenever k > K. It follows that lan k-(I < c: whenever k > K, i.e . , by definition, (a n k) converges to (. 0 Proof.
As a consequence , we can say that a sequence with two convergent subsequences , but with different limits , cannot converge (for if it did, every convergent subsequence would have the same limit, namely, the limit of the original sequence). For example , the sequence (an ) ((- 1 ) n ) does not and a2 , a4,as, are convergent subsequences converge, since alt a3, as, with limits - 1 and 1, respectively. What does it mean to say that a given sequence does not converge? The sequence (bn) does not converge if it is never eventually close to any point in C; that is , for any given ( E C, (bn) is not eventually close to (. This means that there some c:o > 0 such that it is false that (bn) is eventually within co of(. This, in turn, means that no matter what we choose for N, there is always some n > N such that Ib n - c:o. In particular, there is n1 such that lbn 1 -(I > c:o. Similarly, there is n2 > n1 such that lbn 2 -(I > Eo, and so on , giving an increasing sequence n1 < n 2 < n 3 < such that lb n k -( I > Eo for k 1 , 2, . . . . Hence, for any given ( E C, there is some c:o > 0 and some subsequence (b n k) with the property that lbn k -(I > c:o, k E N. The subsequence and c:o will, in general, depend on the chosen _point (. The next theorem tells us that the convergence of a sequence of complex numbers is equivalent to that of its real and imaginary parts , with the real and imaginary parts of the limit being the limits, respectively, of t he real and imaginary parts of the sequence. =
•
.
.
.
•
.
·
·
·
=
Theorem 2.1
The sequence of complex numbers (an ) converges to ( if and only if both Re an --+ Re( and Iman --+ Im( as n --+ oo. ,
Suppose an = an + i/3 n and ( = E + iTf and that an --+ ( as Let c: > 0 be given. Then there is some N E N such that ian - (I < c:, whenever n > N. The inequalities
Proof.
n
--+ oo.
and l iman - Im( l
=
I Im(an - () I :::; lan - (I
show that both I Re a n - Re(! < c: and ! Iman - Im(! < c: whenever n > N, that is, Rean --+ Re( and Iman --+ Im( as n --+ oo.
20
Lecture Notes on Complex A nalysis
For the converse, suppose that both Re a,.,--+ Re ( and Im a,.,--+ Im ( as n--+ oo. Let e > 0 be given . Then there is N1 E N such that I Re an - Re( l < el2 whenever n > N1 and there is N2 E N such that llman - lm(l < el2 whenever n > N2 . Let N = N1 + N2 (or max{N1 , N2}, it will work just as well) . Then n > N implies that e e I a,., - ( I ::; I Rean - Re( l + lima,.,- lm(l < 2 + 2 = e and the result follows.
0
Remark 2.2 A somewhat streamlined proof of this last result can be given as follows. We begin by noting that the convergence of (a,.,) to ( is equivalent to that of the real sequence ( I an - ( I ) to 0. This is equivalent to 2 that of (ian - (1 ) to 0, i.e. , it is equivalent to the convergence of the real sequence ( (an - e) 2 + (fJn - ry) 2 ) to 0 in R But this, in turn, is equivalent to the convergence of both (an) to e (in R) and (fJn) to 77 (in R), and the proof is complete. Theorem 2.2
(i) (ii) (iii) (iv)
Suppose that Zn --+z and (n --+ ( as n--+ oo. Then
a Zn + b (n --+a z + b ( , for any a, bE C; Zn (n --+ z ( as n--+ oo; if (n =f. 0 for all n and ( =f. 0, then 1l(n --+ 11( as n--+ oo; if (n =f. 0 for all n and ( =f. 0, then Zn l(n --+z I ( as n--+ oo .
We can prove these statements directly or we could appeal to the corresponding familiar statements for real sequences. For example, we can prove (iii) directly as follows . Let e > 0 be given. We wish to show that eventually l 1l(n - 11(1 < e. Now, l1l(n - 11(1 = I ( - (n l I l(n(l and so we must say something about both terms on the right hand side. We know that I ( - (n l is eventually small because we are told that (n--+(. We are also told that ( =f. 0 and so 1(1 =f. 0 as well. Taking �1(1 as our "e > 0", we deduce that there is some N1 E N such that I ( - (n l < � I ( I , whenever n > N1 . Hence Proof.
whenever n > N l and therefore (rearranging) � 1(1 < l(n l whenever n > i N1 . Let K = min{! 1(1 , 1( 1 I , . . . , I ( N11}· Then K > 0, since each of the N1+1 terms is positive, and l(n l:;:::: K for all n E N. Hence 1l( i (n l l ( l) ::; 1I(K I ( I ) for all n. ·
21
Sequences and Series
We are now in a position to piece things together. Let c > 0 be given . Then there is N E N such that n >
N implies that
Note that the right hand side is strictly positive. Hence
whenever n > N, as required. Next, we shall prove (ii) via the familiar real versions. To this end, let Zn Xn + iyn , Z X + iy, (n = �n + i'f/n and ( = � + iry where each Xn, Yn , �n, 'f/n, x, y, �, 'f/ belongs toR Then =
=
We are told that Zn --+ z and (n --+ ( and so Xn --+ x, Yn --+ y, �n --+ � and 'f/n --+ 'f/· By "real analysis", we conclude that Xn�n - Yn'f/n --+ x� - Y'f/ = Re z( and Xn'f/n +yn�n --+ xry+y� = Im z(. In other words, Re Zn(n --+ Re z( and lm Zn(n --+ lm z(, and therefore Zn(n --+ z (. Part (i) can be verified in a similar way (or directly) , as can (iv) . We also note that (iv) follows from (ii) and (iii) . 0 2.4
Cauchy Sequences
It makes sense to talk about Cauchy sequences in C, they are defined just as for the real case, as follows. Definition 2.2 The sequence ( zn) is said to be a Cauchy sequence in C if for any given c > 0 there is N E N such that l zn - zm l < c whenever both m, n > N . Proposition 2.2 only if both (Re Zn) Proof.
The sequence (zn) is a Cauchy sequence in C if and and (1m Zn) are Cauchy sequences in R
The proof follows from the observations that
22
Lecture Notes on Complex A nalysis
so that
l im Zn - Im Zm l The left hand inequalities in each pair imply that both (Re zn ) and (Im zn ) are Cauchy sequences if (zn ) is, whereas the right hand inequality implies that (zn ) is a Cauchy sequence whenever both (Re zn ) and (Im zn ) are. (For given c > 0 there is N1 such that IRe zn - Re zml < c/2 when ever m, n > N1. Similarly, there is N2 such that l im Zn - Im zm l < c /2 whenever m, n > N2 . Putting N = max{Nt, N2}, we find that l zn - Zm l :::; 0 I Re Zn - Re zm l + l im Zn - Im zm l < c whenever m, n > N.) The set of complex numbers C , in common with the set of real num bers IR, possesses the property that Cauchy sequences necessarily converge (the completeness property). In fact, this important property is inherited from the real number system, as we show next. (It does not hold, for example, for Q, the set of (real) rationals. ) A sequence of complex numbers (zn ) converges in C if and only if it is a Cauchy sequence.
Theorem 2.3
Proof. Suppose that Zn � z as n � oo. We shall show that (zn ) is a Cauchy sequence. Let c > 0 be given. Then there is N E N such that l zn - z l < c/2 whenever n > N. Hence, for any m, n > N, we have that
l zn - Zm l = l (zn- z) + (z- Zm) l :::; l zn- z l + l z- Zm l
and so we see that ( zn ) is a Cauchy sequence, as claimed. Conversely, suppose that (zn ) is a Cauchy sequence in C. Then both (Re zn ) and (Im Zn) are Cauchy sequences of real numbers. It is a basic property of the real number system lR that any Cauchy sequence of real numbers has a limit (IR is complete) . Hence there is some a in lR and some (3 in lR such that Re Zn � a and Im Zn � (3 as n � oo. But then Zn � a + i/3 as n � oo. 0
23
Sequences and Series
2.5
Complex Series
We can set up a theory of complex series just as is done for their real counterpart. A series is simply a limit of a sequence of partial sums. Of course, it is necessary to worry about the existence of any such limit, that is, we must pay attention to the convergence, or otherwise, of the sequence of partial sums . Definition 2 . 3 Suppose that wo, w1. w2, ... is a sequence o f complex numbers. The series E�o Wk is said to converge provided that the sequence of partial sums Sn = L::�=O Wk converges in Cas n--+ oo; otherwise, it is said to diverge. If Sn--+ S as n --+ oo, then we write L::�=O Wk for S, the limit of the series.
By considering the real and imaginary parts, we see that the series converges if and only if both series E�o Re wk and E�o Im Wk converge. If this is the case, then
E�o wk
00
=
00
00
LWk L:Re wk+ i L:Im wk . k =O k =O k =O Indeed, for any n E N, Sn ReSn +i Im Sn and Sn converges if and only both ReSn and ImSn converge . Writing Wk uk+ivk, we see that ReSn Re L::�=O Wk = L::�=O Uk and, similarly, Im Sn L::�=O Therefore Sn converges if and only if both E�o Uk and L::�=O Vk converge; in which case =
=
00
00
=
Vk.
=
00
L:wk L:uk+iLvk, k =O k =O k =O =
as
claimed.
Remark 2.3 If E�o Wk converges , then Sn = L::�=O Wk converges and so is a Cauchy sequence; that is , for any given c > 0, there is N E N such that ISn- Sml < c whenever m, n > N. In particular, if m = n - 1 and n > N + 1 (so that m > N) , then IBn -Sn-11 < c. B ut Sn -Sn-1 = Wn, and so lwnl < c whenever n > N + 1; that is Wn --+ 0 as n --+ oo. This observation tells us that if it is false that Wn--+ 0, as n--+ oo, then E�o Wk does not converge. Notice that Wn --+ 0 as n --+ oo is a necessary but not a sufficient condition for the convergence of E�o Wk. Indeed, taking wk = 1/(k + 1), k = 0, 1, 2, . . . provides a counterexample (familiar from real analysis) .
24
un
Lecture Noles
Complex A nalysis
Suppose that both the series 2:;�0 wk and 2:;�0 (k are convergent. Then, for any a, b E C, the series 2:;�0 (awk + b (k) converges and
Proposition 2 . 3
00
00
00
� ) awk + b (k) = a LO Wk + b LO (k · k= k=O k= In other words, convergent series can be added term by term, and multipli cation by any complex number can be performed termwise. Proof.
2.6
This follows from l:: �=O awk
+ b (k
=
a l:::�= O Wk + b l:::�= O (k .
0
Absolute Convergence
The following definition of absolute convergence is j ust as for real series. Definition 2.4 The series 2:;�0 Wk is said to converge absolutely if the series 2:;�0 lwk l (with real non-negative terms) converges. The series 2:;�0 Wk is said to converge conditionally if it converges but does not converge absolutely. 00
Example 2 . 1
The series L
·k
t
k = O (k +
l ) converges conditionally. (The real
and imaginary parts are convergent, by the alternating series test.) Proposition 2.4
Every absolutely convergent series is convergent.
Proof. Suppose that 2:;�0 wk is absolutely convergent. Then 2:;�0 lwk l is convergent and its partial sums Tn l:: � = O lwk I form a Cauchy sequence. The generalized triangle inequality l 2:: �= m+J wk l � 2:;�= m+ J lwk l implies that the partial sums Sn = l:: � =O Wk form a Cauchy sequence in C and 0 therefore converge. =
It turns out that absolutely convergent (power) series play a central role in complex analysis, so any indications as to whether or not a given series is absolutely convergent are welcome. The nth_root test is one such. To discuss this, we recall some terminology. For the moment we consider only real sequences. Let ( o:n ) be a sequence of real numbers which is bounded from above; that is, there is some M E lR such that O:n � M for all n. For each k E N, set /A = supn�k O:n - Since O:n � M, it follows that Ak {O:n : n � k} is a bounded set of real numbers and so f3k is well-defined. Furthermore, since =
25
Sequences and Se:ries
M is an upper bound for A k , it follows that f3k
$
M. Next, we observe that
Ak + l � Ak and so any upper bound for Ak is certainly an upper bound for Ak + I · Hence f3k is an upper bound for Ak + l and therefore f3k + l $ f3k , since f3k + l sup Ak+ l · We see that (f3k ) k e N is a decreasing (i .e. , non-increasing) sequence of real numbers . It follows that either (f3k) is bounded from below , in which case there is some {3 E R such that f3k --+ {3 as n --+ oo (in fact, {3 = inf k f3k ) or else (f3k ) is not bounded from below and so neither is (a n ) · In the first case, we call {3 the limit superior of the sequence (an ) and write =
{3
=
lim sup an n -> oo
or, alternatively,
If (f3k ) is not bounded from below, we write lim sup an = - oo. Similarly, if (an ) is not bounded from above (so that no f3k is defined) , then we indicate this by writing lim sup an = oo. Note that these last two "equalities" are no such thing, but simply convenient and suggestive shorthand pieces of notation . Examples 2 . 2
( 1 ) I f an = � then lim supn an = 0 . (2) I f an = ( - 1 ) n then lim SUPn an = 1. (3) If an = 2 n then lim supn an = oo (i.e . , (an ) is not bounded from above) . (4) If an = - n + ( - 1 ) n n then lim supn an = 0 (but notice that (an ) is not bounded from below) . Remark 2.4 The value of lim supn an is unchanged by the alteration of the values of any finite number of the an s. Indeed, suppose that (a� ) is another sequence in R such that a� = an for all n > N. Then {3k = supn > k a� supn > k an f3k for all k > N and so the sequences ({3k ) and (f3k ) either both c�nverge to the same limit or are both unbounded from below . =
2.7
=
nt h -Root Test
We can now discuss the so-called nth _Root Test for absolute convergence of a complex series. Let ( wn ) n� o be a given sequence of Theorem 2.4 (nt h _Root Test) complex numbers and suppose that lim supn E N lwn i i / n = L . Then the series
26
Lecture Notes on Complex Analysis
l:�= O Wn converges absolutely if L <
1 and diverges if L > 1 . n
For any k E N, let f3k = supn� k lwn i i/ . Suppose first that L < 1 and let r E IR be such that L < r < 1 . Since f3k --+ L, it follows that for all sufficiently large k, we have r > f3k � L . In n other words, there is K E N such that lwnl l/ < r whenever n > K. Taking n nth powers , we deduce that lwnl < r whenever n > K. It follows that the series l:�= O l w n I converges by comparison with the convergent geometric series l:�= O r n . Suppose now that L > 1 . Then f3k � L > 1 for all k. It follows from the definition of the supremum that for any given /j > 0, there is some m > k n m such that lw m l 1 / > supn� k lwnl l/ - 8 . In particular, if we take /j to be L - 1, then it follows that, for any k, there is some m > k such that Proof.
lwml l / m > f3k - /j � L - /j = L - ( L - 1 ) But then lw m l � 1 for such m and it is false that conclude that l:�=O Wn cannot be convergent.
=
1.
lwnl --+ 0 as n --+ oo.
We 0
Remark 2.5 Note that the theorem says nothing about the situation when L = 1 . This is because the information that L = 1 is simply insuffi cient to determine the convergence or otherwise of the series . For example, n if Wo 1 and Wn 1/n, for n � 1, then lwnl l/ = 1/n 1 / n --+ 1 as n --+ oo and L = 1. The series l:�= I 1/n is not convergent . However, putting Vn w� , we still have lim sup lvn i i / n 1 but now l:�=O Vn is absolutely convergent. =
=
=
2.8
=
Ratio Test
We can recover the Ratio Test
as
a corollary.
Suppose that Wn -=/=- 0, for all n = 0, 1 , . . . and that the ratios l w n+ l / lw n l --+ L , as n --+ oo . Then if L < 1 the series l l:�= O Wn converges absolutely, and if L > 1 , it diverges.
Theorem 2.5 (Ratio Test)
Proof. To simplify the discussion, we introduce a harmless adj ustment to the series l:�= O Wn. Indeed, if K -=/=- 0 and Vn = KWn , then l:�=O Wn converges absolutely (or diverges) if and only if the same is true of the series l:�= O Vn.
27
Sequences and Series
For any given c > 0, there is some N E N such that lwn+ 1 1 < L+€ lwn l whenever n > N . For given J.L, let K J.L N + 1 / l wN + t l so l vN+ 1 1 Then, for n > N + 1 , L
_
€
<
=
=
J.L N+ 1 .
_ .12:L l vn - 1 1 . l vN +2 1 l vN + l l J.L N + l I Vn I l vn - 1 1 l vn - 2 1 . . l vN + 1 1 J.L N + J � lwn - 1 1 . lw N+2 1 J.L N + 1 (**) lwn - d l wn - 2 1 . . lw N+ 1 1 . Suppose L < 1 . Then we may choose L < J.L < 1 and we may suppose that c > 0 is so small that L + c < J.L. From (*) and (** ) , we see that l vn l < J.Ln 1 n for all n > N + 1 and therefore lim supn l vn l / ::; J.L < 1 . Hence the series l:�=O Vn and consequently l:�=O Wn converges absolutely, by the nt11-root test . On the other hand, if L > 1 , then we may choose 1 < J.L < L and we may suppose that c > 0 is so small that J.L < L c . But then ( * ) and ( * * ) imply that n l vn l > J.L n for all n > N + 1 and so lim supn l vn i i / 2 J.L > 1 and we conclude that l:�= O Vn and therefore l:�= O Wn diverges. 0 _
-
As before, the series with Wn Wn 1/n 2 for n 2 1 (and w0 L = l. =
=
=
1/n for n 2 1 (and w0 1 , say) or with 1) show that nothing can be said when =
Chapter 3
Metric Space Properties of the Complex P lane
3.1
Open Discs and Interior Points
In this chapter we introduce some terminology and discuss some properties of various types of subsets of C. The ideas readily extend to the more general setting of metric spaces. The open disc with centre z0 and radius r > 0 is the set { z E c : I z - zo I < T } .
Definition 3 . 1
D ( zo , T )
=
Notice that the circumference of the disc, the set { z : lz - zo l = r } , is not included in the set D(zo , r) . Intuitively, we might think of a typical set in C as having an "edge" or boundary together with an "inside" or interior. An interior point of a given set should be completely surrounded by points of the set. The following definition captures this idea. Definition 3 . 2 A point w in a given set S is said to be an interior point of S if there is some r > 0 such that D(w, r) � S.
Thus, w is an interior point of the set S if and only if it is the centre of some open disc made up entirely of points from S itself. Examples 3 . 1
( 1) Let S = { z : Im z > 0 } . Every point w E S is an interior point; indeed, for any w E S, D(w, r) � S whenever 0 < r < Im w . (If ( E D(w, r), then Im ( = Im w + Im ( ( - w) � I m w - I C - wl > Im w - r > 0. ) ( 2) Let S = { z : Im z � 0 }. Then w is an interior point of S whenever Im w > 0, as above. If Im w = 0, then w is not an interior point of S, since D(w , r) will always contain points z with Im z < 0 no matter 29
30
Lecture Notes on Complex Analysis
what r > 0 is chosen (for example, the point w - ir /2) and so cannot be contained in S. (3) The set { z : Re z 0} contains no interior points. (It is a line, and a line cannot contain discs.) More formally, let w E S { z : Re z = 0 }. Then, for any r > 0, the disc D(w, r) contains the point w + ir /2 which does not belong to S, so there is no disc, D(w, r) , wholly contained in S . Therefore w is not an interior point of S. (4) Every point in C is an interior point of C. (For any given point w E C, any r > 0 of your choice will be such that D(w, r) � C . So w is an interior point of C.) =
=
Definition 3 . 3 A set G � C is called open if for any point w E G there is some r > 0 such that D(w, r) � G. In other words, G is open if and only if it consists entirely of interior points. Examples 3 . 2
( 1 ) C is open. (2) G = { z : Im z > 0} is open. (3) A = { z : Re z � 0} is not open. (For example, w = i E A, but it is impossible to find r > 0 such that D ( i, r) � A. Indeed, the disc D ( i, r) contains the point - r /2 + i ¢:. A.) (4) 0 is open. This is a rather peculiar situation. If 0 were not open, then there would be some point w E 0 such that , for any r > 0, the disc D(w, r) contained some point not in 0. Since there is no such point w, we conclude that 0 cannot fail to be open , i.e., it is open essentially by default. Whilst this result may appear a little bizarre, it is nonetheless very convenient , since it means that the intersection of two open sets is also open, even if they have no points in common-see below. First let us show that the so-called open discs D ( z, r) really are open. Proposition 3. 1
For any z E C and any r > 0 , the set D(z, r) is open .
Let w E D ( z, r) be given. Then lw - z l < r. Let p > 0 be such that 0 < p < r - l w - zl (note that the right hand side is positive) . Then lw - zl < r - p. We claim that the disc D(w, p) is contained in D(z, r) . Proof.
Metric Space Properties of the Complex Plane
·· · ·
··· · · · ·· · · ··· ··· ··
31
// ��:-:")\ ··
···
·· ·· ·
i
[
\�
·· ··
·
i
·.
.... .
. .... . . ..
· ·· . ...
r .-/" z
·· .
...
.
.. ..
Fig. 3 . 1
.. ..
..
..
....
... .
· ··· · · · · · ···· · · ···
· ·· · .
. . ..
...
.
...
..
...
..
..........
The set D ( z , r) i s open.
( E D( w, p) . Then we have I ( - z l = I ( ( - w ) + (w - z ) l � I ( - w l + lw - z l , by the triangle inequality, < p + ( r - p)
To see this, let
=r, as required. Since w E D( z , r) i s arbitrary, w e conclude that D( z , r) is open. D
Remark 3 . 1 This proposition justifies the terminology "open disc" . The set { z : l z - w l � r } consists of the "inside" (interior) together with · the circumference (boundary) of the disc, and is called the closed disc, denoted by D( w , r) . We shall discuss closed sets shortly, as well as this "overbar" notation. Theorem 3. 1
(i) Let { Ga } aE/ be an arbitrary collection of open sets in C, indexed by
the set I. Then the union U aE/ Ga is an open set. G 1 , , Gm be any finite collection of open sets in C. Then their intersection G 1 n G 2 n n Gm is an open set.
( ii) Let
.
.
.
·
·
·
Let { Ga} be given. If they are all empty, then so is their union and we know that 0 is open. Otherwise, let ( E U a Ga . Then there is some ao E I such that ( E Gao· By hypothesis, Ga0 is open and so there is some r > 0 such that D((, r) � Gao· But then it follows that D(( , r) � U a Ga and so U a Ga is open. Proof.
32
Lecture Notes on Complex A nalysis
Suppose now that G1 , . . . , Gm are open sets in C. If n; 1 Gi 0 , then we are done, since 0 is open. Otherwise, let w E n� 1 Gj · Then w E Gj for each j = 1 , 2, . . . , m . By hypothesis, each Gj is open and so there is some 1 , 2, . . . , m. Let r = min{r1 , r2 , . . . , rm } · rj > 0 such D (w , rj ) � Gj , j 1 , 2, . . . , m . Hence , r) � D ( w , r1 ) � G1 , j Then r > 0 and clearly D ( w D( w , r) � n� 1 G1 and we conclude that the intersection G 1 n n Gm is 0 open. =
=
=
·
·
·
Remark 3 . 2 These are basic and crucially important properties of open sets. Abstractions of these are the building blocks for a topological space . Remark 3 . 3 Note that i t is false in general that a n arbitrary intersection of open sets is open. For example , for each n E N, let Gn be the open disc D (O , 1/n) { z : l z l < 1/n } . Then each Gn is open, but their intersection is n�= 1 Gn { z : z E Gn for every n } { z : I z I < 1 In all n E N } = { 0} which is not an open set (there is no r > 0 such that D(O , r) � {0}). =
=
=
Remark 3 .4 We can express convergence of sequences using discs. The sequence (zn ) converges to z in C provided that for any given e > 0 it is true that Zn is eventually within e of z, i.e. , there is some N E N such that n > N implies that l zn - z l < e. This is the same as saying that Zn E D(z , e) , whenever n > N. Thus, Zn --+ z if and only if (zn ) is eventually in any given disc D(z, e ) , e > 0 (centred on z ) . The mental image of a convergent sequence in C is one of a collection of dots (representing the points of the sequence) eventually moving inside (and staying inside) any given disc around the limit. 3.2
Closed Sets
Next, we define closed sets. Definition 3 .4 A set F in C is closed if its complement C \ F is open. In other words, closed sets are precisely the complements of open sets, and vice versa. Examples 3 . 3
( 1 ) The empty set 0 is closed , because its complement, C , is open. The
whole complex plane, C , is closed, because its complement , 0 , is open. = { z : Im z ;::: 0 } is closed. To see this, we j ust have to show that its complement is open. But if w ¢:. F, then we must have
(2) The set F
33
Metric Space Properties of the Complex Plane
Im w < 0. P utting r = � I Im wl we see that D(w, r) deduce that C \ F is open, i.e. , F is closed.
�
C \ F and we
The following is a very useful characterization of closed sets.
The non-empty set F in C is closed if and only if every sequence in the set F which converges in C has its limit in F; i. e. , F is closed if and only if whenever (zn) is a sequence in F such that Zn --+ z , for some z E C , as n --+ oo , then it is true that z E F . Proposition 3.2
Proof. Suppose that F is closed and suppose that (an) is a given sequence in F such that an --+ z, as n --+ oo. We must show that z E F. Suppose , on the contrary, that z ¢:. F. Then z belongs to the open set C \ F. Hence there is some r > 0 such that D(z , r) � C \ F. In particular, this means that D(z, r) n F = 0 . However, an --+ z means that an E D(z, r) for all sufficiently large n. This is a contradiction and we conclude that z E F, as required. For the converse, suppose that an --+ z with an E F implies that z E F. We must show that C \ F is open. If F = C , there is nothing to prove, so suppose that F =/:- C. Let z E C \ F. We claim that there is r > 0 such that D(z, r) � C \ F. If this were not true , then, no matter what choice of r > 0 we make , we will have D ( z, r) n F =/:- 0. In particular, for each n E N, we would have D(z, �) n F =/:- 0 . So suppose an E D(z, � ) n F. Then an E F and l an - z l < � ' for all n E N. It follows that an --+ z, as n --+ oo. By hypothesis, this means that z E F, a contradiction. We conclude that there is indeed some r > 0 such that D(z , r) � C \ F. Therefore C \ F is open and so F is closed. D
Remark 3.5 So the set F fails to be closed if there exists some sequence of points in F which converges to a point not belonging to F. In fact, we could therefore define a set F to be closed if (an) in F and an --+ z in C implies that z E F.
Example 3 .4 The disc D ( w, r) is not closed. To see this, we can take , for example , a sequence of points converging radially outwards towards a point on the circumference of the disc. Specifically, consider the sequence Zn = w + (1 - �)r, n E N. Then Zn E D ( w, r), for each n, but Zn --+ w + r which does not belong to D(w, r) . Remark 3 . 6 A given set A � C need be neither open nor closed. For example , the set S = { z : 0 < Re z � 1 } is not open, nor is it closed. Indeed, 1 E S, but there is no r > 0 such that D ( 1 , r) � S, so 1 is not an
34
Lecture Notes on Complex A nalysis
interior point, and therefore S is not open. On the other hand, the sequence Zn = � , n E N, belongs to S and converges to 0 ¢:. S. So S is not closed. So there are sets which are neither open nor closed , but is it possible to have sets which are both open and closed? We have seen that C and 0 have this property. It is no use trying to find further examples: we will see shortly that these are the only possibilities. 3.3
Limit Points
Closed sets can also be characterized in terms of limit points, which we now consider. Definition 3 . 5 The point w is said to be a limit point of a given subset A of C if every disc D ( w , r), r > 0, contains some point of A other than w. Limit points are also called cluster points or accumulation points. Note that the point w may or may not actually belong to A. If D ' ( w , r) denotes the punctured disc { z : 0 < l z - w l < r } , then the definition says that w is a limit point of A if D' ( w , r) n A =/:- 0 for all r > 0. The idea is that there should be points of A not equal to, but arbitrarily close to w. By setting r = n.!. for each n in N we see that w is a limit point of A if and only if there is some sequence ( an ) of points in A, all different from w, such that a n _, w, as n _, oo . Examples 3.5 ( 1 ) If G is a non-empty open set, then every point of G is a limit point of G. Indeed, if w E G, then there is some p > 0 such that D ( w, p) � G. Hence , for any r > 0 ,
D' ( w , r ) n G ;;::! D' (w , R) =/:- 0 (2)
where R = min{r, p} . The point w = i is a limit point of the unit disc D(O, 1 ) . To see this, let Zt = i - it = (1 - t ) i for 0 < t < 1 . Then Zt E D( O , 1) and l zt - il = t. It follows that Zt =/:- i and Zt E D'(i, r) whenever 0 < t < r. Hence i is a limit point of the unit disc. In fact , every point on the circumference of the disc is a limit point of the disc (take a sequence inside the disc which converges radially outwards to the point of the circumference) . The set of limit points of the disc D( (, r) is the set { z : l z - ( I :=::; r } .
Metric Space Properties of the Complex Plane
35
(3) If A contains a finite number of points , then it has no limit points. This is clear if A = { a } , a singleton set. Otherwise, for any w E C , let p = min { lw - a l : a E A and a =/:- w } . Then p > 0 and the punctured disc D' ( w, p) contains no points of A, so w cannot be a limit point of A.
Proposition 3 . 3
limit points.
The set F is closed if and only if F contains all its
Proof. Suppose first that F is closed, i.e. , Fe , the complement of F, is open. Suppose that ( is a limit point of F. We must show that ( E F. If this were not the case , then we would have ( E Fe . But then this would mean that there was some r > 0 such that D((, r) � Fe (because Fe is open, by hypothesis) . In particular, D ( ( , r) n F = 0 . But for ( to be a limit point of F, we must have that D'(( , r) n F =/:- 0 , which is a contradiction. We conclude that ( E F, as required. Next , suppose that F contains all its limit points. We must show that Fe is open. If Fe = 0 , then we are done. So suppose that Fe =/:- 0 and let w E Fe . We wish to prove that there is some r > 0 such that D(w, r) � Fe . If this were not true, then no disc D(w, r) would be wholly contained in Fe . This is j ust the statement that D(w, r) n F =/:- 0 for every r > 0. But w ¢:. F, by hypothesis, and so we may say that D'(w, r) =/:- 0 for all r > 0. This means that w is a limit point of F and so must belong to F, if F is assumed to contain all its limit points. We have a contradiction. We conclude that , indeed , Fe is open and so F is closed. D
Remark 3. 7 The statement that a set A contains all its limit points is equivalent to the statement that whenever (a n ) is a convergent sequence in A then its limit is also in A. To see this, we first note that if w is a limit point of A, then it is the limit of some sequence (an ) in A and so the first statement above follows from the second. Suppose now that (an ) is a sequence in A and that an --+ w. If w = ak for some k, then w E A (because ak E A) . On the other hand, if an =/:- w for all n, then this means that w is a limit point of A . Hence the second statement follows from the first. We therefore have three equivalent statements , namely: • • •
the complement of F is open; F contains the limit of any of its convergent sequences (that is , if ( zn ) belongs to F and Zn --+ z for z E C, then z actually also belongs to F) ; F contains all its limit points.
36
Lecture Notes on Complex Analysis
What's going on ?
The equivalence of these statements means that a closed
set in IC can be defined
( equivalently)
in several ways.
Which formulation one
chooses amounts largely to personal preference. To each definition adopted , one then has two propositions of the form "F is closed if and only if . . . " , where " . . . " denotes ei ther of the other two equivalent statements.
It turns out that
the mos t appropriate definition is the one we have given , namely that in terms of the complement being open . This generalizes to metric spaces and also to the theory of topological spaces, where the use of sequences is generally not the best approach to take .
3 .4
Closure of a Set
Definition 3 . 6
For any
A � C , the closure o f A i s the s et
A = { z : there is some sequence
(an)
in
A such that an
--+
z }.
In words, A consists of all those points of C which are the limits of some a, for sequence from A. In particular, for any a E A, we can set an all n E N. Then, trivially, an --+ a and so a E A and therefore A � A. Furthermore , if A is closed , then (an) in A and an --+ z implies that z belongs to A. Therefore A A if A is closed. =
=
Remark 3.8 Alternatively, one could define the closure of a set A to be the set A itself together with all its limit points. This follows because any point w ¢:. A is a limit point of A if and only if w is the limit of a sequence of points of A . Example 3 . 6 What is the closure, K, say, of the disc D(w, r)? We know that D(w, r) � K. Let ( E K. Then, by definition, there is some sequence (zn) in D(w, r) such that Zn --+ ( , as n --+ oo. In particular , l zn - wl < r for each n E N. But Zn --+ ( implies that Zn - w --+ ( - w and therefore lzn - wl --+ I ( - wl , as n --+ oo. Since lzn - wl < r for each n E N, it follows that I ( - wl ::::; r ( the real sequence ( lzn - w l ) lies in the interval [0, r] , so, therefore , must its limit ) . We deduce that {z
: lz - wl
<
r } � K � { z : l z - wl ::::; r } .
Now let z be any point in C with lz - w l = r. We shall show that z E K. Indeed , let (zn) b e the sequence of points "moving along the radius of D(w, r) from w towards z" given by Zn = w + (1 - � ) (z - w) . We have I Zn - w I = I (I - �) ( z - w) I = ( 1 - �) I z - w I < r, so that Zn E D ( w, r) ,
Metric Space Properties of the Complex Plane
37
for each n. Clearly Zn --+ w + ( z - w) = z as n --+ oo and we conclude that z E K , as claimed. It follows that K = { z : lz - wl :::; r } ; which justifies our notation for this set , namely, D(w , r) , as given earlier (we shall also shortly j ustify the terminology "closed disc" ) . Proposition 3 .4
For any set A , its closure A is a closed set.
Proof. If A = 0 , we are done. So suppose that ( zn ) is a sequence in A and that Zn --+ ( as n --+ oo. We must show that ( E A. By definition of A, this follows if we can show that there is some sequence ( an ) in A with an --+ (. To see this, we note that for each n, Zn E A, which means that Zn is itself the limit of some sequence of points from A. In particular, any such sequence is eventually within 1 /n of Zn , that is, there is certainly some element of A within distance 1 /n of Zn . Let an be one such element of A; an E A and lzn - an i < 1 /n. Doing this for each n yields a sequence ( an ) of elements of A. We claim that a n --+ ( , as n --+ oo. Let c > 0 be given. Then there is N0 E N such that n > N0 implies that l zn - ( I < c/2. Let N = max{N0 , 2/c} , so that, in particular , 1 /n < c/2 whenever n > N. We have
i an - ( I
i an - Zn l + l zn - ( I < � + � c , whenever n > No , :::;
< l2 c + l2 c = c whenever n > N. It follows that that A is closed. Remark 3.9 set D(w, r) .
an
--+
( giving ( E A and we conclude D
This result j ustifies the terminology "closed disc" for the
Remark 3 . 1 0 The above result implies that the set A together with its limit points is closed. This means that this "enlarged" set has no new limit points. Any limit point of A is already a limit point of A. This can be seen directly as follows. Let ( be a limit point of A and let r > 0 be given. Then there is some w E A with w =/:- ( and I ( - wl < � r . Now , w E A, and so for any given p > 0 there is some a E A with a =/:- w such that lw - a l < p. In particular, we may take p = I ( - wl so that a =/:- ( (it is closer to w than ( is) so that I ( - a l :::; I ( - w l + lw - a l < r and it follows that ( is a limit point of A.
38
LfXture Notes on Complex A nalysis
Since a set is closed if and only if it contains its limit points , we see that the above discussion constitutes, in fact, an alternative proof that the closure of any set is closed. 3.5
Boundary o f a Set
Next, we define the boundary of a set. It is usually quite clear from a diagram what the boundary of a set is, but we still need a formal definition. Definition 3. 7
The boundary of a set
A
� C is the set
aA = { z : for all r > 0 , D(z, r) n A =/:- 0 and D(z, r) n ( C \ A) =/:- 0 } . I n other words, a point z belongs to the boundary o f the set A provided every open disc around z contains both points of A and points not in A. The next result is a useful characterization of the boundary of a set. Proposition 3 . 5 For given A � C , the point z belongs to aA if and only if there is some sequence (an ) in A and some sequence (bn) in C \ A such that an --+ z and bn --+ z, as n --+ oo. Proof. First suppose that z E 8A. Then, for any n E N, the sets D(z, �) n A and D(z, �) n (C \ A) are non-empty. Let an be any point of D(z , �) n A and let bn be any point of D(z, � ) n ( C \ A) . In this way, we have constructed sequences (an) in A and (bn) in C \ A. Furthermore , lz - an i < � since an E D(z, �) , and so an --+ z as n --+ oo. Similarly, l z - bn l < � , and so bn --+ z as n --+ oo. Now suppose that there are sequences (an) , (bn) such that an E A, bn ¢:_ A, for all n and an --+ z and bn --+ z, as n --+ oo . Let r > 0 be given. We must show that both D(z, r) n A and D(z , r) n ( C \ A) are non-empty. But an converges to z and so (an) is eventually in D(z, r) . This implies that D(z , r) n A =/:- 0 . Similarly, bn --+ z and therefore (bn) is eventually in D D(z , r) and so D(z, r) n ( C \ A) =/:- 0 .
From this result, we deduce that
Example 3 . 7
The sets
D(w, r) and D(w, r) have the same boundary,
8D(w, r)
=
8D(w, r)
=
{ z : lz - w l
=
r }.
Metric Space Properties of the Complex Plane
39
Example 3.8 Let A be the horizontal strip A = { z : - 1 < Im z < 1 } . It is clear from a diagram that the boundary is made up o f the two lines £ 1 = { z : Im z = - 1 } and £2 = { z : Im z = 1 } . This can be seen directly from the definition, but it also follows quite easily from the proposition above. Indeed, if ( = a - i E L1 , then setting an a - (1 - � )i and bn = a - (1 + � )i, we see that an E A, bn ¢:_ A and both an and bn converge to ( . Hence ( E 8A. A similar argument shows that any point of £2 belongs to 8A. (In fact, if w = a + i E £2 , then an E A , bn ¢:_ A and both an and bn converge to w.) If I Im z l < 1, then clearly no sequence from Ac can converge to z . (This is because if Zn --+ z, then Im Zn --+ Im z . But Zn ¢:_ A implies that lim Zn I 2: 1 and the imaginary part of any limit of Zn would also have to satisfy this inequality. ) Similarly, if I Im z l > 1 , then no sequence from A can converge to z . =
The following is the counterpart of theorem 3. 1 for closed sets-but notice that now the union must be finite whereas it is the intersection which can be arbitrary. Theorem 3.2
For any finite collection F1 , . . . , Fm of closed subsets of C, the union F1 U F2 U U Fm is a closed set. (ii) Let {F,. } a EI be an arbitrary collection of closed sets in C, indexed by the set I . Then the intersection n ,.E I F" is a closed set. (i)
·
Proof.
·
·
This follows from theorem 3.1 by taking complements. Indeed, if
F1 , . . . , Fm are closed , then the sets C \ F1 , . . . , C \ Fm are open. Hence , by theorem 3. 1 ,
C \ (F1 U
·
·
·
U
Fm ) = ( C \ F1 ) n
·
·
·
n (C \ Fm)
U Fm is closed. is open and we deduce that F1 U For any collection {F,. } of subsets of C, we have ·
·
·
Now, if each F,. is closed, each (C \ F,.) is open and so, by theorem 3 . 1 , is open. Hence C \ ( n ,. F,. ) is open and so n ,. F,. is closed. D
U,. C \ F,.
40
3.6
Lecture Notes on Complex A nalysis
Cantor's Theorem
Example 3.9 For n E N, let closed, but Un Fn = { z : Re z >
Fn
0}
=
{ z : Re z � � } . Then each which is not closed.
Fn is
The next theorem sheds some light onto questions such as "what do you end up with if you take a set and keep cutting it in half ( and throwing away the half that has been cut off) ?" It will make a crucial appearance in the middle of arguably the most important theorem in the subject, namely, Cauchy's Theorem. Definition 3.8 A sequence ( An ) n EN of sets is said to be nested if it is true that An + l � An , for all n E N.
Fig. 3 . 2
A
nested sequence
of
sets in C .
Definition 3.9 A non-empty set A � C is said to be bounded if there is some M > 0 such that l z l < M for all z E A . If A is bounded, its diameter is defined to be the number diam A = sup{ lz - (I : z, ( E A } . So the set A is bounded if and only if it is contained inside some disc, A � D(O, M) . Notice also that if l z l < M for all z E A, then it is true that l z - (I � l z l + 1 ( 1 < 2M for all z, ( E A and so diam A is well-defined ( and is not greater than 2M) .
Suppose that (Fn) is a nested sequence of (non-empty) bounded, closed sets such that diam Fn --+ 0 as n --+ oo. Then n:=l Fn consists of exactly one point.
Theorem 3 . 3 (Cantor's Theorem)
Metric Space Properties of the Complex Plane
41
Proof.
First we shall show that n n Fn can contain at most one point . To see this, suppose that Zl , Z2 E n n Fn . Then Zl , Z2 E Fn , for every n E N , and l z1 - z2 l � diam Fn . Now, diam Fn --+ 0 and so for any given c > 0 there is N E N such that diam Fn < c, whenever n > N. In particular, for any n > N , That i s , 0 � l z 1 - z2 1 < c for any given c > 0 , which i s only possible if l z l - Z2 1 = 0. We must have Zl = Z2 which means that nn Fn can contain at most one point. The proof is complete if we can show that n n Fn is not empty, for then it must contain precisely one point. For each n E N, let Zn be some point in Fn (any point at all) . We claim that the sequence (zn) thus obtained is a Cauchy sequence . For this, let c > 0 be given. Since diam Fn --+ 0, there is N E N such that diam Fn < c whenever n > N. Let m, n > N . Then Zm , Zn E FN + l , since both Fm � FN+ l and Fn � FN+ l (because the Fns are nested) , and therefore l zm - zn l � diam FN+ l < c. Hence (zn ) is a Cauchy sequence. Now, we know that any Cauchy sequence converges, that is, there exists ( in C such that Zn --+ ( as n --+ oo. We shall complete the proof by showing that ( E n n Fn . Let k E N be given. For m = 1, 2, . . . define (m = Zk+ m · Then clearly (m --+ ( a s m --+ oo . (For given c > 0 there i s N such that I ( - zn l < c whenever n > N. In particular, for any m > N, we have that m + k > N and so I ( - (m l I ( - Zk+m l < c . ) Furthermore, (m Zk+m E Fk+m � Fk for each m and, by hypothesis, Fk is closed. Hence ( E Fk . This holds for any k E N and so we finally deduce that 0 ( E nn Fn , as required. =
=
Example 3 . 1 0 For each n E N, let Fn be the "half- plane" defined by Fn = { z : Re z � -n } . Each Fn is a closed set and evidently Fn+ l � Fn . However, n n Fn = 0. This does not contradict Cantor's Theorem because Fn is not bounded.
3.7
Compact Sets
A further very important concept is that of compactness. Definition 3 . 10 A set K � C is compact if every sequence in K has a convergent subsequence with limit in K. Thus, K is compact if and only
42
Lec ture Notes on Complex A nalysis
if whenever (zn) is a sequence in K, there is a subsequence and some z E K such that Znk --+ z as k --+ oo .
(znk )kEN ,
say,
Evidently, a set A fails t o b e compact i f i t possesses a sequence which has no convergent subsequence with limit in A. It follows that the empty set, 0 , is compact because it cannot fail to be compact.
Example 3 . 1 1 The set A = { z : 0 < l z l $ 1 } is not compact; for example, the sequence (zn = � )nEN converges to 0 and so every subsequence will also converge to 0. However, 0 ¢. A.
Compactness of sets in C is characterized by the following very useful criterion. Theorem 3 .4
bounded.
A subset K
s;;; C
is compact if and if K is both closed and
Proof. Suppose first that K is closed and bounded . Let (zn) be any given sequence in K. Since K is bounded, there is M > 0 such that l z l < M for all z E K . In particular, lzn l < M, for all n. Writing Zn = Xn + iyn , it follows that l xn l < M and also I Yn l < M , for all n, i.e. , (xn) and (Yn) are bounded sequences o f real numbers. From "real analysis" , we know that every bounded sequence of real numbers has a convergent subsequence (Bolzano-Weierstrass theorem) . Hence, there is some subsequence (xnk ) of (xn) and x E lR such that Xnk --+ x as k --+ oo . We are unable t o say anything about the convergence o f the subsequence (Ynk ) of (Yn) · However, we do know that IYnk I < M , for all k . For notational convenience, write Uk = Xnk and Vk = Ynk . Then ( Uk) converges to x, as k --+ oo. Again by the Bolzano-Weierstrass theorem, (vk) has a convergent subsequence; say, (vk; )iEN converges to y , as j --+ oo . B ut then (uk; ) is a subsequence of ( u k) and so also converges to x, as j --+ oo . It follows that Uk; + ivk; --+ x + iy as j --+ oo. Furthermore , by hypothesis, K is closed and Uk; + ivk; E K for each j. Hence x + iy E K. Since ( uk; + ivk; )iEN is a subsequence of (zn), we conclude that K is compact. For the converse, suppose that K is compact. First we shall show that K is closed. Let (zn) be any sequence in K such that Zn --+ z. We must show that z E K. By compactness, there is a convergent subsequence; Zn k --+ (, with ( E K, as k --+ oo . But Znk is a subsequence of the convergent sequence (zn) and so must have the same limit, that is, ( = z and so z E K. It follows that K is closed. To show that K is bounded, we suppose the contrary, i.e . , suppose that K is not bounded. Then for any M > 0 there is some z E K satisfying
43
Metric Space Properties of the Complex Plane
lzl � M. In particular, for each n E N , there is some Zn, say, in K such that l zn l � n . Since K is compact, (zn) has a convergent subsequence, (znk ) , oo, for some ( E K. We shall show that this say, with Znk --+ ( , as k is incompatible with the inequalities l znk I � n k . Indeed , convergence to ( implies that certainly \ ( - Znk I is eventually smaller than 1 . Hence, for all --+
sufficiently large k, k � nk � �
\ znk \ l (znk - () + (I lznk - (I + 1 (1 =
< 1 + I CI ,
which is impossible. This contradiction shows that, in fact, K is bounded 0 and the proof is complete. There is yet a further characterization of compactness. Theorem 3 . 5 Suppose that K is a compact subset of C and suppose that {Ga } aE / is some family (indexed by I) of open sets such that K � U a E / Ga . Then there is a finite set a1, . . . , am E I such that K � Ga, U U Gam ; that is, any open cover of K has a finite subcover. Conversely, if K � C has the property that every open cover of K has a finite subcover, then K is compact. · ·
·
Suppose that K is compact and let {Ga } be a collection of open sets such that K � U a Ga (i.e., {Ga } is an open cover of K) . Let us suppose that there does not exist a finite collection of the Gas whose union contains K and we shall show that this leads to a contradiction. (This is a two-dimensional version of the Reine-Borel Theorem. ) B y theorem 3 .4, we know that K i s bounded. Hence there i s some d > 0 such that K is contained in the closed square 81 , centred at 0 and with side equal to d (see Fig. 3.3); Proof.
K � St
=
{ z : IRe zl � �
and llm z l �
� }.
1
By bisecting the sides of 81 , we construct 4 similar closed squares 8 � ) , 8 �2) , 8 P) , 8 �4) each of side d/2. For example , 8 P ) could be the (top right hand) closed square 8(1} 1
and
8� 2)
=
{z
:
0 -< Im z <- !!.2
and
0 <- Re z <- !!.2 } .
could be the (top left hand) closed square
8 � 2)
=
{z
:
0 � Im z � �
and
-�
�
Re z � 0 } .
44
Lecture Notes on Complex A nalysis
Fig. 3 . 3
Constr uct squares.
Notice that these squares always contain their boundaries, so that some (common) edges will overlap. Now, K � 81 and so
By hypothesis, it is impossible to cover K with any finite number of the Ga s and so the same must be true for at least one of the four sets (K n S � i) ) , i = 1 , 2 , 3 , 4 . Call this set F2 (so F2 = (K n S� i) ) for some i = 1 , 2 , 3 , 4 ) and, for notational convenience, let F1 = K. Then F2 � F1 , F2 is a closed set, is contained in a closed square of side d/2 and cannot be covered by any finite collection of the Ga s. We can now divide this square into 4 similar subsquares and repeat the construction. In this way, we construct a nested sequence F1 , F2 , F3 , of closed sets such that Fn is contained in a square of side d/2n- l (the side is halved at each step of the construction) and such that no Fn can be covered by a finite collection of the Ga s. The diameter of a square of side r is equal to r v'2, and so it is clear that diam Fn ---+ 0, as n ---+ oo . By Cantor's theorem, theorem 3 .3, it follows that n�= l Fn { (} , for some (. In particular, ( E F1 = K. The family {Ga }a E J is a cover for K and so there is at least one a, say ao , such that ( E Gao · Now, Gao is open and so there is some r > 0 such that D((, r) � Gao · But ( E Fn , for every n and diam Fn ---+ 0 and so, for all sufficiently large n (in fact, for n with d/2n - l < r ) , we have that Fn � D((, r) � Gao · This contradicts the impossibility of covering any Fn with a finite collection of Gas-we can do it with j ust one! •
•
•
=
45
Metric Space Properties of the Complex Plane
We conclude that there is some finite collection, say a 1 , . . . , am , such that K � Ga1 U · · · U Gam ' as required. For the converse, suppose that any open cover of K possesses a finite subcover. For each z E K, let Gz be the disc Gz = D(z, 1 ) . Evidently, { G z } z EK is an open cover of K and therefore , by hypothesis, has a finite subcover. That is, there is z 1 , . . . zp such that K � D(z1 , 1) U · · · U D(zp , 1) . It follows that K is bounded (in fact, K � D(O, r) , where the radius given by r = max{ l zi l : 1 $ i $ p } + 1) .
r
is
/ Fig. 3.4
K is closed.
To prove that K is closed, we will show that C \ K , the complement of the set K, is open. For this, let ( E C \ K (note that C \ K is not empty because K is bounded) . For each n E N , let Gn = { z : l z - (I > � } . Then Gn is open, and Gn � Gn+ l · Furthermore, every z =/:- ( is contained in some Gn (in fact, z is a member of every Gn for which n l z - (I > 1) and so certainly { Gn }nEI\! is an open cover of K (because ( r:f:. K) . By hypothesis, there is a finite subcover,
=
for some n 1 , . . . , nm E N. Let k max{n 1 , . . . , nm } · Then Gn; � Gk for each i 1 , . . . , m and we have K � Gk · It follows that D((, f;) � C \ K D and we conclude that C \ K is open and so the proof is complete.
=
46
Lec ture Notes on Complex Analysis
Remark 3 . 1 1 We therefore have three equivalent formulations, each one expressing the compactness of a given set K. • •
•
Any sequence in K has a subsequence convergent to some point in the set K. K is both closed and bounded. Every open cover of K has a finite subcover.
In the more general context of topological spaces, the third version is taken as the definition of compactness. It should be noted that in this more general situation ( i.e . , in a general topological space ) these three statements about the set K are no longer necessarily equivalent. ( In fact, boundedness may not even be defined . )
Let G be a non-empty proper open subset of C . Then there is a nested sequence Kn � Kn+l of compact sets in G such that G = U�= l Kn - (The sequence (Kn ) is a compact exhaustion of G.)
Proposition 3.6
Let Fn = { z : z E G and l z - wl � � for all w E Gc }. It is clear that Fn � Fn+l · We claim that Fn is closed. To show this, suppose that (zk ) is a sequence in Fn and Zk --+ ( , as k --+ oo . Then for each w E Gc , l zk - w l --+ I ( - w l . But l zk - wl � � for all k and so I( - wl � � - In particular, ( � Gc and so ( E G. But then this means that ( E Fn and so Fn is closed. Let Kn = Fn n D(O, n) . Certainly, Kn � G and Kn is bounded . Kn is also closed because it is the intersection of closed sets. Hence Kn is compact. It is also clear that Kn � Kn+l · Finally, suppose that z E G. Since G is open, there is r > 0 such that D(z , r) � G. In particular , for any w E Gc , l z - w l � r . Choose n E N such that n > 1 /r and n > l z l . Then z E Fn and also z E D(O, n) , that is, D z E Kn - It follows that G = U�= l Kn , as required. Proof.
By considering the discs D(O, 1) and D(2i, 1) or even D(O, 1) and the closed disc D(2i, 1) , for example, we see that it is quite possible for two sets A and B to be disj oint but be "almost touching" in the sense that there are points a E A and b E B as close together as we wish. This cannot happen for closed sets if one of them is compact, as we now show. ( We shall give two proofs, illustrating different aspects of compactness. )
47
Metric Space Properties of the Complex Plane
Proposition 3. 7 Suppose that K and F are disjoint subsets of C with K compact and F closed. Then there is some r > 0 such that l z - w l 2: r for e very z E K and w E F. In other words, there is some r > 0 such that the distance between any point in K and any point in F is not less than r. Proof. Suppose the contrary, that is, for any r > 0 there is some z E K and w E F with l z - w l < r. In particular, for any n E N, there is Zn E K and Wn E F such that l zn - wn l < � (taking r �). By compactness of K, the sequence (zn) has a convergent subsequence (znk h EN , say, Znk -+ (, as k -+ oo , with ( E K. We claim that the (sub )sequence ( Wnk h EN also converges to (. This is to be expected . After all, for large n, the WnS are close to the zns, so the Wnk s get dragged along with the Znk s towards (. To show this, let c > 0 be given. There is ko E N such that
=
whenever k > k0 • Choose N E N such that N > ko +
I( - Wnk l $ I( - Znk l + l znk 1 < I( - Znk l + nk
�·
We have
- Wnk l
whenever k > N (since nk ;::: k and t < 4 c , in this case) . It follows that Wnk -+ (, as k -+ oo , as claimed. However, each Wnk E F and, by hypothesis, F is closed. It follows that ( E F, that is, ( is a common point of both K and F. This contradicts the hypothesis that K and F are disjoint. We conclude that there must be r > 0 such that l z - w l � r for any z E K and any w E F.
Alternative Proof By hypothesis, K and F are disjoint. This means that
K � C \ F. Since F is a closed set, G = C \ F is open. Hence, for every z E K, there is some r z > 0 such that D (z , r z ) � G. The family { D(z, 4 rz) : z E K } of open discs is clearly an open cover of the compact set K (each point of K is the centre of one of the discs) . By compactness, this family contains a finite subcover, so it follows that there is a finite set of points z1 , . . . , Zm in K such that the discs D (z1 , 4 r z 1 ) , , D (zm , 4 r z m ) cover K. .
•
.
48
Lecture Notes on Complex Analysis
Set r = min { rz1 1
•
•
•
, r z,.. } so that r > 0 and consider the collection
{ D(z, � r) : z E K } of open discs, labelled by the points of K. We claim that for any z E K, the disc D(z, � r) belongs to one of the m discs D(zj , TzJ ) , j = 1 , . . . , m . To see this, first note that for given z E K there is 1 $ j $ m such that z E D(z3 , � rzj ) (since these m discs cover K). Let w be any point in the disc D(z, � r) . Then
which means that D(z, �r) � D(z; , TzJ ) , as claimed . Each disc D(z; , rzj ) lies in G and so we see that, for every z E K, the disc D(z, � r) also lies in G. (This is also true of their union, which contains K; K C U z E K D(z , � r) � G . ) Let w E F. Then w ¢:. G = C \ F and so, in particular, it follows that w ¢:. D(z, � r) and therefore lz - wl � � r, for D any z E K. Example 3 . 1 2 Let F1 = { z = x + iy : x � 1 and xy � 1 } and let F2 = { z = x - iy : x � 1 and xy � 1 } . (F2 is the set of complex conj ugates of points in F1 , that is, its mirror image in the real-axis. We can also think of F1 as all those points (x, y) in the plane with x � 1 and lying above the hyperbola y = 1/x. ) Evidently, F1 n F2 = 0. We claim that F1 and F2 are closed. To see this, suppose that (zn ) is a sequence in F1 such that Zn --+ w . Let Zn = Xn + iyn and w = u + i v . Then we know that Xn --+ u and that Yn --+ v and so XnYn --+ u v . But XnYn � 1 , for all n , and therefore u v � 1 . Furthermore, Xn � 1 and so u � 1 . It follows that w u + i v E F1 and we conclude that F1 is closed . Similarly, we see that F2 is closed. Now, (n = n + � i E F1 and (n E F2 and the distance between these two points is I Cn - Cn l = � ' which can be made as small as we wish by choosing n sufficiently large. It follows that inf{ l z - wl : z E F1 , w E F2 } , the distance between the two closed , disjoint sets F1 and F2 , is zero.
=
49
Metric Space Properties of the Complex Plane
3.8
Polygons and Paths in C
Definition 3 . 1 1
The line segment from z0 to z1 is the set
[zo , z l ] = { z : z = zo + t(z1 - zo) ,
0 ::; t ::;
1}
It consists of all those points in the complex plane lying on the straight line between zo and z 1 .
y
/
Z1 zo + t(z, - zo)
zo X
Fig. 3 . 5
The line segment joining
zo
to
z1 .
Remark 3 . 1 2 By looking at the real and imaginary parts, we can express this in terms of ( familiar ) cartesian coordinates. Let z = x +iy E [z0 , z l ] and write zo = xo + iyo and z1 = x1 + iy1 • z can be written as z = zo + t(z1 - z0) for some 0 ::; t ::; 1 , which gives the pair of equations
x = xo + t(x1 - xo ) Y = Yo + t (y l - Yo ) · Eliminating t shows that (x, y) lies on the straight line
( Y l - Yo) (X - Xo ) . Y = Yo + ( X 1 - Xo ) Moreover, the restriction 0 ::; t ::; 1 means that x varies between x0 ( when t = 0) and x1 ( when t = 1 ) . Definition 3 . 1 2
for some
n
A polygon i n C i s a set o f the form
E N and points zo , z1 , . . . , Zn ·
50
Lecture Notes on Complex Analysis
In other words, a polygon is a finite collection of line segments placed end to end. We say that the polygon joins the points zo and Zn · Notice that Zn need not be the same as zo .
zo
Fig. 3.6
A polygon j oining
zo
to
Zn .
Definition 3 . 1 3 A path in C is a continuous function cp : [a, b] � C, where [a, b] is some closed interval in JR. We say that cp joins its initial point, cp ( a ) , to its final point, cp(b ) . The points cp(a) and cp(b) are its end points. The trace (also callec the track, or the impression) of the path cp is defined to be the set { z : z = cp(t) , t E [a , b] } and is denoted by tr cp. (Note that, by definition, a map 1/J from a subset S of IR to C is continuous at the point xo E S if for any given € > 0 there is 6 > 0 such that l x - xo l < 6, with x E S, implies that 1 1/J(x) - 1/J(xo) l < € . 1/J is said to be continuous on S if it is continuous at each point of S. This is the natural definition of continuity of a complex-valued function of a real variable.)
So a path is a continuous complex-valued function of a real variable, defined on some closed interval in IR, and its track is j ust its range in C.
cp ( b)
Fig. 3. 7
A path joining cp( a) to cp(b) .
51
Metric Space Properties of the Complex Plane
Remark 3.13 A polygon can be thought of as a path, as follows. For given points (vertices) z0 , , Zn E C, define cp : [0, n] ---+ C by cp(n) Zn and .
cp(t)
=
.
=
•
Zk + (t - k) (zk+ I - Zk )
for k � t < k + 1
for k = 0 , 1 , 2 , . . . , n - 1 . Then as t increases from k to k + 1 , cp(t) moves along the line segment [zk , Zk+ I ] from Zk towards Zk+ l · Evidently, the trace of cp is given by tr cp [zo , z1] U [z1 , z2] U · · · U [zn - 1 1 Zn] =
Remark 3 . 1 4 Let cp : [a, b] ---+ C be a given path. Since cp is continuous, it follows that tr cp is compact in C. To see this, let (zn ) be any sequence in tr cp. Then Zn cp(tn ) for some tn E [a, b] . B ut (tn ) has a convergent subsequence tnk ---+ T, say, as k ---+ oo , with T E [a , b] . The continuity of cp implies that cp(tnk ) ---+ cp(T) . Thus, Znk = cp(tnk ) ---+ cp(T) E tr cp, as k ---+ oo and so tr cp is compact. We note, in particular, that any line segment [zo , z 1 ] i s compact i n C . =
Remark 3 . 1 5 Suppose that cp : [ a , b ] ---+ C and 'ljJ : [c, d ] ---+ C are paths with cp(b) '1/J(c) , i.e. , 'ljJ starts where cp ends. We can join these together into a single path as follows. Define 1 : [a, b + d - c] ---+ C, by the formula
=
t I( ) -
{
cp(t) , '1/J(c + t - b) ,
a�t
c.
=
We see that 1 is a path starting from cp(a) and ending at '1/J (d) . If c b, then 1 is equal to cp on the subinterval [a, b] of [a , d] and equal to 'ljJ on [b, d] . In any event, we shall write 1 cp + '1/J .
=
3.9
Connectedness
Definition 3 . 1 4 A subset S � C is said to be pathwise connected if for any pair of points zo , z 1 in S there is some path, cp, say, lying entirely in S (i.e. , such that tr cp � S) and joining z0 to z1 Definition 3 . 1 5 A set A in C is said to be polygonally connected if and only if for any two points z', z" E A there is some polygon lying entirely within A which joins z' and z". Definition 3 . 1 6 A set A in C is said to be stepwise connected if and only if for any two points z', z" E A there is some polygon with line segments
52
Lecture Notes on Complex Analysis
parallel to either the real or the imaginary axes which lies entirely within A and which joins z' and z".
Fig. 3.8
Stepwise, polygonally and pathwise connected sets.
Figure 3.8 illustrates these three notions of connectedness. (Each set comprises two "blobs" (without their boundary) attached by a line.) Notice that s3 is not polygonally connected, and s2 is not stepwise connected. Remark 3 . 1 6 The notion of connectedness is supposed to convey the idea of a set being "all in one piece" . Clearly, the circle { z : l z l = 1 } is "all one piece" and so should qualify as being connected. However, evidently it is not polygonally connected-because it contains no line segments at all (it is everywhere curved!) . One could therefore wonder at the usefulness of the notion of polygonal connectedness. It turns out that this is very convenient in the consideration of open sets-which is of special importance in complex analysis. (The precise point here is that pathwise connectedness is equivalent to polygonal connectedness for open sets in C, as we will soon show.) There is a further notion of connectedness as follows.
=
Definition 3 . 1 7 A subset S of C is said to be disconnected if there are disjoint open sets A and B (so A n B 0) such that A n S -# 0 , B n S -# 0 and S � A U B. A set is said to be connected if it is not disconnected. For open sets, the situation is a little less complicated, as the following proposition shows.
Let G be an open subset of C. Then G is disconnected if and only if there are non-empty disjoint open sets G1 and G2 , say, such that G = G1 u G2 .
Proposition 3.8
53
Metric Space Properties of the Complex Plane
. . . . .B . . . . .A
Fig. 3.9
The set S (shaded) is disconnected (the open sets A and B are dotted) .
If G1 and G2 exist as stated , then by definition, G is disconnected . Conversely, suppose that G is disconnected; G � A U B, with A, B disjoint, open sets such that A n G =f:. 0 and B n G =f:. 0. Setting G1 A n G and G2 B n G gives the open sets required . D Proof.
=
=
The next result is geometrically obvious, but nevertheless, requires proof. It rests on the completeness property of the real number system (which plays a central role in the Intermediate Value Theorem) . Theorem 3.6 Suppose that A is a proper non-empty subset of C. Then any path joining a point of A to a point not in A contains a point of the boundary of A. In other words, for a path to "escape " from a set, it must cross its boundary.
Fig. 3 . 1 0
Proof.
Suppose that
The path must cross the boundary of A. cp :
[a, b] -+ C is a path joining
z0
E A to
z1
rfc A .
Lecture Notes on Complex Analysis
54
Let g : [a, b]
-+
lR be the function
g( t )
={ ,
0 cp (t) E A 1 , cp ( t ) � A .
The function g maps [a, b] onto the points {0, 1 } and so, by the Intermediate Value Theorem, g cannot be continuous. We will show that g is continuous at every point s in [a, b] for which cp ( s ) � 8A. If cp did not meet 8A then g would be continuous on the whole of [a, b] , which we know to be false. To proceed , then, suppose that s E [a, b] and that cp ( s ) � 8A. Suppose first that cp ( s ) E A , so that g ( s ) = 0. Since cp ( s ) � 8A, there is r > 0 such that D ( cp ( s ) , r) s;;; A. By continuity of cp, there is 6 > 0 such that
It - s l < 6, that is, g ( t )
=
0
=
t E [a, b]
�
cp ( t ) E D ( cp ( s ) , r ) s;;; A ,
g( s ) . Hence, for any given £ > 0, l g (t) - g( s ) l
=
IO - O l
=0<£
whenever It - s l < 6 , t E [a, b] . In other words , g is continuous at s. Now suppose that cp ( s ) � A. Then g ( s ) 1 . Arguing j ust as before, we conclude that there is r' > 0 such that D ( cp ( s ) , r' ) s;;; C \ A and that there is some 6 ' > 0 such that =
cp ( t ) E D ( cp ( s ) , r' ) s;;; C \ A
=
whenever It - s l < 6' , t E [a, b] . In other words, g( t ) 1 whenever we have I t - s l < 6 ' , t E [a, b] , which again shows that g is continuous at s . We have shown that g i s continuous at every point s i n [a, b] such that cp (s) � 8A. As we have already noted , the Intermediate Value Theorem implies that g cannot be continuous on the whole of the interval [a, b] and so we conclude that there must be some s for which cp ( s ) E 8A. In other words, tr cp n 8A =f. 0 and the proof is complete. D We can now prove the equivalence, for open sets, of the various notions of connectedness. Theorem 3 . 7 (i) (ii) (iii) (iv)
Let G be an open set. The following are equivalent.
G is stepwise connected. G is polygonally connected. G is pathwise connected. G is connected.
55
Metric Space Properties of the Complex Plane
If G = 0 there is nothing to prove, since 0 is "connected" in each of the four senses anyway. Suppose that G =F 0. Clearly, (i) ==} (ii) ==} (iii) . We first show that (iii) ==} (iv) . Suppose then, that G is a non-empty pathwise connected open set in 0. In particular, D (z* , r) n A = 0 , which contradicts z* E aA. We deduce that G is connected , as required. Notice that by replacing "path" by "step-path" or by "polygonal path" , respectively, it follows that (iv) is a direct consequence of any one of the properties (i) , (ii) or (iii) . To show that (iv) ==} (i) , and hence complete the proof, suppose that G is open and connected. Let z0 and z1 belong to G and set Proof.
A = { ( E G : ( is stepwise connected to B = G \ A.
z0
in G }
We shall show that A is open and non-empty. Since G is open, there is r > 0 such that D ( zo , r) � G. Evidently, D ( zo , r ) � A, so A =F 0 . Let w E A. Then w E G and s o D ( w , p) � G for some p > 0. Now, any point in D ( w , p) is stepwise connected to z0 , via w , and so D ( w , p) � A and it follows that A is open. Next we show that B is also open (in fact, we will see that B is empty) . I f z ' E B, then z ' E G and so D ( z ' , r' ) � G , for some r' > 0. Any point w , say, i n D ( z', r' ) i s stepwise connected t o z ' . I f w were stepwise connected to z0 , then z ' would be as well (via w ) . We conclude that no point of D ( z ' , r' ) can be stepwise connected to z0 , i.e. , D ( z ' , r' ) � B, showing that B is open. But G = A U B and G is connected, by hypothesis. It follows that B 0 . D Hence G = A and the proof i s complete. =
going on? The equivalence of these various notions of a set being all in one piece is only claimed to hold for open sets in C. They are not the same in general. However, our main concern will be with open sets, so we can appeal to whichever version we find appropriate at any particular time, to suit our own convenience.
What's
56
Lecture Notes on Complex Analysis
We can now easily answer the question "which sets are both open and closed " ? Theorem 3.8 The only subsets of
3.10
Domains
Definition 3 . 1 8 An open connected set is called a domain (also some times known as a region) .
Fig. 3. 1 1
Examples of a star-domain and a convex domain.
Definition 3 . 1 9 A domain D is (star-like or) a star-domain if there is some zo E D such that for each z E D the line segment [zo , z] lies in D. Any such point z0 is called a star-centre.
57
Metric Space Properties of the Complex Plane
A domain D is convex if for any pair of points z , ( E D , the line segment [z , (] lies in D. Evidently, if D is convex, then it is star-like and each of its points is a star-centre. The converse is false, in general. Example 3 . 1 3 The set D { z : z + l z l =f. 0 } is star-like, but not convex. In fact, D is the whole complex plane with the negative real-axis (including the point 0) removed. (The complement of D is the set { z : z = - l z l } , which is the set of complex numbers which are real and negative (or 0 ) . ) D i s not convex because the line segment joining, say, the point ( - 1 - i) - 1 which does to the point ( - 1 + i) crosses the negative real-axis at z not belong to D. On the other hand , we see that, for example, z0 = 1 is a star-centre. (For any z E D , the line segment [1 , z] lies entirely in D.) =
=
Example 3 . 1 4 Let £ 1 and £2 be the semi-infinite line segments given by £1 {z : z r, r :::: 1 } and £2 { z : z ir, r :::: 1 } and let D = C \ (£1 U £ 2 ) . Evidently D is a star-domain with star-centre zo 0. Moreover, the point z0 = 0 is the only star-centre for D. =
=
=
=
=
Example 3 . 1 5 For 0 < r < R, the ring (annulus) A = { z : r < lzl < R } is pathwise connected . To see this, let z1 and z2 be any pair of points in l z2 l · Then z1 can be connected the annulus A. Let p 1 = l z1 l and P2 to w 1 p 1 (on the positive real axis) in A by the path given by cp 1 (t) = p 1 (cos( ( 1 - t) Arg z l ) + i sin( ( 1 - t) Arg z l ) ) , 0 � t � 1 . The point w 1 can be joined to the point w2 p2 (also on the positive real axis) in A by the path cp2 (t) = w 1 + t(w2 - wl ) , 0 � t � 1 . Finally, the point w2 can be joined to z2 in A by means of the path cp3 (t) P2 (cos(t Arg z2 ) + i sin(t Arg z2 ) ) , 0 � t � 1 . Combining these three paths, we get a path which lies in the annulus A and joins z1 to z2 , as required . =
=
=
=
Example 3 . 1 6 Suppose D1 and D2 are domains and that D1 n D2 =f. 0 . Then D1 U D 2 is also a domain. Clearly D1 UD2 i s open. To see that D1 UD2 is pathwise connected, let z1 and z2 be any two points in D1 U D2 and let w be any point in D1 n D2 . By hypothesis, there is a path cp 1 : [0, 1] --+ C joining z1 to w in D1 and a path 'P2 : [0, 1 ] --+ C joining w to z2 in D2. For 0 � t � 1 , set
cp(t)
=
{'Pl(2t),
0�t� � 'P2 (2t - 1 ) , � < t � l .
58
Lecture Notes on Complex Analysis
Then
=
Notice that, in general, the union D1 U D2 need not be star-like even if both D1 and D2 are. For example, let Dt C \ £1 and D2 C \ £2 where £1 = { z : z x + iy, x � 0, y 0 } is the non-negative real-axis and £2 = { z : z 0 } is the non-positive real-axis. Then x + iy, x :S 0, y D1 and D2 are both star-like but D1 U D2 C \ { 0 } , the punctured plane, which is not star-like. =
=
=
=
=
=
It is natural to ask whether D1 n D2 is a domain if both D1 and D2 are. This need not be the case. For example, let D1 be the ring (annulus) given by D1 { z : 10 < l z l < 1 1 } and let D2 be the vertical strip given by D2 { z : - 1 < Re z < 1 } . Evidently both Dt and D2 are domains but D1 n D2 consists of two disconnected parts. =
=
C hapter 4
Analyt ic Funct ions
4. 1
Complex-Valued Functions
A complex-valued function of a complex variable is a mapping /, say, from (a subset of) C into C. The mapping f is real-valued if its range lies in lR (considered as a subset of C ) . Note that f need not be defined on the whole of C; for example, we may wish to consider the function z � f ( z ) = � which is undefined at z = 0 . We will be concerned almost exclusively with functions defined on domains in C, but as we have already seen, we will also need to consider complex-valued functions of a real variable, such as paths. 4.2
Continuous Functions
The definitions of continuity and of differentiability are straightforward extensions of the corresponding real versions. Definition 4 . 1 A function f : A -+ C is continuous at zo E A if and only if for any given c > 0 there is J > 0 such that l z - zo l < J and z E A imply that l f( z ) - f( zo) l < c . In words, f is continuous at zo if and only if f ( z ) is as close to f (zo) as we wish, provided z is sufficiently close to z0 . Remark 4 . 1 The set A i s assumed t o b e a given subset o f C , but by assuming it to be a subset of !R, we get the definition of continuity of a complex-valued function of a real variable. If, in addition, f happens to be real-valued , then we recover precisely the definition of continuity of a real function of a real variable. (This is because the modulus on C extends that on JR.) 59
60
Lecture Notes on Complex Analysis
The following result is sometimes useful. (One might prefer to think in terms of sequences rather than c-cis. ) Proposition 4 . 1 The function f : A -+ 0 be given. Then there is ci > 0 such that l f(z) - f( zo ) l < c whenever l z - zo l < ci, z E A. However, there is N E N such that l (n - z0 1 < ci whenever n > N. It follows that l f ((n ) - f(zo ) l < c whenever n > N, i.e., (f((n ) ) converges to f ( zo ) , as n -+ oo . Conversely, suppose that f((n ) -+ f (zo) whenever (n -+ zo , with (n E A. Suppose that f is not continuous at z0 • This means that there is some co > 0 such that no matter what ci > 0 is, there is some ( E A with I( - zo l < ci such that I f (() - f (zo) l 2:: co . In particular, for each n E N, there is some element of A, (n , say, such that i (n - zo l < I jn but l f((n ) - f(zo ) l 2:: co - Evidently, we have a sequence ((n ) in A such that (n -+ zo , as n -+ oo, but such that (f((n ) ) does not converge to f(zo ) . This contradicts our hypothesis, and so we conclude that f is continuous at z0 , and the proof is complete. 0
Some basic facts about continuous functions can be readily established . Proposition 4.2 (i) Suppose that f : A -+
61
Analytic Functions
Furthermore, by proposition 4. 1 , f ((n ) -+ f(zo ) implies that h (f ((n ) ) -+ h( f (z0) ) , as n -+ oo. Again, by proposition 4. 1 , we see that the composition D h o f is continuous at z0 , which proves (iii) . Of course, these results could have been proved directly from the c-6 definition of continuity. We will need the following result later. Proposition 4.3 Suppose K is a compact set in C and f : K -+ C zs continuous. Then the image f ( K ) is compact. In particular, if"( is any path and if f is continuous on tr "(, then there is M > 0 such that I f ( w ) I � M for all w E tr "( . Let ( (n ) b e any sequence in f( K ) . Then for each n , there is Zn E K with f ( zn ) = (n · Since K is compact, there is a convergent subsequence Zn k -+ z E K as k -+ oo. By continuity of f , it follows that f (zn k ) -+ f( z ) as k -+ oo , that is, (n k -+ ( = f( z ) E f( K ) and so f(K) is compact. To prove last part, we note that the trace tr "( of any path "( is compact (see remark 3 . 14) . Hence the set f (tr "() is compact and so, in particular, D it is bounded and the proof is complete. Proof.
4.3
Complex Differentiable Functions
Definition 4.2 Let D be an open set and suppose f : D -+ C. f is differentiable at z0 E D if and only if there is (0 E C such that for any given c > 0 there is some 6 > 0 such that
I
f(z) - f (zo ) z - zo
_
l
(o < c
(4. 1 )
for all z E D satisfying 0 < l z - zo l < 6. The complex number (o is the derivative of f at the point z0 E D and is written f' (zo ) (0 . In other words, f is differentiable at z0 if and only if there is some (0 E C such that (f( z ) - f ( zo)) / ( z - zo ) -+ (o as z -+ zo with z =/: zo . =
Note that z = z0 is not allowed because of the term z - z0 which appears in the denominator. We also observe that since D is open, there is some r > 0 such that the disc D ( z0 , r ) is contained in D and so the left hand side in Eq. (4. 1 ) is well-defined for all 0 < l z - zo l < r. Remark 4.2 This definition of differentiability of a complex function of a complex variable is the straightforward extension of that for a real
Lecture No tes on Complex Analysis
62
function of a real variable. On the basis of this, one might expect that the theory of complex differentiable functions is much the same as that of real differentiable functions. This is far from true. The theory of complex differentiable functions is a much "tighter" theory. A hint of this can be glimpsed directly from the definition. In order for the function f to be differentiable at z0, the limit (f( z ) - f(zo ) ) / (z - z0 ) must exist, as z -+ z0 , no matter how z approaches the complex number z0 . The point is that z can approach z0 in many ways (for example, from various directions, in a spiral, or quite haphazardly) . Nonetheless, the quotients (f( z ) - f (zo)) / ( z - z0) are required to always have a limit, and this limit is always to be the same, namely (0 . In the case of a real variable, the real number x can only approach the real number x0 from the left or from the right (or a combination of both) . So, in a sense, it seems that, in the real case, it might be easier for differentiability to be realized. Put another way, we might expect that in order for a complex function of a complex variable to be differentiable, it ought to be better-behaved than its real counterpart. This is, indeed, the case, and we will see as the theory unfolds, that complex differentiability has far reaching consequences. The next result is no surprise. Proposition 4.4 Suppose f is differentiable at the point zo E C . Then f is continuous at zo . Proof.
that
Let c > 0 be given. Then we know that there is some 6 > 0 such
I
f (z) - f (zo) z - zo
_
I
f' (zo ) < c
whenever 0 < \ z - z0 \ < 6. It follows that
\ f(z) - f( zo ) - J ' (zo ) ( z - zo) \ <
t:
\ z - zo \
whenever 0 < \ z - z0 \ < 6. Hence, for any such z ,
\ f(z) - f(zo) \ = \ f(z) - f ( zo ) - J ' ( zo ) ( z - zo ) + f' (zo) ( z - zo) \ :::; l f(z) - f(zo) - f' ( zo ) ( z - zo) l + l f' (zo ) (z - zo ) \ < c \ z - zo \ + \ f ' (zo) \ \ z - zo \ · Let 6' = min {6, 1 /2, c/ ( 2 ( \f ' ( z0) \ + 1 ) } . Then \ f (z) - f (zo ) \ < c whenever D \ z - zo \ < 6' and the proof is complete.
63
Analytic Functions
The usual basic properties of differentiation hold , as shown in the next proposition. Proposition 4.5 (i) Suppose f : A -+ C and g : A -+ C are differentiable at zo E A. Then so are f + g and fg with derivatives
(f + g) ' (zo )
=
f' ( zo) + g ' ( zo )
and (fg) ' ( z o)
=
J ' (zo)g ( zo) + f ( zo )g ' ( zo) ,
respectively. If f (z) =/: 0, then the quotient 1 / f is differentiable at zo and (1/ f)' (zo) = - f' (z0) / f ( zo) 2 (Quotient Rule). (ii) (Chain Rule) If f : A -+ C is differentiable at z0 E A, ran f � B and g : B -+ IC is differentiable at f(zo) , then the composition g o f is differentiable at zo with derivative (g o J ) ' ( zo ) Proof.
I
=
g ' (f ( zo ) ) J ' (zo) .
For any z =/: z0 , we have
; = �� + g) ( zo ) - (f'(zo ) + g' (zo ) ) I
(f + g ) (z
f(z) z � f(z) z
=
Il
-
f (zo) zo f (zo) zo
_
_
I
g(z) - g(zo ) g' (zo) z - zo g( z ) - g(zo) g ' (zo) · J ' (zo) + z - zo
J ' (zo) +
l l
_
_
l
The right hand side is arbitrarily small provided l z - zo l is sufficiently small (and not zero) and so we deduce that f + g is differentiable at zo with derivative ( f + g)'(zo ) = f' ( zo ) + g'( zo ) . Next, consider
z= (z-"-'(z--'--)_ (z....:. f (,_ --=-f__,__ o)= o) )g__,__ .: g-'-=-z - zo
=
f (zo) (g(z) - g(zo ) (f(z) - f(zo))g(z) . + z - zo z - zo
Using the continuity of g at z0 , we see that the first term on the right hand side approaches f' (zo)g(zo ) whilst the second term approaches f (zo)g' ( zo) , as z -+ zo , with z =/: zo , as required.
64
Lecture Notes on Complex Analysis
Assuming f(z) =/: 0, we see that for z =/: z0 ,
( 1 / f) (z) - ( 1 / f) (zo ) z - zo
as
--
f' (zo) f(zo ) f ( z) --''--'--'- --'---"---'...,.--- -+ f(zo) 2 (z - zo) f(zo ) f(z)
z -+ zo (since f is continuous at z0 ) . This proves the Quotient Rule. To prove the Chain Rule, let
{
=
=
g( w ) - g(f( zo)) , w - f(zo) g' (f(zo ) ) ,
s:
10r w
r ..J.
f ( zo )
for w = f(zo) .
By hypothesis (namely, that g is differentiable at f ( zo ) ) , it follows that
g(f(z) ) - g (f (zo)) =
(f ( z) - f (zo) ) g (f(z)) - g(f(zo) ) =
( 1) Let f (z) = (, for all z E C, i.e., f is constant. Clearly f is differentiable with derivative f' ( z ) 0, z E C. ( 2) Let f (z) z , for z E C. Let zo E C. Then, for any z =/: zo , =
=
z - zo f ( z ) - f ( zo ) = z - zo z - zo and so
'
f(z ) - f(zo) _1 z - zo
,
=
=
1
0
and we conclude that f is differentiable at every zo , with f'(zo ) = 1 . ( 3 ) For any n E N, the function f ( z ) z n i s differentiable at every z E C , with derivative f' ( z ) = nz n - l . This can be proved directly using the binomial theorem, or else proved by induction, using the product rule. =
65
Analytic F'u.nctions
(4) The function f(z)
f' (z0 )
=
-
1
2.
zo
= �z is differentiable at every z =/= 0, with derivative
Indeed , for z =/= 0, zo =/= 0 and z =/= zo , 1 Iz
zo - z - 1 I zo = z zo ( z - zo ) z - zo
- 1 -+
=
z zo
1 - z5
as z -+ zo . (5) For any n E N , the function f ( z ) z- n is differentiable at any z =/= 0 with derivative f' ( z) -nlz n + l . This may be proved by induction. ( 6) For z E C , set f(z) z and let zo E C be given. Then for any complex number z x + iy =/= zo = xo + iyo , we have =
=
=
=
f(z) - f ( zo) z - zo
a - ib a + ib
z - zo z - zo
where we have put z - zo = a + ib so that a = x - x0 and b y - yo . We are interested in the behaviour of the quotient ( * ) when l z - zo l becomes small. Let z be such that Im z = Im zo . Then b 0 and we see that the quotient ( * ) is equal to 1 ( as long as z =/= zo , i.e. , a =/= 0 ) . On the other hand, if z is such that Re z Re z0 , then a 0 and the quotient ( * ) is equal to - 1 ( as long as z =/= zo , i.e. , b =/= 0 ) . In other words, whenever z lies on the line through z0 parallel to the real axis, ( * ) assumes the value 1 , whereas whenever z lies on the line through zo parallel to the imaginary axis, the value of the quotient ( * ) is - 1 . We conclude that the quotient ( * ) does not have a limit as z -+ z0 . ( We have already exhibited two "limits" , namely, ± 1 .) (7) Let f (z) = l z l 2 , for z E C, and let zo 0. Then =
=
=
=
=
f (z) - f ( zo ) z - zo
:...� ..:... ___:__;..._;..:... =
l z l2 - 0 z-0
= zzz -
=
=
z -+ 0
as z -+ z0 = 0. Hence f is differentiable at z 0 with f' ( O ) = 0. This function f is not differentiable at any other value of z . Indeed, suppose the contrary, namely, that f (z) l z l 2 is differentiable at some zo =/= 0. Then the product fg would also be differentiable at z0 , where g is the function g( z ) 1 lz . But f( z )g( z ) z, which we have seen is nowhere differentiable. This contradiction establishes our claim that f is not differentiable at any z0 =/= 0.
=
=
=
66
Lecture Notes on Complex Analysis
4.4
Cauchy-Riemann Equations
We can relate complex-valued functions of a complex variable to real-valued functions of two real variables by looking at the real and imaginary parts: Z
= X + iy f(x + iy)
= lm f(x + iy) .
(------+
(x, y) E JR 2 Re f(x + iy) + i lm f(x + iy) = u (x , y) + iv(x , y) .
=
That is, given f(z) , we define
v (x, y)
u,
v : JR2 -+ JR by u (x , y)
=
Re f(x + iy) and
Conversely, given two functions u (x , y) and v(x , y) , we can construct the function f(z) u (x, y) + iv(x, y), where z = x + iy. The complex differentiability of f implies that u and v, the real and imaginary parts of f, are not completely independent of each other, as we shall now discuss. Suppose that f is differentiable at zo = xo + iyo . Then we know that the quotient (f(z) - f(zo))/(z - zo ) -+ f ' (zo) as z -+ zo . We consider two cases, the first where z = z0 + s, with s E JR. To say that z -+ zo is to say that s -+ 0. Then =
f(z) - f(zo) z - zo
u(xo + s , Y o) - u (xo , Y o) + i(v(xo + s , Yo) - v(xo , Yo)) s -+ f ' (zo) ,
as s -+ 0 , with s =/: 0. This means that the real and imaginary parts of the left hand side separately have limits, which is to say that the partial derivatives Ux and Vx exist at (xo , Yo) and, moreover,
u x (xo , Y o) + ivx (xo , Yo) = f ' (zo ) . Next, consider z
=
z0 + it with t E JR. Once again, we know that
f(z) - f(zo) u(xo , Yo + t) - u (xo , Y o) + i (v (xo , Y o + t) - v (xo , Y o)) it z - zo -+ f ' (zo) , _
as t -+ 0 , with t =f. 0. Again, this means that the real and imaginary parts of the left hand side separately have limits. Hence the partial derivatives u y and Vy exist at (xo , Yo) and
- iu y (xo , Yo) + vy (xo , Yo) = f ' (zo) .
67
Analytic Functions
}
Equating these two expressions for f' (zo ) leads to the following relations.
u x (xo , Yo) : vy (xo , Yo) uy (xo , Yo) - - v x (xo , Yo)
Cauchy-Riemann equations
If we denote complex differentiation by Dz , then the discussion above tells us that
Dz = Ox = i Oy -
where it is understood that Dz is applied to f(z) and the partial derivatives are applied to u (x , y) + iv(x, y) . Thus
OxU + i Ox V
=
=
Dz f = J ' - i oy (u + iv) = oyv - i oyu .
Equating real and imaginary parts gives the Cauchy-Riemann equations. We have proved the following theorem.
If the function f is differentiable at zo = xo + iyo then f satisfies the Cauchy-Riemann equations at (xo , y0 ) .
Theorem 4.1
Remark 4.3 If f does not satisfy the Cauchy-Riemann equations at some point (xo, Yo) , then certainly f fails to be differentiable at zo = xo + iyo . Example 4.2 We have already seen that the function f(z) = z is nowhere differentiable. Let us consider the Cauchy-Riemann equations for this f. We have f(z) = z = x - iy and so u(x, y) x and v (x, y) = -y. Evidently both u and v possess partial derivatives everywhere in !R 2 , but Ux = 1 , U y 0, Vx 0 and Vy = - 1 . We see that Ux is never equal to Vy and so the Cauchy-Riemann equations are never valid. We can therefore conclude that f is nowhere differentiable ( as we already knew ) .
=
=
=
Remark 4.4 I s the converse o f this last theorem true, that is, if f satisfies the Cauchy-Riemann equations at some point (xo , Yo) , is it true that f is differentiable at z = xo + i yo ? The answer, in general, is no! Example 4.3 Let f be the function f(x + iy) = lxyl 1 1 2 . Then we see that u (x, y) = l xy l 1 1 2 , v (x, y) = 0, and
u x ( O , 0) = uy ( O , 0) = 0 = Vx ( O, 0) = vy ( O , 0) . u ( s , O) u (O, O) = 0. ) From this, it follows (For example, u x ( O, 0) lim8_.0 s that the Cauchy-Riemann equations hold for f at (0 , 0) .
=
-
68
Lecture Notes on Complex Analysis
However, consider s + is
as
s -+ 0. We have, for s =f. 0,
l s l_ __ s + is
f(s + is) - f (O) s + is
-+
{
l s i (s - is) l s l ( 1 - i ) 2s 2 2s � ( 1 - i ) , if s -+ 0 through positive values, - � ( 1 - i) , if s -+ 0 through negative values.
We conclude that f is not differentiable at the Cauchy-Riemann equations there.
z
= 0 even though f satisfies
Under continuity conditions on the partial derivatives, we can prove the converse. We shall use the following result from the theory of real functions of two real variables. Lemma 4. 1 Suppose that '1/J (x , y) is defined in some disc around the point (x 0 , y0) E JR 2 and has continuous partial derivatives in this disc. Then, for
all sufficiently small h and k in
lR
'l/J(x o + h, Yo + k) - '1/J (xo , Yo ) = h '1/Jx (xo , Yo) + k '1/Jy (xo , Yo) + R ,
where I R I hfh 2 + k 2 -+ 0 as h, k -+ 0 (with h 2 + k 2 =/: 0) . The idea is to find an expression for R and then use the Mean Value Theorem to obtain the required estimate. Indeed, we have
Proof.
'l/J(xo + h,yo + k) - '1/J (xo , Yo) = 'l/J(x o + h, Yo + k ) - 'l/J(xo + h, Yo) + 'l/J(xo + h , Yo) - '1/J (xo , Yo) k 'l/Jy (xo + h, Yo + B k ) + 'l/J(x o + h, Yo ) - '1/J (x o , Yo) =
for some 0 :::; () :::; 1 , by the Mean Value Theorem applied to the function y f--+ 'l/J(xo + h, y) for y between Yo and Yo + k,
where R
=
a 1 + a 2 with a1
=
k ('l/Jy (xo + h, Yo + B k ) - '1/Jy (xo , Yo ) )
and
a 2 = 'ljJ(xo + h, Yo ) - '1/J (xo , Yo ) - h '1/Jx (xo, Yo) .
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Analytic Functions
Now, if h , k -+ 0, with h 2 + k 2 =/= 0, then
I vh�� p I = I )h2 + k2 1 1 '¢y (xo + h , Yo + Ok ) - ,Py (xo , Yo) I k
______.,
-+ 0
:S: l
by continuity of '¢y · Also a2 /v'h2 + k 2
I v'h2 +2 k2 I = I v'h2 + k2 I I ,P (xo + h
a
______.,
:::; 1
-+ 0
=
0 if h
=
0, otherwise ( i.e . , h =/= 0)
h , Yo ) - ,P (xo , Yo) h -+ Wx ( x o ,Yo )
_
'¢x (xo, Yo )
I 0
as h , k -+ 0, with h =I= 0 . The result follows.
T heorem 4.2 Let f u + iv and suppose that the partial derivatives u x , u y , Vx and Vy ex ist for all (x , y ) with x + iy in some open disc around x 0 + iy0 and that these partial derivatives are continuous at (xo , Yo ) . Suppose further, that f satisfies the Cauchy-Riemann equations at (xo , Yo ) . Then f is complex differentiable at the point zo xo + i Yo · =
=
Let A u x (xo , Yo ) By lemma 4. 1 , we can write
Proof.
and
=
= vy (xo , Yo) and p,
u (xo + h, Yo + k ) - u (xo , Yo) v (xo + h , Yo + k ) - v (xo , Yo)
=
=
=
- u y (xo , Yo)
=
vx (xo , Yo) .
hA - kp, + R 1 hp, + kA + R 2
for small h , k and where R d v'h2 + k 2 and R 2 /v'h2 + k 2 tend to zero h, k -+ 0 ( not both zero ) . Hence, with ( h + ik =I= 0,
as
=
f ( zo + () - f (zo) (
=
hA - kp, + i ( hp, + kA) + R 1 + iR 2 h + ik h ( + ik) (A + ip, ) + R 1 + iR 2 h + ik R 1 + iR 2 A + ip, + h + ik ..____._., -+ 0 as h+i k -+ 0 .
It follows that f is differentiable at z0 given by f ' ( zo) = A + ip,.
=
x 0 + iy0 and that its derivative is
0
70
4.5
Lecture Notes on Complex Analysis
Analytic Functions
Definition 4.3 Let D be a domain and f : D -+ C. The function f is said to be analytic at the point zo E D if and only if there is some r > 0 such that f is differentiable at every point in the disc D (r, z0) . If f is analytic at every point o f D, we say that f i s analytic i n D . The set of functions analytic in a domain D is denoted H(D) . I f the function f : C -+ C is analytic at every point in C, then we say that f is entire.
Analytic functions are also called holomorphic functions. Note that to say that a function f is analytic at every point of an open set G is the same as saying that f is differentiable at every point in G. (This is because every point in G is the centre of a disc lying in G.) Remark 4 . 5
Examples 4.4
( 1 ) For any n E N, the function f (z) z n , z E C, is entire. The function g(z) z- n , z =/= 0, n E N is analytic in the punctured plane, C \ {0}. ( 2 ) For any fixed ( E C, the function f(z) = 1/(z - () is analytic in C \ { (}. =
=
Suppose the function f is real-valued. Then either f is not differentiable at zo or f'(zo) 0.
Proposition 4.6
=
Suppose that f is differentiable at zo. Then
Proof.
lim
z - zo
f (z) - f (zo) z - Zo
=
f ' (zo) .
Set ( s + it =I= 0 and let z = zo + (. f(zo + s) - f(zo) -+ f ' (zo) as s -+ 0. But the Taking t 0, we have s left hand side is real-valued, and so we must have that f'(zo ) E JR. f(zo + it) - f(zo) -+ f ' (zo ) . However, the left Now, take s = 0. Then it hand side is purely imaginary and so if ' (z0) E JR. This is only possible if 0 f ' (zo) 0. =
=
=
Let f E H(D) and suppose that f ' (z) Then f is constant on D.
Theorem 4 . 3 Proof.
z(t)
=
=
0 for
every z
E,
D.
Suppose first that [z ' , z " ] c D where z ' =/= z " . For 0 ::::: t :::; 1 put z ' + t(z " - z ' ) , and let cp Re f and '1/J = Im f. =
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Analytic Functions
Fix t0 E (0, 1). Then t -:/: to implies that z(t) -:/: z(to ) and
f(z(t)) - f(z(to)) t - to
( f(z(t)) - f(z(to)) ) ( z(t) - z(to) )
z(t) - z(to) ' -+ J (z(to)) (z " - z ' )
=
=
t - to 0 , by hypothesis,
as t -+ t0. Taking real and imaginary parts of the left hand side, we conclude that
,P(z(t) ) - ,P(z(to)) -+ 0 t - to as t -+ t0 . If we set a(t) =
=
f(z ' )
=
=
=
=
=
=
Now let ( and � be any pair of points in D. Since D is connected, there is a polygon with vertices ( zo , z 1 , . . . , Zn = � joining ( to { in D. By the above argument, applied to the line segments [zo , zi ] , . . . , [zn - l , zn ], one by one, we get =
/( ( )
=
f(z l )
=
· ·
·
=
f (zn - 1 )
=
!(�) .
It follows that f is constant on D. Remark 4.6
equations,
as
0
Another proof can be given using the Cauchy-Riemann follows. We have seen that
f ' (x + i y)
=
=
- i
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Lecture Notes on Complex Analysis
But f' = 0 on D, by hypothesis, and therefore 'P x r.py = '1/Jx = '1/Jy = 0 throughout D. Suppose that the line segment [xo + iy0 , x 0 +h+iy0] lies in D. Applying the Mean Value Theorem to the real-valued function x f---> r.p (x , y0 ) for x E [xo , xo + h] , we see that r.p (xo , Yo ) = r.p (x0 + h, y0 ) . In other words, r.p has the same value at each end of the (horizontal) line segment. The same argument applies to 'ljJ and similar reasoning shows that this is also true for (vertical) line segments of the form [x0 + iy0 , x0 + i ( y0 + k)J . Now let ( and � be any two points i n D. Since D is open and connected, it is stepwise connected (by theorem 3. 7) and so ( and � can be joined by a step-path. Applying the above discussion to each of the line segments making up such a path, we deduce that r.p has the same value at the start and end of the whole path (in fact, is constant throughout) . The same is true of 'ljJ and so f r.p + i'ljJ has the same value at each end , that is, =
=
! ( ()
=
! ( 0.
Corollary 4 . 1 constant on D .
Suppose that f E H(D) and f i s real-valued. Then f is
Proof. Since f E H(D) , we know that f is differentiable at every z E D. But then the fact that f is real-valued means that we must have f'(z) 0, 0 for all z E D. Hence, by the theorem, f is constant on D. =
Corollary 4. 2 Suppose that f E H(D) and that lfl is constant on D . Then f i s constant on D .
Suppose that lf(z) l 2 = a for all z E D. If a = 0, then f vanishes on D and we are done. So suppose that a =/:. 0. Then f is never zero on D and so 1 / f E H(D) . But a = I f I 2 = f f, which means that -f a/ f is analytic in D. Hence the real-valued function f + 7 belongs to H(D) and so it is const ant. Similarly, i(f - 7) belongs to H(D) and , since it is real-valued , it must also be constant. Therefore Proof.
=
1
is constant on D.
=
H u + 7) - i ( i (f
-
7) ) )
0
Example 4 . 5 Let D be a domain and let f E H(D) . Suppose that there is some straight line L in C such that f(z) E L for every z E D. Then f is constant on D. Indeed, any straight line L in C has the form L { z : z = z0 + t (, t E IR } for suitable z0 and ( =/:. 0. If f(z) E L, then (f(z) - zo ) / ( E IR for all z E D and so is constant (because it is analytic and real-valued on D) . But this means that f is also constant on D. =
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Analytic Functions
4.6
Power Series
A complex power series is a series of the form L::= o a n ( z - z0) n , with z, zo and a n E C ( it is a series of "powers" , ( z - z0 ) n ) . The absolute convergence of such a power series is, by definition, determined by the convergence of the real power series L::= o l an l lwl n , where we have set w (z - zo ) . Clearly, the power series converges ( also absolutely) when z zo . If we let S denote those points z E C for which this power series converges, then it is natural to ask what S can look like. Evidently, zo E S and so S is not empty. Suppose that ( E S with ( =/= zo . Then L::= o a n ( ( - zo ) n converges and so l an (( - zo) n l -+ 0 as n -+ oo. In particular, the collection { l a n ( ( - z0 ) n l : n E N } is bounded, i.e. , there is some M > 0 such that l a n l pn � M for all n E N, where we have set p I ( - z0 1 . For any point z in the disc D (z0 , p) , we see that =
=
=
l an ( Z - zo) n l
=
I a n i Pn
( l (z � zo ) l ) n � M rn
where r = l z - zo l fp < 1 . It follows ( by the Comparison Test) that the power series converges absolutely for all z in the open disc D(z0 , p ) . If S contains any point ( =/= z0 , then S must also contain the whole open disc D ( z0 , I ( - z0 1 ) ( and in this open disc, the power series converges absolutely) . If the set S is unbounded, then clearly, the power series will converge ( and also absolutely) for all z E C. On the other hand, if S is bounded, it could consist of just the single point z0 or it could contain other points. In this second case, there will be some R > 0 such that the power series converges absolutely for all z in the open disc D ( z0 , R) but diverges at every point z outside the closed disc D ( zo , R) . Indeed , R is given by sup { l z - zo l : z E S } . The discussion so far says nothing at all about the behaviour of the power series on the circle l z - zo l k This will depend very much on the details of the series in question and will vary from power series to power series. =
Examples 4.6
( 1 ) The power series L::= o z n converges ( absolutely) for all z with l z l < 1 . It diverges for all other values of z . In particular, it diverges o n the circle l z l 1 ( since then z n does not converge to zero) . (2) The power series L::= l z n /n 2 converges ( absolutely) for z with l z l � 1 but diverges for all other z ( since zn / n 2 does not converge to 0 when lzl > 1) . =
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Lecture Notes on Complex Analysis
(3) The power series I:: ::=:' 1 z n fn converges ( absolutely ) for all z obeying l z l < 1 . It diverges for z = 1 ( where it becomes the divergent harmonic series 1 + � + k + -! + . . . ) and so it necessarily diverges for all z with l z l > 1 . One can show that it converges ( but not absolutely ) for all z with l z l 1 except for the point z 1 . ( For fixed z =f 1 with z + z 2 + . + zk lzl 1 , the ( moduli of the ) partial sums ak k 1 + (z - z ) / ( 1 - z ) are bounded ( by 2/ ( 1 1 - z l ) ) . An application of Dirichlet's Test ( applied separately to the real and imaginary parts ) gives the stated convergence. ) =
=
=
=
·
=
·
The value of R introduced above is called the radius of convergence of the power series I:: ::=:' o a n (z - zo ) n ; where we say that R = 0 if the series converges absolutely only for z zo , i.e., only for w 0, and that the series has an infinite radius of convergence if it converges absolutely for all values of z - z0 , i.e. , for all values of z . The disc D ( z0 , R) is called the disc of convergence of the power series.
=
4. 7
=
The Derived Series
The ( complex ) derivative of the typical term a n (z - z0 ) n in the power series is na n (z - zo) n - 1 • This leads to a new power series, called the derived series. We wish to show that a power series is differentiable everywhere inside its disc of convergence, and, moreover, that its derivative is got by simply differentiating term by term. To avoid notational complications, we shall consider the case zo 0 and then apply the chain rule to recover the general situation. First, however, we must establish convergence of the derived series. =
Proposition 4.7 Suppose the power series f(z) I:: ::=:' o a n z n converges absolutely for l z l < R. Then the derived series g(z) = I::::=:' 1 n a n z n - l also converges absolutely for i z l < R. =
Let z with lzl < R be given and let r satis fy lzl < r < R. Then L:: := o I a n i rn is convergent and so certainly there is some constant K > 0 such that I an i r n � K for all n ( the terms of a convergent series actually converge to zero, so the sequence of terms is bounded ) . Hence Proof.
1 na n z n - 1 1
But if we set t
=
=
n i a n i rn - 1
1 � r-
1
�
n
� 1 � r- . 1
l z/ rl, then t < 1 and so the series 2:: ::=:' 1 n t n - 1 converges
75
Analytic Functions
(to 1/ ( 1 - t ) 2 ) . It follows, by the Comparison Test, that the derived series 0 g converges absolutely, as claimed.
Now let us tackle the question of the differentiability of f . Theorem 4.4 Suppose the power series f (z) = L::= o an z n converges ahsolutely for l z l < R. Then f is differentiable at any z with l z l < R, and its derivative, f' (z) , is given by the absolutely convergent power series I::= 1 n an z n - 1 . In other words, the derivative of a power series is its derived series (inside its disc of convergence).
We have seen that the power series g( z ) = I::= 1 n a n z n - 1 con verges absolutely. We must show that (f( w ) - f( z ) ) / ( w - z) - g(z) converges to zero, as w -+ z . To do this, we use the fact that convergent series can be manipulated termwise and so this expression can be written also as a series. Next, we· split this into two parts and estimate each one separately. Our first observation, then, is that for w =/:. z Proof.
t
w n - zn f (w) - f ( z ) an - na n Z n - 1 - g(z) = w-z w z n= 1 00
=
=
L an (w n - 1 + w n - 2 z + wn - 3 z 2 + . . .
n= 1
. . . + wz n - 2 + z n - 1
_
nz n - 1 )
¢ 1 ( w ) + ¢2 ( w )
where ¢1 (w)
N
=
L an (w n - 1 + wn - 2 z + w n - 3 z 2 + . . . + wz n - 2 + z n - 1 - nz n - 1 )
n= 1
and 00
¢2 ( w )
=
L
n= N+1
a n (w n - 1 + w n - 2 z + w n - 3 z 2 + . . .
We will say something about N shortly. Let l z l < R and
c >
0 be given
76
Lecture Notes on Complex Analysis
and let r satisfy l z l < r < R. Then, for any w with l w l < r, we have
l an ( w n - 1 + w n - 2 z + . . . + wz n - 2 + z n - l - nz n - 1 ) I _.:::: l an l ( l w n - 1 1 + l w n - 2 z l + . . . . . . + l wz n - 2 1 + l z n - 1 1 + n l z n - 1 1 ) 1
_.:::: l a n l 2 nrn - .
The series I::= 1 n I an i r n - l converges (the series for g converges absolutely for I z I = r ) and so we may choose N sufficiently large that I ¢ 2 ( w ) I , the modulus of the second term on the right hand side of ( *) (the tail) , is less than c/ 2 (it is bounded by the tail of a convergent series) . Fix N so that this is so . Next, we consider ¢ 1 ( w ) , the first term on the right hand side of ( * ) · This i s a sum o f N terms, each of which tends t o zero a s w --+ z . I t follows that there is �' > 0 such that l ¢ 1 ( w ) l < c/2, whenever lw· - z l < �'. Now we piece these two arguments together. Let � = min{�', r - l z l } . Then, if 0 < l w - z l < �. we have that 0 < l w - z l < �' and also that l w l _.:::: l w - zl + l z l < ( r - l z l ) + l z l r . Therefore =
l
2 = �(z ) - g (z ) l
f(w
=
l ¢ 1 ( w ) + ¢2 ( w ) l
_.:::: l ¢ t ( w ) l + l ¢ 2 ( w ) l
< 21 c + 21 c
=
c, 0
whenever 0 < l w - z l < �' and the proof is complete.
L::=o an ( z - zo ) n Corollary 4.3 Suppose that the power series f ( z ) converges absolutely for l z - zo l < R. Then f is differentiable at each z E D ( zo , R) with derivative f ' ( z ) = I::= 1 n a n ( z - z0 ) n - 1 . =
Let h ( w ) L::= o a n w n . By hypothesis, the power series for h ( w ) converges absolutely for all l w l < R. In particular, h is differentiable with derivative h ' ( w ) I::= 1 nan w n - 1 , for l w l < R. Let ,P ( z ) = z - zo . Then '¢ is differentiable and ,P ' ( z ) 1 for all z . By the chain rule, h o 'ljJ h ( '¢( z )) is differentiable for z with l '¢( z ) l < R and, for such z, its derivative i s given 0 by ( h o ,P) ' ( z ) h'( ,P ( z ) ) ,P '( z ) , as required . Proof.
=
=
=
=
=
L::= o a n ( z - zo ) n Corollary 4.4 Suppose that the power series f ( z ) converges absolutely for z E D ( zo , R) . Then for any k E N, f is k-times =
77
Analytic Functions
differentiable with k th _derivative given by f ( k ) (z)
00
=
L an n ( n - 1 ) . . . ( n - ( k - 1) ) ( z - zo) n - k ,
n= k
where this last series converges absolutely for z E D ( z0 , R) . In particular, f ( k ) (zo)
=
k! a k .
The proof is by induction on k . We know, by corollary 4.3, that the result is true for k 1 . Suppose it is true for k = m. Write
Proof.
=
g(z)
=
f (m) ( z )
00
=
L an n ( n - 1 ) . . . (n - (m - 1 ) ) (z - zo t - m
n= m 00
=
L am +j ( m + j ) (m + j - 1 ) . . . (j + 1 ) ( z - zo)i .
j= O
By corollary 4 . 3, g is differentiable at z E D ( z0 , R) with derivative given by the absolutely convergent power series
g' ( z )
00
=
L am +J ( m + j) ( m + j - 1 ) . . . (j + 1 )j ( z - z0 )i - 1 •
j= l
Relabelling, the result follows for k = m + 1 , and, by induction, the proof of the formula fo r f ( k ) (z) is complete. Setting z z0 completes the proof since only the first term in the series for f ( k ) (z0) survives. 0 =
What ' s going on? These results tell us that power series behave very much like polynomials--as long as we stay inside their discs of convergence. The behaviour on the boundary of these discs varies from power series to power series and can be very complicated.
4.8
Identity Theorem for Power Series
We will need the following result later on. Theorem 4 . 5 (Identity Theorem for Power Series) Suppose that n the power series f ( z ) = I:;�=O a n ( z - zo) and g( z ) = I:�=O bn ( z - zo) n both converge absolutely for all z E D ( zo , R) . Suppose, further, that there is some sequence ( (k ) in D ( zo , R) , with (k =/:. zo for all k, such that (k --+ zo as k --+ oo, and such that f ( (k ) g( (k ) for all k. Then a n bn for all n , that is, f g . =
=
=
78
Lecture Notes on Complex Analysis
a n - bn · Then Write h (z) L;�= O en (z - zo) n , where Cn h ((k ) 0 for all k. Suppose that Cm is the first non-zero coefficient of h, i.e., Cn 0 for n < m and Cm =/:. 0. In this case, we can write h as h(z ) = ( z - zo ) m ( Cm + Cm+ l ( z - zu ) + ). cp( z ) Proof.
=
=
=
=
·
· ·
0, for 0 but (k =/:. zo and so we must have that
=
=
What 's going on?
The theorem above simply amounts to the statement that if the centre of the disc of convergence of a power series is a limit point of zeros, then all the coefficients of the power series are zero. The series is identically zero. It is the zero power series.
Chapter 5
The Complex Exp onential and Trigonometric Functions
5.1
The Functions exp z , sin z and cos z
We take, as our starting point, the definitions of the exponential and the trigonometric functions as complex power series . We will see that they have the expected properties. The complex exponential function, exp z , the complex sine function, sin z , and the complex cosine function, cos z, are defined by the power series as follows:
Definition 5 . 1
. sm z
cos z
= nL=O ( -(2n1 ) n+z21n)+! l oo
oo
=
( - 1 n z 2n
L ( 2n) ) .' n= O
= z
=
z 3 z5 z7 - 3 ! + 5 ! - 7! + z2
z4
·
·
·
6
z6
1-I+4 , - -, + . . . . . 2. .
The Ratio Test shows that each of these series is absolutely convergent for all z E C, that is, they each have infinite radius of convergence. Thus, the functions exp, sin and cos are entire functions. If z is real, we recover precisely the real series expressions for these functions. One readily calculates the various derived series, and the result is that exp1 z = exp z sm I z cos z COS I Z Sl n Z •
=
•
=
79
-
80
Lecture Notes on Complex Analysis
for all z E C. Evidently, exp 0 1 , sin 0 0 and cos 0 = 1 . Furthermore, sin is an odd function and cos is an even function, that is, sin ( - z ) - sin z , and cos ( - z ) = cos z , for all z E C. If x is real, then each o f exp x, sin x and cos x is also real. The relationship between these three functions is not particularly transparent in the real context, but direct substitution shows that, for any z E C , =
=
=
exp iz
cos z + i sin z .
=
From this it follows that exp iz + exp ( -iz ) cos z =
2
5.2
. z sm
and
=
exp iz - exp ( -iz ) . i
2
Complex Hyperbolic Functions
Definition 5 . 2 The complex hyperbolic sine and cosine functions are defined as the series oo z3 z s z 2n + l z1 sinh z L z+I+' 5 , + 7, + . . . . 3. . n= D n + 1 ) oo z 2 z4 z6 z2n cosh z = 1+1+1+ 1 +... . ' n ). 4. 6 . n= O =
We see that
(2 Z::.:: (2
=
.
=
2.
-
4 (exp z - exp ( - z )) and cosh z = 4 {exp z + exp ( -z )) and, moreover, that sinh z = - i sin ( iz ) and cosh z cos ( iz ) . Thus, in the sinh z =
=
complex variable context, the properties of the hyperbolic functions can be readily obtained from those of the complex trigonometric functions, which shows that this is really the natural context for these functions. This is in sharp contrast to the real variable situation, where the behaviour of the hyperbolic functions on the one hand , and that of the real trigonometric functions on the other, are quite distinct. It is the extension from a real to a complex variable that exposes the otherwise hidden connections. 5.3
Properties of exp z
The basic properties of the complex exponential function are given in the following proposition.
81
The Complex Exponential and Trigonometric Functions
The exponential function has the following properties:
Proposition 5.1
(i) ( ii) (iii) (iv) (v)
exp O = 1 , exp(z + w ) exp z exp w , for any z, w E C , exp z =/:. 0, for all z E C, exp( -z) = 1 / exp z , for all z E C, exp(x + iy) exp x (cos y + i sin y) , for x, y E C (and, in particular, for any x, y E IR).
Proof.
set f (z)
=
=
Setting z 0 immediately gives (i) . To prove (ii) , fix w E C and exp(z + w) exp( -z) . Then we find that, for any z E C, =
=
J '(z)
=
(exp(z + w) ) exp( -z) + exp(z + w) ( exp( - z) ) 1
1
exp(z + w) exp( - z) - exp(z + w) exp( -z) = 0.
=
It follows that f
is
constant, and therefore f(z) exp w
=
=
f(O) = exp w , that is,
exp(z + w) exp( -z).
Now set w = a + b and z = -b. Then we obtain exp(a + b) = exp a exp b, for any a, b E C, as required. Using (ii) with w = -z, we find that exp z exp(-z) = exp O = 1 , by (i) , and so (iii) follows, and so does (iv) . To prove (v) , let z x + iy and then apply (ii) to obtain =
exp z
exp(x + iy) = exp x exp ( iy ) = exp x (cos y + i sin y) . =
This holds for any x, y E C and so, in particular, also for real x, y . Corollary 5 . 1
Let e be the real number given by e
Then exp n
=
=
1+1+
1
21
.
en for any n E Z.
+
1
1
3.
1
+ I + . . . = exp 1 . 4.
0
82
Lecture Notes on Complex Analysis
For each n E Z, let P( n) be the statement that exp n = en . For n E N, we shall prove the claim by induction. By definition, exp 1 = e and so P( 1 ) is true. Now let n E N and suppose that P(n) is true. Then Proof.
exp ( n + 1)
= =
=
exp n exp 1 , by the previous proposition, e n exp 1, by induction hypothesis, n e e, since exp 1 = e ,
= e
n+l
so that P(n + 1) is true. By induction, P(n) is true for all n E N. Clearly P(O) is true, since both exp O and e0 are equal to 1 ( e0 definition ) . Now let n = -m with m E N. Then exp m = e m and exp n
=
exp ( -m )
=
1
-exp m
=
1
m e
=
1 , by
= e - m = en ,
0
as required.
It is because of this relationship that one writes e z for exp z. This notation is often very convenient. Remark 5 . 1
Remark 5 . 2 Suppose f is entire and satisfies f' (z ) = f(z) for all z and f ( 0) = 1 . Then f ( z ) exp z . To see this, we consider the entire function g (z) = f(z) exp ( -z ) . We see that g '(z) = 0 for all z E C and so g is constant, g ( z ) = g ( O ) = 1. But then we find that f(z) exp z, as required. For z = x + iy, let f(z) = e x (cos y + i sin y) so that Re f(x + iy) = e x cos y and Im f(x + iy) = e x sin y . These functions have continuous partial derivatives throughout IR2 and obey the Cauchy-Riemann equations and so f(z) is analytic at every z E C. Furthermore, we know that its derivative is given by =
=
f ' (x+iy)
=
Ox Re f(x+iy)+iOx lm f (x+iy)
= e x cos y+i e x sin y
= f (x+iy) .
Since f(O) = 1 we conclude that f(z) = exp z. Sometimes this approach is used to define the complex exponential function-assuming that the real functions e t , sin () and cos () are somehow already known.
The Complex Exponential and Trigonometric Functions
Proposition 5 . 2 Proof.
that
83
The number e is irrational.
For any k E
N,
the power series expression for e = exp 1 implies
Since 1 / ( k + 1 ) ( k + 2) . . . (k + m ) < 1/2m for any m E N, m > 1 , we see that the term in brackets on the right hand side above is bounded above by L::'=l 1 /2m = 1 and so k 1
L� j=O J .
for any k E
N.
<e<
k 1
1
L: � + k '
j=O J .
.
Multiplying through by k ! and rearranging, we get
-:y) < 1
k 1
0 < k ! (e - L
j =O J
for any k E N . If e were rational, then we could write e as e = pjq for some p, q E N. Setting k in the inequality above to be any integer greater than q and then taking k! inside the bracket, we see that the middle expression is a positive integer strictly less than 1 , which is impossible. We conclude D that e is irrational. In fact, it is known that e is transcendental (that is, it is not a root of any real polynomial with integer coefficients, unlike V'i, for example) . Numerical considerations give e = 2. 71828 . . . . Remark 5 . 3
5.4
Properties of sin z and cos z
We turn now to basic properties of the complex trigonometric functions, sin and cos. These must be established from their power series definitions as given above. For real a, {3, we have that exp i (a + {3) = cos(a + {3) + i sin( a + {3) . However, exp i ( a + {3)
exp ia exp if3 (cos a + i sin a) (cos {3 + i sin {3) = cos a cos {3 - sin a sin {J + i(sin a cos f3 + cos a sin f3) . =
=
84
Lecture Notes on Complex Analysis
Equating real and imaginary parts leads to the addition rules sin(a + {3) = sin a cos {3 + cos a sin {3 and cos( a + {3) = cos a sin {J - sin a sin {J, for any a, {3 E JR. It is natural to ask whether these relationships are also valid for complex a and {3. In fact they are, as we now show. 5.5
Addition Formulae
For any a, b E C,
Theorem 5 . 1 (Addition Formulae)
sin( a + b) = sin a cos b + cos a sin b cos( a + b) = cos a cos b - sin a sin b. Proof. The easiest way of proving these relations is to write the right hand sides in terms of the exponential function and do a bit of algebra. We find
sin a cos b + cos a sin b ( eia e-ia ) ( eib + e-ib ) 2 2i ( eia + e-ia ) ( eib e-ib ) + 2 2i ( eiaeib + eia e-ib e-iaeib e-iae-ib ) 4i ( eiaeib eiae-ib + e-iaeib e-iae-ib ) + 4i ( ei(a+b) + ei(a-b) - e-i(a-b) - e-i(a+b) ) 4i ( ei(a+b) ei(a-b) + e-i(a-b) e-i(a+b) ) + 4i ( ei(a+b) _ e-i(a+b) ) 2i = sin( a + b) . _
_
_
_
_
_
_
_
85
The Complex Exponential and Trigonometric Functions
Similarly, we calculate . a s . b = ( eia + e-ia ) ( eib + e-ib ) cos a cos b + sm m 2 2 ( eia e-ia ) ( eib e-ib ) + 2i 2i ( ei(a + b) + e-i(a + b) ) _
_
2
= cos( a + b) , as
required.
D
Corollary 5 . 2
For any z E C, (sin z) 2 + (cos z) 2 = 1 .
Proof. The identity cos( a + b) = cos a cos b - sin a sin b, with a b = - z, gives
=
z and
cos O = cos z cos( - z) - sin z sin( - z) cos z cos z + sin z sin z, =
where we have used cos( - z) = cos z and sin( - z) follows since cos 0 1 .
=
- sin z. The result now
=
D
An alternative proof is to set '1/l(z) = sin2 z + cos2 z and then to calculate '1/J'(z) . One finds that '1/J' (z) = 0, for all z E C, and so it follows that '1/J is constant. Therefore '1/J(z) = '1/J(O) = 1 , as required. In the same spirit, we can prove the addition formulae by considering the functions Remark 5 . 4
and
f (z) = sin(w - z) cos z + cos(w - z) sin z g(z) = cos(w - z) cos z - sin(w - z) sin z .
One finds that f'(z) = 0 and g' (z) = 0 for all z, so that both f and g are constant: f(z) = f(O) = sin w and g(z) g(O) = cos w. Letting w = a + b and setting z = b, we get the required formulae. Corollary 5.3 For any z E C, we have =
sin 2z = 2 sin z cos z and cos 2z = cos 2 z - sin2 z = 2 cos2 z - 1 = 1 - 2 sin2 z . Proof. We simply put a = b = z in the addition formulae above and use D the identity sin 2 z + cos 2 z = 1 .
86
L&ture Notes on Complex Analysis
Corollary 5 .4
For any t E
IR,
lexp(it) l = 1 .
Proof. We have exp( it) = cos t + i sin t . Both cos t and sin t are real if t is, and therefore lexp(it ) 1 2 = cos 2 t + sin2 t = 1 . D
Remark 5 . 5 For real x, sin x and cos x are also real, and so the relation sin2 x + cos2 x = 1 shows that I sin xl :-::; 1 and also I eos xl :-::; 1 . These inequalities do not extend to the complex case. For example, if z = it, with t E IR, then sin z sin it = ( exp( - t) - exp t) /2i. If t is large (and positive), then exp t is large, exp( -t) is small and so I sin it I is large. A similar remark applies to cos z . =
5.6
The Appearance o f 1r
We wish to discuss various properties of sin x and cos x for real x. In particular, we would like to introduce the number 1r . Theorem 5 . 2 Proof.
If x E (0, 2) , then sin x > 0 .
Suppose that 0 < x :-::; 1 . Then
x 2 x4 x6 cos x = 1 - I + I - 6 ,. + . . . 4. 2. x 2 x4 xB > 1 - 2! - 4! - 6! - . . . 1 1 1 >1- -... 2 23 25 ' x4 1 1 x2 < 1 1 < < 3 , . . . etc. , since - 2 ' 4 . - 4. 3 .2 2 2! --
=1-
1
2
(1 -
�)
=
1
3 .
Hence sin' x = cos x > i on [0, 1] and so (by the Mean Value Theorem) , sin x is strictly increasing on [0, 1] . But sin O = 0 and therefore sin x > 0 for 0 < x :-::; 1 . It follows that sin x = 2 sin � cos � > 0 whenever 0 < x :-::; 2 , since, in this case, sin � > 0 and cos � > i ·
D
The Complex Exponential and Trigonometric Functions
Corollary 5 . 5
87
The function cos x is strictly decreasing on [0, 2] .
We have cos ' x = - sin x which is strictly negative on (0, 2 ) , by the theorem. By the Mean Value Theorem, it follows that the function D cos x is strictly decreasing on [0, 2] .
Proof.
Theorem 5 . 3 Proof.
sin 4 < 0.
From the definition of sin x, 43 45 47 49 sin 4 = 4 - - + - - - + 9! 5! 7! 3! 2 11 13
<4-
43
- �:� ( ; ) - ��� ( 1 - 1 :.: 7 ) - ��� ( 1 - 2;.: 1 ) - . . . 45
47
49
7! + 9! 3! + 5!
and the result follows.
=
-
�g�
D
There is a unique real number 1r satisfying 0 < 1r < 4 such that sin 1r = 0. Furthermore, cos � = 0, sin � = 1 and cos 1r = - 1 .
Theorem 5 . 4
Proof. We have seen that sin x > 0 for x E (0, 2 ) and so, in particular, sin 1 > 0. We have also shown that sin 4 < 0. Now, the map x �----> sin x is continuous on IR, and so, by the Intermediate Value Theorem, there is some real number, which we will denote by 1r, with 1 < 1r < 4 and such that sin 1r = 0. We must now show that 1r is the only value in (0, 4 ) obeying sin 1r = 0. To this end , suppose that sin a = 0 with 0 < a < 4. Then
0
=
sin a = 2 sin � cos � .
Hence either sin � = 0 or cos � = 0 ( they cannot both vanish because the sum of their squares is equal to 1 ) . But if 0 < a < 4, then 0 < � < 2 and we know that sin x > 0 on (0, 2), so sin � cannot be zero. Hence, we must have that cos � = 0 . In particular, cos � = 0 . However, cos x is strictly decreasing on [0, 2] and so there can be at most one solution to cos x = 0 in this interval. Hence � = � so that a = 1r , and the uniqueness is established . Next, sin2 z + cos2 z = 1 implies that sin 2 � = 1 , since cos � = 0. Hence sin � = 1 because sin � > 0 ( since 0 < � < 2 ) . Finally, we have cos 1r = 2 cos2 � - l = - 1 . D
88
Lecture No tes on Complex Analysis
What 's going on? blue
as
it
were, by
We have defined the trigonometric functions, out of the means of power series. This approach avoids any app eal to
geometry and right-angled triangles. However, having chosen this route, stick with it. In particu lar , it is necessary to get to
1r
we
must
via these definitions rather
than by drawing trian gles or circles . As we have seen , t h is is all perfectly possible. Our view here is that the trigonometric fu nctions are A ny results must be deduced
as
as
we have defined them.
consequences of these power series definitions .
Remark 5 .6 Numerical investigation yields 1r 3 .14159 . . . . It is known that 1r is irrational ( in fact, transcendental ) . It is something of a sport ( involving some fascinating numerical analysis ) to calculate the value of 1r to a large number of decimal places and this has been done to over a million decimal places. Such programs have been used to test the computational integrity of supercomputers by checking to see whether they get these digits correct or not . =
Theorem 5 . 5
For any z E C and n E Z , sin ( z + n1r ) cos ( z + n1r )
Proof.
=
=
( - 1 ) n sin z ( - 1 ) n cos z .
We use the trigonometric formulae; sin ( z + 1r )
= =
sin z cos 1r + cos z sin 1r - sin z + 0,
and cos ( z + 1r )
cos z cos 1r - sin z sin 1r = - cos z - 0.
=
For n > 0, the result now follows by induction. Substituting z the result then follows for n < 0.
=
w - n1r,
D
Remark 5 . 7 Particular cases o f the above formulae deserve mention. If we set z = 0, then we see that sin( n1r ) = 0 and cos ( n1r ) = ( - 1 ) n , for any n E z . The functions sin and cos are periodic ( with period 27r ) . In particular, for any x E lR and any n E Z,
sin ( x + 2n7r )
cos ( x + 2n7r )
=
=
sin x cos x.
and
The Complex Exponential and Trigonometric Functions
89
If we piece together all the information gained above, we recover the familiar picture of sin x and cos x, for x E IR, as periodic 'wavy' functions. In fact, sin x is an odd function, so it is determined by its values on x � 0. It is periodic, so it is determined by its values on [0 , 21r] . But, sin(x + 1r) sin x cos 1r + cos x sin 1r = - sin x and so sin x is determined by its values on the interval [0 , 1r] . Now, =
(
sin �
±
x
)
=
sin � cos x ± cos � sin x
=
cos x,
so we see that sin x is symmetric about x = l It follows that sin x is completely determined by its values on [0 , � ] . Furthermore, sin (x + �) = cos x and so the graph of cos x i s got by translating the graph of sin x by � to the left. 5.7
Inverse Trigonometric Functions
From the analysis above, we see that sin x is strictly increasing on the interval [- � , �] , cos x is strictly decreasing on the interval [0, 1r] and is strictly increasing on the interval [ -1r, 0] . In particular , this means that sin is a one-one map of [- � , � ] onto [- 1 , 1 ] , cos is a one-one map of [0, 1r] onto [- 1 , 1 ] and also of [-1r, 0] onto [-1 , 1 ] . For t E [- 1 , 1] , let ¢( t ) be the unique element of [- � , � ] such that sin ¢(t) t, let 'lj; (t) be the unique element of [0 , 1r] such that cos 'lj; (t) = t and let p(t) be the unique element of [-1r, OJ with cos p(t) = t. Thus, ¢ is the inverse of sin : [- � , � ] --+ [- 1 , 1 ] , 'lj; is the inverse of cos : [0, 1r] --+ [- 1 , 1] and p is the inverse of cos : [-1r, OJ --+ [ - 1 , 1 ] . The standard inverse trigonometric functions sin- 1 and cos- 1 are given by sin- 1 (t) = ¢(t) and cos- 1 (t) = '!f; (t) for t E [- 1 , 1] so that sin - 1 takes values in [- � , �] whilst cos- 1 takes values in [0 , 1r] . =
Suppose that f : [a, b] --+ [c, d] is a strictly increasing {or decreasing) continuous map from [a , b] onto [c, d] . Then the inverse map f - 1 : [c, d] --+ [a, b] is continuous. Theorem 5 .6
Proof. The idea of the proof is straightforward , but it is a nuisance having to consider the end-points c and d of [c , d] and the interior (c , d) separately. To avoid this, we shall first consider the case of f : IR --+ R Suppose, then, that f : IR --+ IR is, say, strictly increasing and maps IR onto R Let Yo E IR and £ > 0 be given. Let x0 E IR be the unique point such that f ( xo ) = Yo , thus, xo = f - 1 (Yo ) . Set Y 1 = f ( xo - £ ) and Y2 = f ( xo + c) ,
90
Lecture Notes on Complex Analysis
X Fig. 5 . 1
Continuity of the inverse function.
as shown in Fig. 5.1 . Then Y 1 < Yo < Y2 · Furthermore, if Y 1 < y < Y2 , then xo - € < f- 1 (y) < Xo + €. Putting o = min{yo - Y 1 , Y2 - Yo } we see that I Y - Yo i < 8 implies that j ! - 1 (y) - f- 1 (yo) j < €. That is, f- 1 is continuous at Yo and therefore on all of JR. The case where f is strictly decreasing is similarly proved. Alternatively, one can note that f- 1 (y) = ( - f) - 1 (-y) and that -f is increasing if f is decreasing. Returning now to the case where f is strictly increasing and maps [a , b] onto [c, d] , we simply extend f to the whole of JR. Define F : lR --+ lR by
F(x)
=
{
x + f(a) - a , x < a a�x�b f(x) , X + f(b) - b, b < X.
Then F maps lR onto JR, is strictly increasing, and is equal to f on [a, b] . As above, we see that F - 1 is continuous. In particular, F- 1 = f - 1 is 0 continuous on [c , d]. Theorem 5 .7 Proof.
Each of the functions ¢,
1/J ,
and p is continuous.
This follows immediately from the preceding theorem.
0
The Complex Exponential and Trigonometric Functions
5.8
91
More o n exp z an d the Zeros o f sin z an d cos z
Having established the familiar properties of the real trigonometric func tions , we return to a discussion of the complex versions. We attack these via the exponential function, which determines them all.
The equality exp z = exp w, for z, w E C, holds if and only if there is some k E Z such that z = w + 27rki .
Proposition 5 . 3
Multiplying both sides by exp( -w) , we see that exp z = exp w if and only if exp(z - w) = 1 . Now, exp(27rki) = cos 27rk + i sin 27rk = 1 and therefore exp z exp w if z - w 27rki. Conversely, suppose that exp(z - w) = 1 , and write z - w = a + i{3, with a, {3 E JR. Then
Proof.
=
=
1 = exp(z - w) = exp(a + if3) = exp a exp if3 = exp a (cos f3 + i sin {J) . It follows that exp a = 1 (taking the modulus of both sides) , and so cos {3 = 1 and sin {3 = 0 (equating real and imaginary parts) . This implies that a = 0 and {3 is of the form {3 27rk, for some k E z . It follows that D z - w = 21rki, for some k E Z, as required. =
We have extended the definition of the functions sin x and cos x from the real variable x to the complex variable z . It is of interest to note that these complex trigonometric functions have no new zeros, as we show next. Proposition 5 .4 For z E C, sin z = 0 if and only if z 1rk for some k E Z, and cos z = 0 if and only if z = (2k + 1 H , for some k E Z. =
Proof.
We have sin z = 0 -<===> -<===>
-<===> -<===>
-<===>
as claimed .
exp( iz) - exp( -iz) =0 2i exp( iz ) = exp( - iz) exp(2iz) = exp O 2iz = 0 + 21rki, for some k E Z, z = 1rk, for some k E Z,
92
Lecture Notes on Complex Analysis
A similar argument is used for cos z. Indeed, cos z = 0 <===> <===>
<===>
<===>
<===>
<===>
exp(iz) + exp(-iz) =0 2 exp( iz) = - exp( -iz) exp(2iz) = - 1 exp(2iz) = exp(irr) , since exp(irr) = - 1 , 2iz = irr + 2rrki, for some k E Z, z = (2k + 1 H , for some k E Z,
and the proof is complete. 5.9
D
The Argument Revisited
We know, informally, that any non-zero complex number can be written as r( cos 9 + i sin 9) = rei8 , where r is its modulus and 9 is some choice of argument (angle with the positive real axis) . We shall establish this formally and also consider the argument mapping in more detail. Suppose, then, that z =/: 0. Write z = a + ib, with a, b E JR. Then r = j z j .../a 2 + b2 =/: 0 and z r(a + i,B) where a = ajr, ,8 bjr. Clearly, a 2 + ,8 2 1 and so ja j � 1 and j,Bj � 1 . We would like to show that it is possible to find 9 such that a = cos 9 and ,8 = sin 9. Since j,Bj � 1 , there is a unique 9 E [- � . �] such that sin O = ,B. Now , cos 2 9 = 1 - sin 2 9 = 1 - ,8 2 = a 2 • It follows that cos 9 = ±a. If cos 9 a, we have z rei8 and we are done. (Note that cos 9 � 0 since - � � 9 � � . ) If not , we must have cos 9 = -a. Let 9' = 71' - 9. Then, using the trigonometric formulae above, we find that sin 9' sin 71' cos 9 - cos 71' sin 9 = sin 9 and cos 9' = cos 71' cos 9 + sin rr sin 9 = - cos 9 = a. Therefore, in this case, we ' can write z as z = re;o . Consider the family of complex numbers given by z (t) e 2 11'it cos 2rrt + i sin 2rrt, for t � 0. Clearly, j z (t) j = 1 and z(O) 1 . As we continuously increase t, the complex number z (t) moves continuously anti clockwise around the circle centred on the origin and with radius one. For -1 , t = t . we find z ( t ) = ( 1 + i) / ,;2. Also , we have z ( -! ) i , z ( ! ) z ( � ) = - i and z(1) = 1 . We have seen that exp z exp w if and only if there is some integer k E Z such that w = z + 2rrki. It follows that we can always write z = rei8 ==
=
==
==
=
=
=
=
=
=
=
=
=
93
The Complex Exponential and Trigonometric Functions
Fig. 5.2
The point z(t)
=
e 2" it moves anticlockwise around the unit circle.
where the argument () E IR is only defined up to additional integer multiples of 21r, i.e . , z rei0 = rei>. if and only if .X = () + 21rk for some k E Z. In particular, we can always find a unique value for () in the range ( -1r, 1r] . (The difference between successive possible values for () is 271", so there must be exactly one such value in any open-closed interval of length 21r.) This value of the argument is given a special name, as already discussed (though somewhat informally) . =
Definition 5 . 3 For any z # 0, the principal value of the argument of z is t li.e unique real number Arg z satisfying -7r < Arg z � 1r and such that z = l z l ei Arg z . (Arg z is not defined for z = 0.)
Notice that if z is close to the negative real axis, then its imaginary part is small and its real part is negative. If its imaginary part is positive, then the principal value of its argument is close to 1r, whereas if its imaginary part is negative then the principal value of its argument is close to -1r. In the example above, the principal value of the argument of the complex number z (t) increases from 0, when t 0, through � ' when t -!, to 1r, when t � and z = - 1 . However , as z(t) crosses the negative real axis, from above to below, the principal value of its argument j umps from 1r to "nearly" -7r (it never assumes the value -1r ) . (The limit from above is 1r , whereas the limit from below is -1r.) It continues to increase, as t increases , until it has the value 0 , when t = 1 and z = 1 . The negative real axis is a line of discontinuity for Arg z . (Recall that Arg z is not defined if z 0.) =
=
=
=
94
Lecture Notes on Complex Analysis
Arg z
=
95
100 rr
'
Fig. 5.3
5.10
'
'
.... - - - - "'
/
/
I
I I I
Arg z is discontinuous across the negative real axis.
Arg z is Continuous i n the Cut-Plane
We shall show next that Arg z is continuous everywhere apart from on the negative real-axis. Theorem 5 . 8 The map z f---> Arg z is a continuous mapping from the cut-plane C \ { z : z + l z l 0 } onto the interval ( -rr, rr ) . =
Proof. Let z E C \ { z : z + l z l 0 } , i.e. , z is any point of C not on the negative real axis (including 0 ) . Write z x+ i y = r(cos Arg z + i sin Arg z ) , where r l z l . Consider first the region where Im z y = r sin Arg z > 0 , that is, the upper half-plane. Then sin Arg z > 0 and s o 0 < Arg z < rr. But r cos Arg z = x and so Arg z = ,P(xjr) where 'ljJ is the inverse to cos : [0 , rr] -+ [ - 1 , 1] . But this map is continuous, by theorem 5 .7, and so Arg z is continuous on the upper half-plane { z : Im z > 0 } Next consider the region x = Re z > 0 (right half-plane) . For such z , we have cos Arg z > 0, so that - � < Arg z < � . However, r sin Arg z y and therefore Arg z ¢(yjr) , where ¢ is the inverse to sin : [- � , � ] -+ [- 1 , 1] . Again, by theorem 5. 7, we conclude that Arg z is continuous on this region, namely the right half-plane { z : Re z > 0 } . Finally, consider the region with y = I m z < 0 . Here, sin Arg z < 0 , so that -rr < Arg z < 0. Since r cos Arg z = x, it follows that Arg z p(x/r) , where p is the inverse to cos : [ -rr, O] -+ [-1 , 1] . Again, by theorem 5.7, we see that Arg z is continuous on { z : Im z < 0 } , the lower half-plane. We conclude that z f---> Arg z is continuous on C \ { z : z + l z l = 0 } · =
=
=
=
=
=
=
95
The Complex Exponential and Trigonometric Functions
If z -:/- 0 does not lie on the negative real-axis, then Arg z -:/- rr and so Arg z E ( - rr , rr ) . On the other hand, for any - rr < () < rr, we have Arg( cos () + i sin B) = (). It follows that z �----> Arg z maps C \ { z : z + l z l = 0 } D onto the interval ( - rr , rr ) . What's going on? plane C
\ {z : z
+
The claimed continuity of the function Arg
lzl
=
0}
is clear from a diagram.
z
on the cut
However, our starting
point has been the power series definitions of the trigonometric functions and the function Arg
z
is formally constructed in terms of these
( via
"obvious"
)
behaviour.
) ( but
suit able inverses .
This has meant that we have had to do some work to wring out the required
Chapt er 6
The Complex Logarithm
6.1
Int roduction
The logarithm is an inverse for the exponential function: if x = e t , then In x t, where t E JR. Note that the natural logarithm In x is very often also written as log x. C an we mimic this construction to get a logarithm for complex variables? If z E C , we want log z w , whatever it turns out to be, to satisfy the relation e w = z . To see how we might proceed, write z as z = x + iy = r e i 0 , with r = Jx 2 + y 2 and note that (} is not uniquely determined-we can always add 27rk, for any k E z. Nevertheless, suppose that we have made a choice for 0. Then we want to construct log z log ( rew ) . We try the formula =
=
=
log z
=
log ( r e ; 0 ) = log r .......,..,
usual log
=
ln r
+ +
log e i0 ...__.,
undefined as yet
iO
as an apparently reasonable attempt. Then we find that whatever our choice for (}
since r > 0 and so
e 1n r
= r. In fact , if w = In l z l + iO + i27rk then ew
= =
=
=
e 'n l z l + i0 +i27r k e 'n l z l e iO e i 27r k l z l eiO e i27rk z,
97
=
z e i27r
k
98
Lecture Notes on Complex Analysis
for any k E Z. In other words, for given z =f. 0, the equation
has infinitely-many solutions, namely,
w
=
ln l z l + i arg z ,
where arg z is any real number such that z l z l ei arg z . For z = 0 , the equation becomes ew = 0, which has no solution. (We know that the complex exponential function is never zero.) These heuristics suggest that setting up a theory of complex logarithms might be a little more involved than for the real case. =
6.2
The Complex Logarithm and its Properties
Definition 6 . 1 For z E C \ { 0 } , we say that a logarithm of z is any particular solution w to ew = z .
I f W I and w2 are solutions to ew = z , then ew1 = z = ew2 and so we see that ew1 -w2 = 1 . Putting WI - w 2 = a + ib, with a, b E JR , this becomes e a e1b = 1 and so a = 0 and b = 21rk, for some k E Z. In other words , the difference between any two possible choices for the logarithm is always an integer multiple of 27ri. The arbitrariness of the complex logarithm reflects the ambiguity in the choice of the argument of a complex number. When there is no chance of confusion, one often j ust writes log r to mean the usual real logarithm of any positive real number r. Definition 6 . 2 The principal value of the logarithm of z =f. 0 is that obtained via the principal value of the argument and is denoted Log z; thus
Log z = ln I z I + i Arg z . We see that -1r < Im Log z � 1r. Moreover, since possible choices o f the argument must differ by some integer multiple of 27r, it follows that any choice of logarithm of z , log z , can be written as log z = Log z + i27rk, for some k E z. Of course, k may depend on z . Various properties o f any such choice , log z , o f the logarithm o f z are considered next.
The Complex Logarithm
99
Suppose that for each z E C \ { 0} a value for log z has been chosen. Then the following hold.
Theorem 6 . 1
(i) elog z = z . (ii) log(e z ) = z + 2 7r ki, for some k E Z . (iii) log(z1 z2) = log Z 1 + log z2 + 21rki, for some k E Z . (iv) log
(;.) = - log z + 27r ki, for some k E
z.
Proof. By definition, any choice w log z o f the logarithm o f z satisfies ew = z , which is (i) . To prove (ii) , let w = log( e z ) be any choice of the logarithm of e z . Then ew = e z . It follows that w - z = 21rki, for some k E Z, as required. Let w 1 = log z1 and w2 = log z2 be any choices of the logarithms of z1 and z2 , respectively, and let w 3 = log(z1 z2) be some choice for the logarithm of the product z 1 z2. By the definition, =
It follows that there is some k E Z such that w 3 = w 1 + w2 + 21rki , which proves part (iii). Finally, suppose that w is a choice of log � and let ( be any choice of log z . Then ew = 1 /z and e( = z. It follows that z = 1 /ew and therefore
We deduce that ( = - w + 27rki , for suitable k E Z, which proves (iv) .
0
Examples 6 . 1
( 1) Possible choices o f the logarithm o f 1 are log 1 = 0, o r 27ri, o r 47ri , or . . . , or -2 7ri, or -47r i , or . . . . ? (2) log 1 = log 1 2 � log 1 + log 1 . This is only true if we make the choice log 1 0. No choice log 1 = 21rki , with k E Z, k -:/- 0, will work. ( 3) Consider the principal value of the logarithm of the product i ( - 1 + i ) . =
We have
Log i( - 1 + i)
ln l i ( - 1 + i) l + i Arg i ( - 1 + i) = ln v'2 - i 3_; . =
100
Lecture Notes on Complex Analysis
Now, Log i = ln I ii + i Arg i = 0 + i � and Log( - 1
+
i) is given by
Log( - 1 + i) = ln l -1 + il + i Arg( - 1 + i) = ln v'2 + i 3,r . We see that Log i + Log( - 1 + i)
=
i � + ln v'2 + i 3,r
# Log i ( - 1 + i) .
=
ln v'2 + i 5;
( 4) For which values of z is sin z ei z , so 3 ? To solve this, put w 1 that sin z (w - w- ) /2i. Then sin z = 3 becomes the quadratic equation w 2 - 6iw - 1 0 with solutions w = (3 ± 2 v'2)i. But ei z = w means that iz is a choice of log w , that is , iz must be of the form iz = Log w + 2k1ri for some k E Z. Hence iz ln 1 3 ± 2 v'2 1 + i � + 2kni and so z = -i ln 1 3 ± 2 J2 1 + � + 2k1r for k E Z . =
=
=
=
=
Remark 6. 1
If z is real and positive, z = x + iy , x > 0 and y = 0, then Log x
=
=
Log z = ln I z I + i Arg z = ln x + iO ln x .
In other words , on the positive real axis, { z : z real and strictly positive } , the principal value o f the logarithm agrees with the usual real logarithm. 6.3
Complex Powers
For real numbers , a and b, with a > 0 , the definition of the power ab is given by ab = exp(b log a) . Now that we have a meaning of the logarithm for complex numbers we can try to similarly define complex powers. Definition 6 . 3
For given z, ( E C, with z # 0 , w e define the power z'
=
exp(( log z ) .
Evidently, z ( depends o n the choice o f the logarithm log z . That is, we must first make a choice of log z before we can define z( . Put another way, different choices of log z will lead to different values for z( . The principal value of zC , for z # 0, is defined to be exp( ( Log z ) .
101
The Complex Logarithm
Examples 6 . 2 (1)
1
What are the possible values of 8 3 ? We have
8t
=
=
e � log S = e � (ln 8+ 2 7rki)
�
e 'n 2+ 1rki
=
2e
� 1rki
for k E Z. Taking k 0, 1 , 2 gives all the possibilities ( further choices of k merely give repetitions of these three values ) . (2) The possible values of i i are =
ii
=
=
=
e i log i
=
ei.,./2
e i ( L og i+2k7ri ) ,
for k E
e i( i f + 2 k7ri ) e - � - 2 k11" ,
which are all real. (3) Taking principal values, e � L og( - 1 )
=
and
for k E Z ,
( -i ) 1 12
i 1 12
=
z,
=
e � L og i
e � L og ( - i ) =
ei.,./ 4
'
=
e - i.,./ 4 , ( - 1 ) 1 /2
so that
This provides an example of complex numbers
w
and ( for which
where the principal value of the square root is taken. Remark 6.2 There are one or two consistency issues to worry about. We seem to have two possible meanings of z m when m E Z , namely, as the product of z with itself m times ( or the inverse of this if m is negative ) or as the quantity exp ( m log z ) , for some choice of log z . In fact, there is no need to worry. Let log z be any fixed choice of the logarithm of z ( where z -:/- 0) . Suppose that m E N. Then
exp ( m log z )
=
exp ( log z ) . . . exp ( log z ) m factors
=
�,
m factors
Lecture Notes o n Complex Analysis
102
which shows that exp(m log z) reduces to the usual "product of m terms" meaning of z m . For negative m, set k -m. Then, as above, =
exp(m log z)
=
exp( - k log z)
1
=
=
=
1 exp (k 1 og z )
k terms z - k , usual meaning z m , usual meaning.
For m 0, we have exp( m log z) exp 0 = 1 = z0 which is the usual meaning of a number to the zeroth power . Another concern is with our agreed notation ez for exp z . Does this conflict with the meaning of the real number e raised to the complex power of z? From the definition, we see that ez = exp(z log e) with some choice of the logarithm being made. Now , any such choice has the form log e = Log e + 2rrki , for some suitable k E z. This means that we can always write exp(z + 2zrrki) = exp z exp(2zrrki) , since ez as exp(z(Log e + 2rrki) ) Log e 1 . For this to equal exp z , we must insist that exp(2zrrki) = 1 , that is, 2zrrki 2rrmi for some m E Z . I n general, this is only possible when k 0 , in which case m = 0. We come to the conclusion that the complex power ez agrees with exp z provided that we always take the principal value of the power . We shall adopt this convention if there is any doubt. In fact , ez is really only usually used as a notational shorthand for exp z. So we can choose not to use it,· and always use exp z instead, or to use it and remember exactly what we are doing, or use it and always interpret it as the principal value of the power. There is unlikely ever to be any confusion. =
=
=
=
=
=
Proposition 6. 1 Suppose that z -1- 0 and for m E N let ( ( m = z, that is, ( is an mth root of z . Proof.
=
z 1 / m . Then
We have that ( = exp( ;k log z) for some choice of log z . But then (m
as required.
=
1 1 exp( - log z) . . . exp( - log z) = exp log z m m m terms
=
z, 0
103
The Complex Logarithm
Example 6.3 Let a be any choice of yCI (i.e . , of ( - 1) ! ) and let b be any choice of y'l (i.e . , of d ) . Then
/-1 VT
{1 Y�
=
=
yCI
= a.
But
yCI y'l
For equality, i.e., for
� and y'l b yCI
=
=
�. a
R = R = Jl f£ r,
y1
11 =
y -1
- , we require that
-1
This, in turn, requires that a 2 b2 o r - 1 = 1 , which evidently can never hold (no matter what choices a and b are made) . We conclude that whilst =
since both are j ust some choice of yCI, nevertheless
This means that the "equalities"
simply cannot ever both be simultaneously true. 6.4
Branches of the Logarithm
We have seen that log z depends on a choice for arg z. A natural question to ask is whether or not there is some consistent choice of arg z so as to make z �---> log z continuous. The answer depends on where exactly we want to define log z , i .e. , its domain of definition. For some regions, for example an annulus around the origin such as { z : 1 < l z l < 2 } , this cannot be done, as we will show below .
104
Lecture Notes on Complex Analysis
Proposition 6 . 2 The choice z 1--+ Log z of the logarithm on the cut plane C \ { z : z + l z l = 0 } .
is
continuous
Proof. For z E C \ { z : z + I z I = 0 } , Log z = ln I z I + i Arg z . Now, we have seen that Arg z is continuous at each such z and so is z �-----+ l z l and 0 hence also z 1--+ ln lz l . The result follows.
Example 6.4 There is no choice of logarithm making the map z 1--+ log z { z : l z l = r } . To see this, continuous everywhere on the circle C suppose that z 1--+ f(z) is such a choice of logarithm, for z E C. We know that z 1--+ Log z is continuous for z not on the negative real axis , and so the map z 1--+ f(z) - Log z is continuous for z E C \ { -r } . e f(z) , by definition of a logarithm. But z = e Log z Now, we have z e L og z for z E C. It follows that / (z) Log z + 21rik(z) and so e f(z) for some k(z) E Z, depending possibly on z E C. However, both f(z) and Log z are continuous on C \ { - r } and so , therefore, is their difference f(z) - Log z = 27rik(z) . r (cos t + i sin t) = reit . Then the map For -1r < t < 1r, set z(t) t �---+ k(z(t) ) is a continuous map from ( - 1r , 1r ) -+ Z and so must be constant (by the Intermediate Value Theorem) . We conclude that there is some fixed k E Z such that =
=
=
=
=
f(z) = Log z + 27rik = ln r + i Arg z + 21rik for all z E C \ { -r} . However, the left hand side of ( *) is continuous at each z E C, by hypothesis, whereas the right hand side has no limit as z approaches - r . (If z approaches -r from above (i.e . , through positive imaginary parts) then Arg z converges to 1r, but if z approaches -r from below, then Arg z converges to - 1r . ) This contradiction shows that such a continuous choice of logarithm on C cannot be made . Definition 6.4 A branch of the logarithmic function is a pair (D, /) , where D is a domain and f : D -+ C is continuous and satisfies exp f(z) z for all z E D . (Note that D cannot contain 0 since the exponential function never vanishes .) The principal branch is that with D C\{ z : z+ l z l 0 } and f(z) = Log z. =
=
=
By suitably modifying Log z in various regions of the complex plane, we can construct other branches of the logarithm . Example 6.5 Let D C \ { z : z - l z l = 0 } , the complex plane with the positive real axis (and {0}) removed. We define f(z) , for z E D, in terms =
The Complex Logarithm
105
of Log z , as indicated in Fig. 6 . 1 . For given z E D, set
f (z ) =
{
Log z ,
Log z + 27ri ,
Im z � 0 Im z < 0 .
Notice that f ( z ) takes the value Log z for z on the negative real axis . Log z
Log z
Log z 0
-
Log z + 27l"i
Fig. 6 . 1
- - - \: �e�oved Log z + 27l"i
A branch of t h e logarithm o n C \ { z
: z
-
lzl
=
0 }.
It is clear that f, as defined here, is continuous at any z not on the real axis . The function Log z j umps by -27ri on crossing the negative real axis from above to below, so the construction of / , via the addition of an extra 27ri in the lower half-plane, ensures its continuity, even on the negative real axis. Example 6.6 Let S be the set S = { z : z = t + it (t - 1 ) , t � 0 } and let D C \ S. By way of example, we seek a branch (D, f) of the =
Log z + i 671"
f( -2)-ln 2 H 771" Log z + i 8 71" Fig. 6.2
Log z + i 8 71"
A branch of log z on D with f( - 2) = In 2 + i 771".
106
Lecture Notes on Complex Analysis
logarithm on the domain such that f( - 2) ln 2 + i 7-rr . Notice that D is the complement of part of a parabola. (Let x t and y = t (t - 1) so that y x ( x - 1) for x � 0 . ) The branch (D, f) is as indicated in the Fig. 6 .2. Note that f(z) i s equal t o Log z + i 61r on the negative real axis (dashed) . =
=
=
Remark 6.3 If (D, f) is a branch of the logarithm, then, for any fixed k E Z, so is (D, g) , where g(z) f(z) + 21rki , for z E D. Indeed, g exp f(z) exp 27rki is continuous, and exp g (z) exp f (z) z . The converse is true as we shall show next. =
=
=
=
Suppose (D, f) and (D, g) are branches of the logarithm on the same domain D . Then there is k E Z such that
Theorem 6.2
g(z)
=
f(z) + 21rik
for all z E D (-the same k works for all z E D) . Proof.
Set h(z) = g(z) - f(z) , for z E D . Then exp h (z)
exp (g (z) - f( z ) ) exp g (z) exp ( - f( z ) ) z z --1 - exp f (z) z =
=
-
·
Hence, for each z E D, there is some k(z) E Z such that h(z) = 27rik(z) . Let z 1 , z2 E D be given. Since D is connected, we know that there is some path 'Y : [a, b] --+ C in D joining z 1 to z2 . Now, h, and therefore k is continuous on D. Hence the map t �---> k('Y(t) ) from [a, b] into Z is continuous. It is therefore constant, by the Intermediate Value Theorem. Hence k("!(a) ) k ("!(b)) , that is, k(z 1 ) = k ( z 2 ) and we deduce that k is D constant on D. =
The next result tells us that any continuous choice of the logarithm is automatically differentiable and that its derivative is exactly what we would guess it to be, namely, 1 / z . Theorem 6 . 3 Let (D, f) be a branch of the logarithm.. Then f E H(D) and f' (z) = 1 / z , for all z E D .
z , for z E D , means that 0 rJ_ D . Proof. First we recall that exp f ( z) Also, i f z -=/- w, then f ( z ) -=/- f( w) (since otherwise, z = ef (z) = ef ( w ) w) . =
=
1 07
The Complex Logarithm
Let z , w E D, with z =I w. Then
f(w) - f(z) w-z
- ef(w) -- ef(z) 1 ( ef(w) ef(z) ) = _
f(z)
f(w)
_
f(w) - f(z)
Now, the continuity of f implies that if w --+ z then f(w) --+ f(z ) . Further more, for any ( E C ,
e� - e< < ----+ exp' ( = e �-(
as � --+ (, with � =I (. Hence, as w --+ z
(*)
--+
( e f(z)) - 1
=
1
ef(z)
=
1 . that 1s, f E H(D) and f'(z) = - , for any z E D.
1 ;' 0
z
What's going on?
The notion of a complex logarithm is straightforward , but
complicated by the fact that there is an infinite number of ways in which it c an be
defined . This is simply a consequence of the ambiguity in the choice of the polar angle, the argument of a complex number . To talk sensibly about a logarithm requires specifying some particular choice. continuity considerations.
This done, one then inquires about
This leads to the notion of branch of the logarithm
where the region of defi nition of the logarithm is highlighted . It is not possible to make a continuous choice of logarithm in some domains - for example, in
an
annulus centred on the origin . The basic definition of a logarithm together with continuity is enough to imply its differentiability. might expect .
Its derivative is
1/ z,
as one
O n c e o n e has some notion o f logarithm, it c a n then be used to construct complex powers. As a corollary, we obtain an alternative proof of the above result on the uniqueness, up to an additive constant multiple of 27l'i , of the branch of the logarithm on a given domain .
Corollary 6 . 1
Suppose ( D, f) and ( D, g) are branches of the logarithm on the same domain D. Then there exists some integer k E Z such that g(z) = f(z) + 27l'ki , for all z E D . Since f and g are both logarithms , it follows that for each z E D there is an integer k(z) E Z such that h(z) = g(z) - f(z) = 27rk(z)i. By the theorem, both f and g are differentiable o n D and so, therefore, is h Proof.
Lecture Notes on Complex Analysis
108
and hence so is k . However , k(z) E JR, for all z E D and so k(z) is constant on the domain D . Note that one could also argue that since f and g both have the same derivative on the domain D, namely 1/ z, then their difference h satisfies h' (z) = 0 on D. This means that h is constant on D and therefore of the D form 21rki for some fixed k E z .
Example 6 . 7 Let S be the set S = { z : z = t e it , where t E JR , t � 0 } . We construct the branch (D, f ) of the logarithm on the domain D = C \ S with f(l) = Log 1 . The idea is to build on the values of Log z by compensating for the j ump in its value as z passes through the negative real axis . the curved regions enclosed between the Denote by R_ 1 , Ro , R 1 , spiral S and the negative real axis, as shown in the Fig. 6 .3, so that D = U�- 1 Rk . The section ( -1r, 0) of the negative real axis is included in R_ 1 , the section ( - 3 7f , 7f) is included in Ro and so on. •
•
•
-
Fig. 6.3
The branch of logarithm on D with f ( l )
= 0.
The Complex Logarithm
109
The function f is defined on the domain D by setting
f(z) = Log z + 2k7ri ,
for z E Rk , k = - 1 , 0, 1 , 2, . . . .
Defined in this way, f is continuous on D and determines a branch of the logarithm. Furthermore, / ( 1 ) Log 1 , since 1 E Ro. This uniquely specifies the branch (D, f). =
Chapter 7
Complex Integration
7. 1
Paths and Contours
In this chapter, we consider integration along a contour in the complex plane. Recall that a path in C is a continuous function 'Y : [a, b] --+ C , for some a ::::; b in JR. The set of points { z : z 'Y(t) , a ::::; t ::::; b } is the trace of 'Y which we denote by tr "(. =
Example 7. 1
Let "f 1 (t) tq1
Now suppose that 'Y2 (t) tq2
=
=
=
=
e 2""i t for 0 ::::; t ::::; 1 . Then we find that
{ z : lzl
=
1}
=
unit circle.
e _4,. i t for 0 ::::; t ::::; { z : lzl
=
1}
=
1.
We sec that again
unit circle.
The paths 'YI and 'Y2 have the same trace but are different paths: the path "( 1 goes round the circle anticlockwise once, whereas 'Y2 goes round twice in a clockwise sense.
Remark 7 . 1 The convention is to take the anticlockwise sense as being positive. This is consistent with the convention that the positive direction is that with increasing polar angle. Definition 7. 1 The path 'Y : [a, b] --+ C is said to be closed if "f(a) 'Y(b) . The path 'Y is said to be simple if it does not cross itself, that is, 'Y(s) =f. 'Y(t) whenever s =f. t with s , t in (a, b) . ( The possibility that 'Y(a) 'Y(b) is allowed . ) =
=
Example 7.2 The path "( 1 in the example above is a simple closed path , whereas 'Y2 is closed but not simple. The path 'Y(t) e -S ,-it ; 0 ::::; t ::::; 1 , has the unit circle as its trace, but 'Y is neither closed nor simple. =
111
112
Lecture Notes o n Complex Analysis
Definition 7.2 Let "( : [a, b] ---+ C be a path. The reverse path ::Y is given by ::Y : [a, b] ---+ C with ::Y(t) = "f ( a + b - t) .
Evidently, ::Y is "'Y in the opposite direction" . Note that the parametric range for ::Y is the same as that for "(, namely [a, b] . It is also clear that tr ::Y tq. =
Definition 7.3 The path "( : [a, b] ---+ C is smooth if the derivative 'Y'(t) exists and is continuous on [a, b] (with right and left derivatives at a and b, respectively) . In other words, if 'Y(t) x(t) + iy(t) , then 'Y is smooth if and only if both x and y are differentiable (real functions of a real variable) and such that the derivatives x' (t) and y' (t) are continuous functions of the parameter t E [a, b] . A contour is a piecewise smooth path, that is, 'Y : [a, b] ---+ C is a contour if and only if there is a finite collection a = ao < a1 < < an b (for some n E N) such that each subpath 'Y : [ai - l ! ai] ---+ C is smooth, 1 :::; i :::; n . + "fn , where 'Yi is the restriction of "( to the We write 'Y 'Yl + "(2 + subinterval [ai- l , ai] · =
·
=
· ·
· ·
=
·
Example 7.3 According to our (reasonable) definition, the path given by t r---> 'Y(t) cos3 (27rt) + i sin3 (27rt) , for 0 :0:::: t :0:::: 1 , is smooth. Its trace is illustrated in the figure, Fig. 7.1 . =
-1
Fig. 7. 1
A
technically smooth path.
1 13
Complex Integmtion
7.2
The Length of a Contour
Next, we wish to develop a means of formulating the length of a path. By way of motivation, suppose that "'( : [a, b] ---+
•
•
=
n
L l zi - Zj - d
j=1
n
=
L b (ti ) - "Y (tj - 1 ) 1
j=1
"Y(tJ ) - "Y(tJ. - I ) can tj - tj - 1 be considered an approximation to the derivative "Y ' (t1 ) , in which case the
with a
=
to < t 1 < · · < tn ·
=
b. Now, the quotient
·
sum on the right hand side becomes an approximation to an integral
Indeed, in view of the assumed smoothness of "'(, one might expect that the left hand side converges to the right hand side as max1 (ti - t1 _ 1 ) ---+ 0. Rather than prove this here, we will j ust take this as motivation for our definition of the length of a path, as follows. Definition 7.4 The length of the smooth path "'( non-negative real number L("'f) given by the formula
L ("Y)
=
1b
[a, b]
---+
l"'f '(t) l dt.
If "'( = "'( 1 + "'(2 + · · · + "Yn is a contour ( with each "Yi smooth ) , we define the length of "'( to be L("Y) L ("'fl ) + L ("Y2 ) + · · + L("Yn ) · =
·
Lecture Notes on Complex Analysis
1 14
Examples 7.4
e 2 " i t , for 0 ::; t ::; 1 . We calculate 1 1 l 27rie 2 " it l dt L('Y) = I 'Y '(t) l dt =
(1 ) Let 'Y(t)
(2) Let 'Y(t)
=
1
=
1
=
27r
1
1
1 dt
=
27r .
e-41rit , for 0 ::; t ::; 1 . Then 'Y ' (t) = -47rie - 47rit , so that L('Y)
=
1 1 47r
=
47r .
( 3 ) Define the path "( by
Then 'Y is a contour and
"f1 (t) = ( At t
t
=
{
O :S t < 1 1 :S t < 2 2 ::; t ::; 3 . 27rie 2 ""it ,
0 ::; t < 1
1
0,
1 , the left derivative is 21ri , the right derivative is 0 and at 2, the left derivative is 0, the right derivative is -27ri .) We find 1 2 27r dt = 47r . 21r dt + O dt + L('Y) = =
1
1
13
(4) Suppose that 'Y(t) zo + t(z 1 - zo) , 0 ::; t ::; 1 . Then "f is the line segment 'Y = [zo , z!] . We have 'Y ' (t) z 1 - z0 and so 1 l z 1 - zo l dt = l z 1 - zo l , L('Y) = =
=
1
as it should . Remark 7.2 We see that every contour has a well-defined length-this is because of its good behaviour. In general, paths need not have a meaningful length. For example, let "( : [0, 1] ---+ C be the path given by
'Y�) =
{
1 15
Complex Integration
=
The path 'Y starts at the point "f (O) 1 and spirals counterclockwise in towards the origin (and encircles the origin an infinite number of times) . Evidently, 'Y is continuous at any t with 0 � t < 1. Furthermore, for any 0 � t < 1 , I 'Y(t) - 'Y(1 ) 1 = (1 - t) --+ 0, as t i 1 , and so we see that "f is also continuous at t 1 . Hence "( really is a path. However, for any 0 � s < 1 , the length of 'Y "from 0 to s" is
=
tl 27ri 1 1 18 271" dt 27r ln (--) > - 0 -1-s 1 -t
t
' (t) l dt = - 1 dt Jo b Jo 1 - t =
--+ oo
as s increases to 1 . One could say that 'Y has infinite length. 7.3
Integration along a Contour
One way of looking at the integral of a real function of a real variable is to think of it as the area under its graph. This is approximated by the area of suitable rectangles (typically very thin and with height determined by the values of the function at the location of the rectangle) .
1b J (x ) dx a
=
area und er graph
ap= prox
""""' f (x ,· ) u A x ,· L.....,
where f(x; ) �x; is the area (height x base) of the rectangle with height f(x; ) and width �x; . This formula makes sense for complex variables and complex-valued functions even though the original idea of area no longer does. Suppose "( : [a, b] --+ C is a smooth path. Let a = t0 < t 1 < < tn = b and set z1 "f (t1 ) . With the preceding comments in mind, we consider
=
·
· ·
provided that the t1s are close together. It is possible to prove convergence of such sums to integrals (provided f is continuous) , but we will simply take the right hand side as our definition of the complex contour integral.
1 16
Lecture Notes on Complex Analysis
Definition 7.5 Let 'Y : [a, b] -+ C be smooth and f a continuous complex function on tr 'Y. The contour integral of f along 'Y is defined to be
Sometimes the notation J f (z) dz is used . If we write f = u + iv and f can be written as X + iY, where X and Y = x(t) + i y (t) , then are real integrals,
h
'Y(t)
X=
1b
{ u (x(t), y (t) ) x' (t) - v (x(t) , y (t )) y' (t)} dt
Y=
1b
{ u (x(t) , y (t ) ) y' (t) + v (x(t) , y (t)) x' (t)} dt.
and
For any contour 'Y, we define
1!=1 !+1 !+..·+1 "'(
where
'Yl ,
"'' I
"'12
"''n
f
. . . , 'Yn are the smooth parts of the contour 'Y·
Examples 7.5
( 1 ) Let 'Y(t) = ei t for a ::; t ::; b, and take f(z)
=
=
1b 1b a
1
.
----:- ie. t e't
=
� · Then
dt
i dt = i (b - a) .
In particular, with a = 0 and b = 27r, we find that
J"Y d:
=
27ri.
1 17
Complex Integration
'Y(t) = a + t (b - a ) , for 0 � t :S 1 , that is , ')' is j ust the line segment [a, b] in JR. Then, for any f, continuous on tr ')',
(2) For given a < b in
IR, let
i = la f
l
f('Y(t)) 'Y'(t) dt
1 1 f (a + t ( b - a ) ) ( b - a) dt = 1 b f(s) ds, =
by the change of variable s = a + t (b - a ) . In particular, if f is real valued, then f is precisely the usual real integral f(x) dx.
I:
I[a, bJ
Remark 7 . 3 This last observation has important consequences. Suppose that ')' = 'Yl + · · · + 'Yn is a contour where one of the smooth parts is a line segment on the real axis, 'Yk [a, b] , say, with a < b E JR. Suppose that f is a continuous complex function on tr ')' and that f is real on tr ')'k , that is, f is real on the line segment [a, b] on the real axis. We have =
1!= 1 !+···+1 !+···+1 'Y
'Yl
'Yk
I:
'Yn
f
which expresses the real integral I'Yk f = f (x) dx in terms of complex integrals. This may not seem much of an advance, but the point is that complex integrals can often be evaluated by general theory based only on rather general properties of the function f. This turns out to be a very powerful method for performing real integration. Proposition 7.1 For any contour ')' and any functions, f, g, continuous on its trace tr ')' , we have
i af + {Jg = a 1 f + {3 1 9 , (ii) � f = - 1 f. (i)
for all a, {3 E
Proof. The integral along a contour is given by the sum of the integrals along its smooth parts, so it is enough to consider the case of a smooth path ')' : [a, b] -+
1 18
Lecture Notes on Complex Analysis
To prove (ii), we evaluate the left hand side.
h
1b f(;:;;( t)) ;y' (t) dt = - 1 b f(-y(a + b - t) ) 'y' (a + b - t) dt 1 a f('Y(s)) -y' (s) ds, putting s = a + b - t, = - 1b f ('Y(s)) 'y' (s) ds
f=
=
=-
i
f,
0
as required.
The next result tells us that the integral along a path is somewhat insensitive to how we choose to parameterize the path.
The integral I'"Y f is independent of the parametrization of the smooth path -y, i. e. , if -y : [a, b] -+ C, cp : [a, ,B] -+ [a, b] with cp(a) = a, cp(,B) = b, cp is differentiable and cp' is continuous, then
Proposition 7.2
where 9 is the smooth path 9 = -y o cp : [a, ,B]
--+
C.
Proof. This i s really just a change of variable. Let f (-y (t) ) -y ' (t) = u(t) + iv(t) , where u and v are real, and let U(s) = I: u (t) dt and V(s) = I: v (t) dt. Put F(s) = U (s) + iV(s) . Then F : [a, b] --+ C and F' (t) = U ' (t) + iV' (t) = u(t) + iv (t) = f ('Y (t) ) -y ' (t) . We have
1 19
Complex Integration
1
f=
1b f('y(t)) 'Y'(t) dt = F(b) - F(a)
=
F(cp({J)) - F(cp(a))
= U(cp({J)) - U(cp(a)) + i (V(cp({J)) - V(cp(a))) =
1:
U(cp(s))' ds + i
1:
V( cp (s))' ds
1{3 U' (cp(s)) cp' (s) ds + i 1{3 V' (cp(s) ) cp' (s) ds = 1{3 F' ( cp(s)) cp' (s) ds = 1{3 f('y(cp(s))) 'Y' (cp(s)) cp' (s) ds = ffJ f('y o cp(s)) ('y o cp )' (s) ds = l f , la
=
�o�
as required.
1,
1
1 ),
0
Example 7.6 For 0 :::; t ::; let 'Y(t) and let ry(t ) = + t(i 1 + t 2 (i - 1) . This corresponds to the reparametrization cp (t) = t 2 , so that ry(t) = 'Y( cp (t)) . By way of illustration, let f be the function given by z r---> f(z) = Re z + 2 lm z . Then
1 1\1 11 f
=
=
i
=
1 ) dt ( 1 + t) dt - 11 ( 1 + t ) dt. -
t
+ 2t) ( i -
On the other hand,
i fo1 ( 1 11 ( 1 = 11 ( 1 f
=
=
i
i
so
f17 f = f
�
- 1 ) dt + t 2 ) 2t dt -11 ( 1 + t 2 ) 2t dt + s) ds - 11 ( 1 + s) ds, setting s = t 2 , e + 2 t 2 ) 2 t(i
f , as it should. The actual value is � (i - 1 ) .
120
7.4
Lecture Notes on Complex Analysis
Basic Estimate
The next result is a basic estimate which is used repeatedly. It is the complex analogue of the result from real analysis that an integral over an interval is bounded by any upper bound for (the modulus of ) the integrand multiplied by the length of the interval. In the complex case, we consider contour integrals, and, in this case, the length of the interval is replaced by the length of the contour. Theorem 7 . 1 (Basic Estimate) Suppose that "( : [a, b] ---+
1 i 1 1 ::; 1b 1/("f (t) h' (t) l dt. In particular, for any contour "(, if 1/(() 1 :S M for all ( E tr "(, then
If J"Y f = 0, there is nothing to prove. So suppose that and let () = Arg J7 f. Then f7 f = ei0 1 f-r !I and therefore Proof.
J"Y f =f. 0,
I i f I = e -iO i f = 1 b e -iO f( "f(t ) h' (t) dt = 1 b Re(e-i0f("f(t))"f'(t)) dt + i 1b I m ( e - i OJ ("f (t) )"f' (t) ) dt ::::; 1b l e -i0 f ("f (t) h' (t) I dt, = 1 b l f("f(t) ) l l"f' (t) l dt ,
=
0 since lhs is real
since Re w :::; l w l , for any w E
121
Complex lntegmtion
claimed. Now, if l f(() l � M for all ( E tq, it follows that l f('y(t) ) l � M for all a � t � b and so as
1 1 1 1 � 1b lf ('y (t ) ) l b'(t) l dt � 1 b M I 'Y' (t) l dt =
M L('y) .
The result for a contour follows by summing over its smooth subpaths;
� M (L('yi ) +
= M L('y) , as
· ·
·
+ L('yn ))
D
required.
7.5
Fundamental Theorem of Calculus
The next result is a version of the Fundamental Theorem of Calculus in the setting of contour integration. Theorem 7.2
Suppose that f is differentiable and that its derivative f ' is continuous on the trace tr 'Y of a contour 'Y : [a, b] ---+ C . Then
1 f' = J('y(b)) - J('y(a)) .
Lecture No t es on Complex Analysis
122
'Y is smooth. Then we have l ! ' = 1 b f ' ('Y (t )) 'Y1( t ) dt = 1b �t (f ('Y ( t ) ) dt = 1b (U' ( t ) + iV' ( t ) ) dt where U ( t ) = Ref ('Y ( t )) and V ( t ) = lmf ( 'Y ( t )) = U ( b ) - U ( a ) + i ( V ( b) - V (a )) = f ('Y ( b)) - f ('Y ( a )). For the general case, suppose that 'Y is the contour 'Y 'Yl + + 'Yk, where C, = 1 , . . . , k, is a smooth path (the restriction each 'Yi = 'Y : [ ti-l> til of 'Y to the subinterval [ti - l> til ) for suitable a = t0 < t 1 < . . < tk = b. Then, as above, we find that k = 1 r j="Ll 1 ! ' k = j='Ll ( ! ('Y (tj)) - J ('Y ( tj -l)) ) = f ('Y( b)) - f('Y ( a )) , Proof.
Suppose first that
=
--+
·
· ·
j
·
"Y
"Yj
0
since the sum telescopes. As a corollary, we recover a familiar result.
Suppose that f E H(D) and that f' = 0 on D . Then f is For any pair o f points, z0 and z 1 , i n D , let 'Y : [ a , b] C be a polygon in D joining z0 to z 1 . By hypothesis, f' = 0 on D and so f' satisfies the hypothesis of the preceding theorem. Hence i f' = f ('Y ( b)) - f ('Y (a )) = f( z i) - f(z0 ) . But f' = 0 which implies that f-r f' = 0. It follows that f ( z0 ) = f ( z i) and we conclude that f is constant on D. constant on D .
Corollary 7 . 1 Proof.
--+
0
1 23
Complex lntegmtion
7.6
Primitive s
Definition 7.6 Let D be a domain and suppose f : D ---+ C is continuous. A map F : D ---+ C such that F' = f on D is said to be a primitive for the function on D.
f
If F is a primitive for f on D, then i f = F('y(b)) - F('y(a)) for any contour lying in D. In particular, if 'Y is closed and f has a primitive in D, then Corollary 7.2
provided tr 'Y Proof.
C
D.
By theorem 7.2, we see immediately that
In particular,
if = i
F' = F('y(b) ) - F('y(a)) .
'Y ( a ) = "f (b) if "( is closed and so J... J = 0.
D
Remark 7.4 The existence of a primitive depends not only on the form of the function, but also on the domain in question. For example, we have seen that Log z has derivative � on the cut-plane. In other words, Log z is a primitive for � on the cut-plane C \ { z : z + l z l = 0 } . However, let 0 � � 'Y( and let D be any domain containing tr "( ( but excluding 0). Then � &as primitive on D. 0. This follows from the observation that "( is closed but J"Y � 2 7r It follows that there is no branch of the logarithm on such D. Indeed, any branch of the logarithm would be a primitive for the function � , and no such primitive exists in this case.
t ) = e21rit,
t 1,
no
=
n
n i- -1,
i i-
Example 7.7 For any E Z, with the function F(z) = is a primitive for zn on C ( or on C \ { 0 } , if < It + zn+ l n follows that J"Y z dz = 0 for any closed contour "( ( lying in C \ {0}, if < Furthermore, considering the line segment [z0 , z! ] , we find that
f(n 1 ) n - 1) .
1
[zo , z d
zn dz
=
-n +1-1 ( z�+ l - z�+ l )
n -1 ).
124
Lecture Notes on Complex Analysis
for n E Z, n i- - 1 . In particular,
1 dz = zl - Zo . [zo , z d
Of course, if n < - 1 in the above, then we must assume that 0 � tr I' , i.e. , that the line segment does not pass through the origin.
[z0, zd Example 7.8 Let f ( z ) = a0 + a 1 z + + an zn be a polynomial. Then the coefficients a k satisfy ak = _27ri1_ 1 zfk( z )l dz ' ·
1
= zn
·
·
+
where I' : [0, 1] --+ C is the circle 'Y(t) r e 2"" i t , for 0 ::::; t ::; 1 and any r > 0. This follows from the fact that has a primitive in C \ { 0 } for any n E Z with n i- - 1 , together with the fact that J1 / 27ri. This type of integral formula is very important and will crop up again in a more general setting.
Example 7.9
Let
I' :
dz z =
[0, 1 ] --+ C be the path 'Y(t)
= 1 + t + i sin ( � 1rt) .
2+i
_
_
.... .
0
Fig. 7.2
We wish to determine
The trace of the path 'Y·
11 �z dz .
125
Complex Integration
To do this, we simply note that the function 1 / z has primitive Log z in the star-domain C \ {z + lzl = 0} , so
1"( dzz = Log ')' ( 1 ) - Log ')' ( O) = Log ( 2 + i ) - Log 1 = Log ( 2 + i ) = � ln 5 + i tan - 1 ( � )
)
•
1
This is somewhat easier than trying to directly evaluate J0 J( 'Y(t)) 'Y' (t) dt. To plot the trace of ")', we note that the point z = x + iy � ( x, y ) belongs to tr ')' provided x E [ 1 , 2 ] and y = sin (� 1r ( x - 1 )) .
Chapter 8
C auchy ' s Theorem
8.1
Cauchy's Theorem for a Triangle
In this chapter, we shall discuss the central theorem of complex analysis, Cauchy's Theorem. This involves integrals around triangles, so first we define exactly what we mean. Definition 8 . 1
T
=
A triangle T with vertices { z1 , z2 , z3 } is the set
{ z E C : z = nz1
+
j3z2 + f'Z3 with
a,
/3, ')' ?: 0 and
a+
j3 + ')' = 1 . }
Thus, by triangle, we mean the interior as well as the edges. Note that T is closed. To avoid unnecessary extra notation, we shall denote the integral around the closed contour [z1 , z2 ] + [z2 , z3] + [z3 , z1 ] , i.e. , the integral around the sides of the triangle, by far . Denote the length of this contour by L ( 8T ) ; thus, L ( 8T ) = L ( [z1 , z2 ] ) + L( [z2 , z3] ) + L ( [z3 , z 1 ] ) = l z2 - z1 l + lz3 - z2 l + lz1 - z3 1 . A moment's consideration reveals that diam T � L ( 8T ) . In fact, diam T is equal to the length of the longest side of the triangle T. Theorem 8.1 ( Cauchy's Theorem for a Triangle)
domain D and any function f E H(D),
r
lar
f
=
o
for any triangle T {wholly) contained in D . 127
For any given
1 28
Lecture Notes on Complex A nalysis
Proof. We emphasize that the triangle T (where we include its whole convex interior) must be contained in D. It is not enough that just the three sides belong to D, which could happen if D has "holes" . Bisect each of the sides of the triangle T to obtain four smaller triangles, T1 , . . . , T4 , as shown in Fig. 8. 1 .
Fig.
8. 1
Construct
4
congruent triangles by bisecting the sides of
T.
Then
r
lar
J
=
r
iar1
J+
r
lar,
J+
r
lar3
J+
r
lar4
J,
since each integral around a triangle is given as the sum of three integrals along line segements, and the integrals along those line segments in the interior of T cancel out. It follows that
Let T1 be one of T1 , . . . , T4 giving the maximum contribution to the right hand side of the inequality ( * ) , so that
We have that T1 c T and L(8T1 ) � L(8T) , since each side of T1 is one half that of the corresponding side of T. =
129
Cauchy 's Theorem
We now repeat this 4 smaller triangles to -cutting-T1 into - procedure 1 1 obtain some triangle T2 with T2 C T� , L(8T2 ) = 7. L(8T1 ) = v: L(8T) and
I faT f I 4 1 hfl f I �
�
42
1 laf2 f I ·
We can do this again for T2 , and so on. In this way, we obtain a nested sequence of triangles T ::J T1 ::J T2 ::J . . . such that L(aT) = 2 n L(8Tn ) and
I faT J I 4 1 hf1 J I �
� ... �
4n
l laTn J I ·
Now, diam Tn = 2- n diam T -> 0, as n -> oo. Furthermore, each Tn is closed, so we can apply Cantor's Theorem, theorem 3 .3, to conclude that there is some zo such that {zo} = nn fn . Next, we note that z0 E D and so f is differentiable at zo because f is analytic in D. Define the function T on D by the formula
{
T (z ) =
o) :... f (:... ...c z )_ ..:. -....:.!--'-.. ( z--'- ! ' ( z0 ) , z - zo 0,
£
or z
for z
-L
-r=
z0 zo.
Then T is continuous on D and, in particular, T ( z ) -> 0 as z -> zo. We can express f in terms of T as follows,
f ( z ) = f( zo) + ( z - zo ) f ' ( zo ) + ( z - zo ) T ( z ) , for z E D. Hence
r�
f larn
18Tn� { J( zo) + (z - zo ) f'(zo) + (z - zo ) T ( z ) } dz = 0 + 0 + r � ( z - zo ) T ( z ) dz, larn =
since fafn dz fafn ( z - zo ) dz = 0 because the integrands have primitives ( on the whole of C ) . We must estimate the ( modulus o f the ) third integral i n the equation above. ( For large n, it is an integral around a very small triangle, located around the point zo, of a function which is close to zero in this region. We therefore expect this integral to have small modulus. However, the earlier estimate for I far ! I in terms of I fafn !I involves the factor 4n , so we must pay attention to the details here. ) Let c: > 0 be given. Since T ( z ) -> 0 as z -> z0, we know that there is some o > 0 such that I T ( z ) l < c: whenever l z - zol < o . Furthermore, we =
130
Lecture Notes on Complex A nalysis
know that diam Tn = 2 - n diam T --> 0 as n --> oo , and so we may choose n so that diam Tn < 8. Then any for z E 8Tn , the boundary of the triangle Tn , it follows that lz - zo l diam Tn < 8, since zo E Tn · Hence IT(z ) l < e and therefore
�
l ( z - zo ) T ( z ) l
for any z E 8Tn - It follows that
� diam Tn
e
I JraTn ( z - zo ) T (z ) dz I � diam Tn
e L ( 8 Tn )
n4 L(8T)
�
e diam T
and therefore
Hence, finally, we have
I !aT J I � 4n l faTn J I < - 4n diam4Tn L(8T) diam T L ( 8T ) . e
=
e
This holds for any e > 0 and so we conclude that
far f 0. =
D
Let D be a star-domain. Then every function analytic in D has a primitive, i. e. , if f E H(D) , then there is F E H(D ) such that F' = f on D.
Theorem 8.2
Proof.
Let zo be a star-centre for D. Define F on D by
F(z)
=
{ f J[zo,z]
for z E D. Note that F is well-defined since [z0, z] c D. We claim that F is differentiable and that F' = f everywhere in D. To show this, let z E D be given. Since D is open, there is r > 0 such that D(z, r) � D. We wish to show that
F (z + () - F (z ) (
_
f(z)
__,
0
131
Cauchy 's Theorem
as ( - 0 . We are assured that F(z + ( ) is defined for all ( with 1 ( 1 < r, since D(z , r ) � D, as noted above. Suppose that 1 ( 1 < r from now on. We will appeal to Cauchy's Theorem to rewrite F(z + () - F (z ) in another form. Indeed, F(z + () - F(z ) =
1
1
f-
[zo ,z +(]
f.
[zo , z]
Let T denote the triangle with vertices z0 , z + ( and z . Since zo is a star centre for D, and since [z + (, z] , the line segment from z + ( to z, lies in D( z , r ) � D, it follows that T C D. Indeed, any point w on the line segment [z + (, z] lies in D and so, therefore, does the line segment [zo , w] . By varying w , we exhaust the triangle T. Now, by Cauchy's Theorem, far f = 0 . However, the contour integral around a triangle is equal ( by definition ) to the sum of the integrals along its sides. Hence, we have
1
f+
1
[z +(,z]
[zo ,z +(]
f+
1
[z , zo]
f
=
0.
Reversing the direction of the contour is equivalent to a change in sign of the integral, and therefore we may rewrite the above equation as
1
f-
[zo ,z +(]
1
[zo ,z]
f=-
In terms of F, this becomes F(z + () - F(z)
=
1
1
!=
1
! ( � ) d� .
[z + ( , z]
[z , z +(]
f
[z , z+(]
Hence F (z + () - F(z )
(
1 1 =
- f (z) = � (
[z, z+(]
[z , z+(]
! ( � ) d� - f (z)
! ( � ) - f (z)
(
�
since f (z ) ft z , z+(] d� = f( z) ( . It remains to estimate this last integral. To do this, we use the continuity of f at z . Let t: > 0 b e given. Then there i s 8 > 0 such that I f ( � ) - f (z) l < t: whenever I � - z l < 8. It follows that if 1( 1 < min { 8, r } , then I � - z l < 8 for
132
Lecture Notes on Complex Analysis
every � E [z, z + (] so that I f ( � ) - f( z) l < c for such �- Therefore
I F(z + (� - F(z) - f(z) I = 1 ; 1 1 1[ , +(] ( /(�) - f (z) ) d� I '>
::::;
zz
j(f c L( [z , z + (J ) 1
= RT1 c 1 (1
= €
0
and the result follows.
What 's going on? The method is constructive in that an explicit formula for a primitive is given, rather than it merely being argued to exist. To achieve this, use is made of the existence of a star-centre for D. The fact that zo is a star-centre means that every line segment (zo , z] lies in D whenever z does. This, together with the continuity of f ensures that F is well-defined. The fact that F actually is a primitive for f is a consequence of the continuity of I and the fact that the integral of f around any triangle is zero. Of course, these two facts are consequences of the (assumed) analyticity of I in D, but the proof that F is a primitive for I carries through if the hypothesis "I is analytic in D" is directly replaced by "I is continuous in D and its integral around any triangle in D is zero" . Stating the theorem this way might seem rather artificial or contrived, but we will see later (see Morera's Theorem, theorem 8.8) that it turns out to be quite relevant.
As a consequence of this last result, we show that if f is does not vanish on a star-domain, then f has an analytic logarithm and also an analytic nth_root. Theorem 8.3 Suppose that f is analytic in the star-domain D and that f (z ) -=1- 0 for any z E D . Then there is g E H(D ) such that f (z) exp g (z) for all z E D . In particular, for any n E N there is h E H(D ) such that hn f in D.
=
=
Proof. Since f does not vanish in D, the function f ' If is analytic in D. But D is star-like and so, by Theorem 8.2, f' If has a primitive there, say F E H(D ) . Then F' f ' If in D. (Experience from calculus suggests that F might be a good candidate for a logarithm of f . This is almost correct. ) Now let 'f/; ( z) f (z ) e - F( z ) for z E D. Then we calculate the derivative =
=
'1/J ' (z)
=
f ' (z ) e- F( z) - F' (z ) f( z ) e - F( z)
=
0
133
Cauchy 's Theorem
because F'(z ) f (z ) = f '(z ) . Since D is connected, it follows that 1/J is constant on D. Fix zo E D. Then
f ( z ) e - F (z)
=
1/J(z )
=
=
1/J(zo)
for any z E D. Rearranging and using f (zo) f (z )
Let g(z )
=
=
f (zo) e - F (z o )
=
e Log / (za ) , we get e F (z) - F ( zo ) + Log f(zo ) .
F(z ) - F(zo) + Log f (zo) . Then g
eg (z)
=
E
H(D) and obeys
f( z ) ,
as
required. It is now easy to construct an analytic nth_root of f in D. Indeed, for any n E N , set h(z ) = exp ( � g(z ) ) . Then hn (z)
=
exp(g (z) )
=
f (z ) D
and the proof is complete. 8.2
Cauchy's Theorem for Star-Domains
Theorem 8.4 (Cauchy's Theorem for Star-Domains) Let f be analytic in a star-domain D. Then J"Y f = 0 for any closed contour 'Y with tr -y C D . Moreover, if cp and 1/J are contours in D with the same initial point zo and the same final point z 1 , then J f = f,p f . "'
Proof. By theorem 8.2, any f E H(D ) has a primitive on D, that is, there is F E H(D) with F' = f on D. Let 'Y : [a, b] -+ C be any contour in D. Then
1 1 l f
=
F'
=
F('Y(b ) ) - F ('Y( a ) ) = 0 ,
if 'Y is closed. For the last part, as above, we have f
as required.
=
F(z 1 ) - F(zo)
=
Lf,
D
1 34
Lecture Notes on Complex Analysis
This central result can be generalized to more general domains, but troubles may arise when D has holes. Example 8.1 Suppose that f E H ( D) and that D, as indicated in Fig. 8 . 2 .
Fig.
8. 2
J""t f
"'f
is a contour lying in
may still vanish i n certain non star-like domains.
We see that J""t f 0 even though D is clearly not star-like. The reason is that we can put in cross-cuts and use the fact that each sub-contour is in a star-domain in which f is analytic. The contributions to the integral from the extra cross-cuts cancel out because they are eventually traversed in both directions. In other words, we may sometimes be able to piece together overlapping star-domains to get a non star-like domain for which the theorem nevertheless remains valid. This kind of argument is often done "by inspection" , that is to say, the precise way the contour is cut up will generally depend on the particular case in hand. =
Example 8.2 It is important to realize that this trick cannot be used if D has holes and the contour "'f goes around such a hole. For example, suppose that the domain D is D =
J""t f = J""t �z dz
and so, in this case, J""t f =/= 0 .
=
27ri
135
Cauchy 's Theorem
Example 8.3
By way of illustration, we evaluate the contour integral
1r z2z+3 1
dz '
where r is the simple closed contour (polygon , in the counter clockwise = sense) whose trace is the square ABCD with corners at A = - 1 - i, B 1 - i , C 1 + i and D = - 1 + i . =
D
'Y
cp
A Fig.
8.3
Applying Cauchy 's Theorem.
The function 1 / z 3 has a primitive on C \ { 0 } and so its integral around any closed contour (not passing through the point 0) vanishes. The function 1/ z is analytic in C \ { 0 } . The domain C \ { z + l z l = 0} is star-like and so the integral of 1 / z around the three sides of the square A -> B -> C -> D is the same as that from A to D along the circular arc -y(t) = J2e i t , -311"/4 � t � 37!"/4, namely
1"Y dzz !-3,./ 4 J2J2iee itit dt =
3 ,. / 4
= 37ri
2
.
Next, we note that { z : Re z < 0 } is star-like and so the integral of 1 / z along the side D -> A of the square is equal to that from D to A along the arc (dotted in Fig. 8.3) cp ( t ) = J2e i t , 37r/4 � t � 57r/4, namely
1'P dzz
Adding, we find
=
{ 5,. / 4 J2 ie i t dt = }3 ,. / 4 J2e i t
7ri . 2
z 2 + -1 dz 1r -1r -z1 dz + = 1"Y -dzz + 1'P -dzz 2 1ri . z3 =
0
=
Lecture Notes on Complex A nalysis
136
8.3
Deformation Lemma
Before proceeding, it is convenient to introduce some terminology. In order to avoid some gratuitous circumlocution, let us agree to call a contour 'Y a simple circular curve, or j ust a circle, if it has the form 'Y(t) ( + p e 2 1rit , 0 � t � 1 , for some ( E C and p > 0 . Here, tr "f is a circle with radius p and with centre at the point ( in C. The path is traced out in the usual positive (i.e., counter-clockwise) sense and makes j ust one single circuit. 'Y is a simple closed smooth path. If w is any point whose distance from the centre ( is smaller than p , then we shall say that "f encircles w (or goes around w ) or that w lies inside the circle 'Y· Note, by the way, that we could just as well have parameterized this circle as 'Y(t ) ( + p e it with 0 � t � 27r. =
=
Lemma 8. 1 (Deformation Lemma) Suppose that ( belongs to the disc D(a, R) and that g is analytic in the punctured set D(a, R) \ {(} . Let 'Y and r be simple circular curves encircling the point ( and lying inside D( a, R) , as indicated in Fig. 8.4 . Then
In particular, if f is analytic in D(a, R) , then
1 wf(w) dw "�
-(
=
r wf (w)( dw .
lr
-
(I + P i e 2 1ri t and r (t) = (2 + The hypotheses are that 'Y(t ) 2 p 2 e 1ri t , for 0 � t � 1 , for some (I , (2 E C and for some PI > 0 and p2 > 0. Furthermore, tq C D(a, R) , tr r C D(a, R) , 1 ( - (I I < PI and I ( - (2 1 < P2 - Without loss of generality, we may assume that the circle "f has a small radius and so is inside r, as illustrated in Fig. 8.4. (If not, we j ust show that each integral is equal to that around one such small circle and so are equal to each other.) The idea is to put in cross-cuts and apply C auchy's Theorem in suitable resultant star-like regions. With the notation of Fig. 8.4, we see that Proof.
=
r
jA+B+C+D+A'+B'+C'+D'
g=
r
lr +;y
g
since the contributions to the contour integral along the parts B and B' cancel out, as do those along D and D'.
137
Cauchy 's Theorem
D ( a, R) _
.
..
· · ·
· · . . · · · .
A
r
· · · · · · · ·
Fig.
8.4
Put in cross-cuts.
A + B + C + D is a closed curve which is contained in a star-domain D 1 � D( a, R ) \ { ( } , as shown in Fig. 8.5 , and g is analytic in D 1 • (We can take D1 to be the star-domain D ( a, R) n (C \ L t ) , where L1 is any straight line (ray) from ( which does not cut A + B + C + D, as shown in Fig. 8.5.)
Now,
Hence, by Cauchy's Theorem, theorem 8.4,
r
jA+B+C+D
g
= 0.
.
Fig.
8.5
..
There is an enclosing star-domai n.
138
Lecture Notes on Complex A nalysis
Similarly, (replacing L1 by another suitable straight line cut (this time up rather than down ) )
r
jA' + B ' +C'+D'
and we conclude that
g=o
Hence as required. For the last ( important ) part, take g (w )
=
f (w ) j (w - () .
0
Remark 8.1 This says that the contour 'Y can be moved or "deformed" into the contour r without changing the value of the integral, provided the change is through a region of analyticity of g. As can be seen from the proof of the Deformation Lemma, the disc D ( a, R) could be replaced by a more general shaped domain, and the closed contours 'Y and r need not be circular. However, the case with circles is of particular interest as we will see later. 8.4
Cauchy's Integral Formula
We now use the Deformation Lemma to obtain another of the basic results of complex analysis. Theorem 8.5 (Cauchy's Integral Formula) Suppose f is analytic in the open disc D ( a , R) and that zo E D ( a, R) . Let r be any simple circular contour around z0 and lying in D ( a , R) . Then
f (zo) =
f (w) � 21r2 lrr w - zo dw.
Let 'Y ( t ) zo + p e 2 1r it , 0 :S: t :S: 1 , where p > 0 is chosen small enough that tr -y C D ( a , R) . Using the Deformation Lemma, we have
Proof.
=
r
f (w) dw w - zo r J
=
=
1 wf-(w)zo dw 1 f (w)w -- f(zo zo ) dw + 1 wf (z-ozo) dw . "�
-y
-y
1 39
Cauchy 's Theorem
Now,
1 wf(zo) dw - zo
1
{ f(zo) 21ri e 2 Tr i t dt Jo p e 2 Tr z t p
=
"�
= 21ri f(zo) · f( w) f(zo) . . 'T' ..LO estimate dw, we use t he contmmty o f f at zo . w zo 'Y Let c. > 0 be given. Then there is 8 > 0 such that lf(z) - f(zo) l < c. whenever l z - zo l < 8. Choose p so that 0 < p < 8. Then for any w E tr ]', we have that lw - zo l = p < 8 so that l f(w) - f(zo) l < c. and therefore f(w) - f(zo) l f(w) - f(zo) l :. I = < . t follows that w - z0 p p •
I
1 I
11 f(w)w -- f(zo) dw l � :_ L(J') p z0 "�
= =
E.
p 27rp
-
21rc. .
Using this, we see that
f(w) - f(zo) I }rr wf(w) dw I dw - 27rif(zo) I 1 1 w - zo - zo "� =
�
21rc..
This holds for any c. > 0. Hence, the left hand side (which, incidentally, D does not depend on p) must be zero. 8.5
Taylor Series Expansion
The next theorem is yet another result of fundamental importance.
Suppose that f is analytic in the disc D(z0 , R) . Then, for any z E D(z0, R) , the function f(z) has the power series expansion
Theorem 8.6 (Taylor Series Expansion)
f (z) = where a n =
r f (w � � 27rt Jr ( w - zo n
+1
00
L a n (z - zo) n
n =O
dw for any circle r encircling zo and lying
in D(zo , R) . The series converges absolutely for any z E D(z0 , R) .
140
Lecture Notes on Complex Analysis
z E D(zo, R) be given. Let r > 0 satisfy lz - zol < r < R and put -y(t) zo + re 2 1rit , for 0 ::=; t :::; 1 . Notice, firstly, that lz - zol / r < 1 and, secondly, that z lies inside the circle -y . Proof.
Let
=
The idea of the proof is to apply Cauchy's Integral Formula
f (z) = � 21Tt
j"' wf (w)- z dw
1 w --z in npowers of ( z - zo) . The formula 1 - a ( 1 - a)(1 + a + a 2 + · · · + an - ) can be rewritten
and expand
=
to give
1 1 -a
--
=
1
1 1 + a + a2 + · · · + an - 1 + an --. 1-a
Hence, we may write
1 w-z
1 (w - zo) - (z - zo) =
=
{
(
1 (w - zo)
) ( ) ( ) ( )
1 z - zo + -z - zo 2 + . . . 1 + -w - zo w - zo (w - zo) n-1 1 Z - Zo n . . . + wZ -- Zo + w - zo 1 - ( �) zo w - zo n 1 (z - zo) n 1 + (z - zo) + + (z - zo) - + (w - zo) (w - zo) 2 · · · (w - zo) n (w - zo) n (w - z) " _
}
Therefore, using Cauchy's Integral Formula, we have
f (z)
=
=
1 (wf (w)- z) dw � 27rt 1
-
27ri
"'
1 (wf-(w)z0) dw + 27ri1 1 (wf-(w)zo) 2 (z - zo) dw + . . . "' "' 1 f . . . + 21Tt. 1 ( (w)zo ) n ( z - zo) n - 1 dw + . . . "' 1 1 f (w) (z - zo) n dw _ . . . + 27ri _ (w - zo) n (w - z) -
-
W -
"'
141
Cauchy 's Theorem
where ak =
1 (w -f (wzo� k+I � 21ft "I
dw , for k = 0, 1, 2 , . . . The value of ak is
independent of 'Y as long as it encircles the point zo . This follows from the f (w ) . . . DeformatiOn Lemma, lemma 8.1, w1 th ( = zo and g (w) = (w - z0 ) k +l We wish to show that R, --> 0 as n --> oo. To see this, we note that the continuity of f and the compactness of tr 'Y together imply that there is some M > 0 such that l f(w) l � M for any w E tr -y ( proposition 4.3) . Also, z ¢:. tr 'Y and so there is o > 0 such that lw - z l � o for any w E tr 'Y· (In fact, o could be any positive real number less than r - l z - zo l - ) This means that 1/ lw - z l � 1/8, whenever w E tr -y. Therefore
M I (w - zof )(w)n (w - z) I <- rno
for any w E tr -y ( since l w - zo l = r if w E tr -y ) . Hence, we find
I R, I =
1 1 f (w) (z - zo) n dw 1I 27ri ( - Zo ) n ( - Z ) "I
W
W
M <- 21!"1 I z - zo I n 21rr rn o = �r ( ' z � zo I r 0 l z - zo l since < 1 . Hence r -
__,
as
n -->
oo ,
00
f (z) = :2: a n (z - zo ) n , n=O
claimed. We still have to establish the absolute convergence of this series. This follows from the convergence of the power series, but can also be shown directly. We estimate
as
- I 1 1 f(w) (z - zo) n I <- 211" rn + l 27rr = M ( ' z � zo l r .
dw I an ( z - zo ) n l - 27ri 7 (w - z0 ) n + l 1 M lz - zo l n
<
But l z - zo l /r 1 and so the series l::'=o l a n (z - zo ) n l converges for any D z E D (zo , R) , by the Comparison Test, as required.
142
8.6
Lecture Notes on Complex Analysis
Cauchy's Integral Formulae for Derivatives
As a direct consequence of this theorem, we shall show that any analytic function is differentiable to all orders and, moreover, that these deriva tives are given by integral formulae. So we see that despite its innocuous formulation, analyticity is a very strong property. Theorem 8. 7 (Cauchy's Formula for derivatives) Let D be a domain and suppose that f E H ( D) . Then f has derivatives of all orders
at every point in D. Furthermore, if D ( z0, R) � D, then f ( k ) (z0) , the k th_derivative of f at the point zo E D, is given by the formula [ f(w) dw f ( k ) (zo) = � 27ri Jr (w - zo) k +l uhere
r
0
D ( zo, R) which encircles zo and k E N. D, there is R > such that D(zo, R) � D.
i s any simple circle in
For any given zo E By theorem 8.6, f has a Taylor series ( power series ) expansion
Proof.
00
f (z) Lan ( z - zo) n , n=O =
valid for any z E D ( zo, R) . The series converges absolutely for every the disc D(zo, R) and an is given by
z in
an = -27ri1- J[r (w -f (w)zo) n+l dw.
0, 1 ,
This means that inside the disc D ( z0, R) , the function f is a power series. Any power series has derivatives of all orders, and so this is true of f ( in this disc ) . In particular, for any k 2, . . . , =
f ( k) (zo) k! a k . Substituting the integral formula for a k , as given by theorem 8.6, we obtain =
the required formula for the kth_derivative of f.
D
Remark 8 . 2 This should be contrasted with the real case where there is no such corresponding result. In fact, there are real functions of a real variable which are infinitely differentiable but whose Taylor series has zero radius of convergence. Indeed, the function
f (x )
=
{ 0� e
t
;-2" ,
x
0
� x > O
Cauchy's Theorem
143
provides j ust such an example. The Taylor Series Expansion Theorem says that any function analytic in any disc is expressible as a power series which is absolutely convergent in that disc. By definition, a domain, D say, is an open set and therefore any point in D is inside a disc also lying in D. Then any f E H(D) is expressible as an absolutely convergent power series in any such disc. We can say that locally f is a power series. Of course, by considering different points in D we are led to different discs and therefore to different power series. We can think of a domain as a (in general, infinite) collection of (often overlapping) discs. In the same way, we can think of an analytic function as a whole collection of power series, each absolutely convergent in some disc in the domain of analyticity of the function. It is important to appreciate that it is part of the Taylor Expansion Theorem that the power series expression for the analytic function f (about the point zo) is absolutely convergent in any disc D (zo, R) centred on zo and lying in the domain D. The coefficients of this power series are the derivatives of the function f (together with extra k! factors) and, in turn, are given by integral formulae. In particular, it is all part of the theorem that the series L� o J( k ) (zo) (z - zo) k /k! converges absolutely in the disc
D(zo, R) .
Finally, we note that these integral formulae for f and its derivatives at the point z0 only involve the values of f on the contour r. This means that the values of an analytic function (and all its derivatives) are determined by the values of the function possibly quite far away. For example, if f is entire, then we can take r to be a circle of arbitrarily large radius about any given point z0 . The value of f (and each of its derivatives) at this point zo is determined by the values of f on this giant circle. In particular, we cannot mess about with the values of f in a small region without spoiling analyticity. To put this another way, if two analytic functions happen to be equal on a circle (and are analytic in some disc which contains this circle) , then they necessarily agree everywhere inside the circle (because their values inside are given by integrals around the circle) . We will pursue this phenomenon later (see the Identity Theorem, theorem 8.12) . Example 8 . 4 We know that the function w f---4 Log w is analytic in the cut-plane C \ lR
144
Lecture Notes on Complex A nalysis
expansion about z = 0 which we are assured, by the theorem, converges 1 absolutely for every z in this unit disc. The derivative of f is -- and so l+z we may compute all the necessary derivatives to deduce that
z 2 z3 z4 zn Log(1 + z ) = z - - + - + . . . + ( - l ) n +l + . . . . 2 3 4 n _
_
An application of the ratio test confirms that this series converges absolutely for all z with l z l < 1 , as it should. An alternative derivation of this expansion of Log(1 + z ) is to note that 1 the power series converges absolutely for l z l < 1 and has derivative -1+z for I z I < 1 . This is also true of Log( 1 + z ) , so their difference is constant on the open unit disc. However, both functions take the value 0 when z = 0 and so they are equal throughout the disc D(O , 1 ) . Corollary 8 . 1 For f E H(D) , the real functions u (x, y ) Re f(x + iy) and v(x, y) = Im f(x + iy) have partial derivatives of all orders at any (x, y) such that x + iy E D. Moreover, these partial derivatives may be taken in =
any o rder {for example, Ux y x = Uxx y = U y xx) ·
Proof. We have seen that f is infinitely differentiable in D as a complex function of a complex variable. The complex derivative can be taken, in particular, in the "real direction" or in the "imaginary direction" . That is,
f' = ax (u + iv) = � ay (u + iv) , 2
which gives
f' = Ux + ivx = - iu y + V y and we get the Cauchy-Riemann equations differentiation by D2 , we have
as
before. Denoting complex
D z = ax = - iay where D2 is to be applied to f(z) and the partial derivatives are applied to the function u(x, y) + iv(x, y). Any occurrence of D z can be replaced by either ax or - iay , and vice versa. I t follows, by induction, that all partial derivatives of u and v exist and that
145
Cauchy 's Theorem
where L1 , j = 1, . . . , k, stands arbitrarily for either ax or - iay . If we choose any m of these terms to be equal to - iay , then we obtain the equality
Lk . . . L 2 L 1 ( u(x , y) + iv( x, y) )
=
=
( D; f) (z)
( - i) ma�-m a; ( u (x , y) + iv( x, y) ) .
Equating real and imaginary parts shows that the order in which the partial derivatives are taken is immaterial. For ex ample (with k 9) , =
ax ( -i ay ) a; ( - iay ) a: (u (x, y) + i v(x, y) )
=
( - i) 2 a�a; (u (x, y) + iv (x, y) )
so that
and 0 What ' s
going on? Under the initial assumption that f is analytic in a star domain D, it follows that f has a primitive there and so has zero integral around any closed contour in D. This leads to the Deformation Lemma, which in turn gives Cauchy's Integral Formula. From this, there follows the power series (Taylor series) expansion in discs. In particular, it follows that if a function is analytic in some domain (star-like or non star-like, with or without holes) , then in fact, it is infinitely-differentiable there. (The point here of course is that as far as analyticity is concerned , one can restrict attention to discs.)
8. 7
Morera's Theorem
Next , we consider a converse to Cauchy's Theorem.
Suppose f : D � C is continuous and that far f = 0 for every triangle T wholly contained
Theorem 8.8 (Morera's Theorem)
on the domain D in D. Then f E H ( D) .
Proof. Let D(zo, R) be any disc in D. Then, by hypothesis, far f = 0 for every triangle T wholly contained in D ( z0, R) . But the disc D ( z0, R) is a star-domain and so we deduce, j ust as in the proof of theorem 8.2 , that f has a primitive there, F, say. Hence f = F' on D ( z0, R) . But the analyticity of F implies, in particular, that F is twice differentiable in D(zo, R) . That is to say, F' is analytic in D ( zo , R) . However, F' f in D ( zo, R) and so f is analytic in D(zo, R) . We conclude that f E H ( D). 0 =
1 46
8.8
Lec ture Notes on Complex Analysis
Cauchy's Inequality and Liouville's Theorem
The following result provides a bound on the derivatives of an analytic function in terms of a bound on the function itself. Theorem 8.9 ( Cauchy's Inequality) Suppose that f is analytic in the disc D ( zo , R) and that l f (z) l :<:::; M for all z E D ( zo , R) . Then
) k l f < l (zo I :<:::;
M k!
Rk '
for any k = 0, 1 , 2, . . . . Let 0 < r < R and put 'Y( t ) = z0 + re 2 1r i t , 0 :<:::; t :<:::; 1 . By Cauchy's Integral Formula Proof.
k f ( ) ( zo ) =
� 27ri
1 (w -f(w)z0) k+ l dw . 7
However, for any w E tr 'Y, we have that l f (w ) l :<:::; so
I
M and
l w - zo l = r and
M f(w) < -k +l . k + l r ) z (w - o
I
Using this, we estimate the integral to get
This holds for any 0 < r < R, and the left hand side does not depend on r. Taking the limit r -> R gives
and the proof is complete.
D
Remark 8.3 There is no analogue of this for functions of a real variable. For example, let f(x) = sin(>.x ) . Then l f(x) l :<:::; 1 for all x E lR and every >., but f'(O) = >., which can be chosen as large as we wish.
147
Cauchy 's Theorem
Theorem 8 . 1 0 (Liouville's Theorem) If f is entire and bounded, then
f is constant. In other words, no entire function can be bounded unless it is constant. Suppose that f is entire and that l f(z) l .S M for every z E C. Then f is analytic i n C and s o has a Taylor series expansion about zo 0 which is valid for z in any disc D(O, R) , i .e., for all z , Proof.
=
f(z)
00
=
L an z n ,
n=U
.
With a n
=
J( n ) (O) n.
--1- .
Applying Cauchy's Inequality, theorem 8.9, to f in the disc D(O, R) , we find that
=
0, 1 , 2 , . . . . This holds for any R > 0 . Fixing n 2: 1 and letting we conclude that f(n ) (O) = 0 . It follows that f (z) = a0, that is, the function f is constant. 0
for any n
R --+
oo,
Example 8.5 Suppose that f is entire and satisfies l f (z ) l � 1 + lz l m for all z E C. It follows that lf (z) l � 1 + R m for all z E D(O, R) and so Cauchy's inequality implies that
for any n > m . So if f(z ) = l:::'= o a n z n is the Taylor series expansion of f about zo 0, then all the coefficients an with n > m vanish. In other words, f(z) is a polynomial of degree at most m .
as R --+
oo,
=
We can use Liouville's Theorem to give a fairly painless proof of the Fundamental Theorem of Algebra.
Every non constant complex polynomial p has a zero, that is, there is zo E C such that p(zo) = 0.
Theorem 8 . 1 1 (Fundamental Theorem of Algebra)
Proof. We may write p as p(z) = a n z n + an - I Z n - l + · · · + a 1 z + ao, where n 2: 1 and a n # 0. To show that p has a zero, we suppose the contrary and
obtain a contradiction. Thus p is entire and, assuming it is never zero, 1/p
148
Lecture Notes on Complex A nalysis
is also entire. However 1
p(z)
=
1
a n z n + a n- l z n - l + 1
(
·
·
·
+ a 1 z + ao
)
a a ao z n a n + n- 1 + + n-1 1 + -n z z z a1 1 a n- I ao As l z l � oo , a n + + + -+ -n � a n and so � 0. In nl p(z) z z z < 1 , whenever lz l > R. particular, i t follows that there is R such that p z) On the other hand, if p is never zero, then 1 /p is analytic and so is certainly continuous . In particular, 1/p is bounded on the closed (and so M compact) disc D (O, R) , that is, there is M > 0 such that p z) � , whenever l z l � R.
(
--
--
·
·
·
· ·
·
) ItI --
Combining these remarks , we may say that
t
p z)
--
ltl
i s entire and obeys
I M + 1 for any z E C. Hence by Liouville's Theorem, 1 /p is lconstant, p tz ) say 1/p = a . But then p = 1 / is also constant, a contradiction. �
a
We conclude that p does, indeed, possess a zero, that is, p(zo) = 0 for some
zo E C.
0
Let p(z) = an z n + + a1z + ao , with a n -1- 0, be a polynomial of degree n . Then there exist (t , . . . , (n E IC and a E IC su c h that Corollary 8.2
·
p(z )
=
·
·
a(z - (I ) . . . (z - (n )
for all z E C. {N.B. The (j s need not be all different.) In other words, any polynomial of degree n has exactly n zeros-counted according to their multiplicity. Proof. We shall prove this by induction. For each n E N, let Q (n) be the statement that any polynomial of degree n can be written in the stated form. Any polynomial of degree 1 has the form p(z) = az + b with a -1- 0. Clearly, such p can alternatively be written as p(z) = a(z - c) , where c -b/a. Hence Q ( 1 ) is true. Next we show that the truth of Q (n) implies that of Q (n + 1 ) . So suppose that Q ( n) is true. Let p be any polynomial of degree n + 1 , =
1 49
Cauchy 's Theorem
By the theorem, p has a zero, zo , say. Then p(zo)
p(z)
=
0 and so
p(z) - p(zo) an+ I Z n + I - an+ I Zon+ I + a n z n - an z0n + · + a i Z - ai zo + ao - ao a n + I (z n + I - z� + I ) + an (z n - z�) + · · · + a I (z - zo) (z - zo) {a n + I (z n + z n - I zo + · · · + z�) + a n (z n - I + z n - 2 zo + · · · + z� - I ) + · · + a 2 (z + zo) + a ! } = ( z - zo)q(z) =
=
·
·
=
=
·
where q(z) is a polynomial of degree n. By induction hypothesis, namely, that Q (n) is true, we can write q(z) as
q(z)
=
,B(z - (I ) . . . (z - (n )
for some ,B and (I , . . . , (n E C. It follows that Q (n + 1 ) is true. Hence, by D induction, Q (n) is true for all n E N. A further direct consequence of Liouville's Theorem is the interesting observation that the range of any non-constant entire function permeates the complex plane, C.
Suppose that f is entire and not constant. Then for any w E C and any c. > 0 there is some ( E C such that f (() E D(w, c.) . In other words, f assumes values arbitrarily close to any complex number.
Proposition 8 . 1
Suppose that w E C and c. > 0 are given. To say that there is no ( with f(() in the disc D(w, c.) is to say that l f(z) - wl 2": c. for all z E C. In particular, f(z) - w -I- 0 and so g 1 / (f - w) is entire. But then g obeys lg (z) l :<::::: 1 /c. for all z E C and so is constant, by Liouville's Theorem. It follows that f - w and hence also f is constant. The result follows. D
Proof.
=
8.9
Identity Theorem
We have seen, in theorem 4.5, that if power series agree at a sequence of points converging to the centre of a common disc of convergence then they are, in fact, identical. In view of the results above, to the effect that analytic functions can be thought of as families of power series based on overlapping discs of convergence, it will come as no surprise that this theorem has an extension to functions analytic in a domain.
Lecture Notes on Complex A n alysis
150
Recall that the complex number ( is a limit point of a set S if and only if there is a sequence (zn ) n EN in S with Zn -:J- (, such that Zn � (, as n � oo. That is, ( is the limit of some sequence from S \ {( } .
Suppose that D is a domain and that f, g E H(D) . Suppose that there is some set S <; D such that S has a limit point which belongs to D and such that f = g on S. Then f = g on the domain D .
Theorem 8 . 1 2 (Identity Theorem)
Proof. Set h = f - g . Then h E H(D) and h(z ) = 0 for all z E S. We must show that h(z) 0 for all z E D. By hypothesis, there is some point (o E D which is a limit point of S. Hence there is a sequence (zn ) n EN in S with Zn -:J- (0 , n E N, and such that Zn � (0 , as n � oo. Each h(zn ) = 0 and , by continuity, h( (o) = 0. Let Z denote the zeros o f h, Z { z E D : h(z) = 0 } . Clearly, S <; Z and we have j ust shown that (o E Z and, moreover, (0 is a limit point of Z . We will appeal to the connectedness of D. To this end, let =
=
A = { z E D : z is a limit point of Z } and
B = D \ A.
As noted above, (0 is a limit point of Z and so belongs to A. In particular, A -:J- 0 . It is also clear (by continuity) that h vanishes on A. We shall show that A is open. Indeed, suppose that ( E A. Since ( E D, there is r > 0 such that D( (, r ) <; D. By theorem 8.6 , h has a Taylor series expansion, absolutely convergent in D((, r) . Since ( E A, it follows from theorem 4.5 that h vanishes in the whole disc D( ( , r) . But then each point w of the disc D (( , r) is clearly a limit point of Z (for example, take a sequence moving radially outwards towards w) which implies that D( ( , r) � A. It follows that A is open. We wish to show that B 0, so that A = D. Suppose that B -# 0 . We shall show that B i s open. Let w E B . Then w E D and s o there is R > 0 such that D( w, R) <; D. However, w is a member of B and so is not a limit point of zeros of h. This means that there is some 0 < p < R such that h(z) -:J- 0 for all z E D with 0 < lw - zl < p , i.e. , h cannot vanish in some punctured disc around w. (Otherwise, h would vanish at some point, distinct from w , in every disc D(w , r) , 0 < r < R, which would force w to be a limit point of zeros of h.) In particular , D(w , p ) <; B and it follows that B is open. =
151
Cauchy 's Theorem
Now, we appeal to the connectedness of D. We have seen that A and B are both open, they are disjoint and D A U B. Hence one or the other must be empty. We know that A =J. 0 and so we deduce that B 0 . Hence A = D and so h vanishes on D ( because it vanishes on A). =
=
Alternative Proof This version uses directly the pathwise connectedness of D. As above, we wish to show that h f - 9 vanishes on the domain D . As before, let Z denote the set of zeros of h and let A denote the set of limit points of Z in D. By hypothesis, A is not empty, so let z0 E A and let w be any point of D. We shall show that h(w) = 0. To see this, first note that h(z) 0 for any z E A. This is simply because h is continuous and so Zn -+ z means that h ( zn ) -+ h(z) and therefore h(z) 0 if each h(zn ) = 0. Next, we note that since D is pathwise connected, there is some path cp : [a, bj -+ C joining zo to w in D. Let 9 : [a, bj -+ { 0, 1 } be given by =
=
=
9 (t)
=
{
0, if cp(t) E A 1 , if cp(t) ¢ A.
Evidently 9(a) 0 because cp(a) = zo E A. We shall show that the real valued function 9 is continuous on the interval [a, bj . Let to E [a , bj be given and let wo cp(to) . There are two possibilities. =
=
C ase ( i ) : w0 E A. This means that 9(t0) 0. Since wo E D there is some r > 0 such that D(wo , r) � D. Now, h(z) is analytic in the disc D(wo , r) and so has a Taylor series expansion about wo valid for all z E D(wo, r) . But wo E A, so by the Identity Theorem for power series, we conclude that h vanishes throughout D(wo , r) and therefore every point of D(w0 , r) is a limit point of Z. Now, cp is continuous at to, and therefore there is some b > 0 such that cp(t) E D(wo , r) whenever l t - to l < b and t E [a, b] . That is, cp(t) E A and so 9(t) 0 for all such t. It follows that 9 is continuous at to. (9 is constant in a neighbourhood of to.) =
=
Case ( ii ) : wo ¢ A. This means that wo is not a limit point of zeros of h ( but this does not preclude the possibility that h(wo) 0) . It follows that there is some p > 0 such that h is never zero in the punctured disc D' (wo , p) ( otherwise wo would be a limit point of zeros of h ) . Furthermore, by construction, 9(to) 1 . Once again the continuity of cp at to means that there is b' > 0 such that cp ( s ) E D( w0 , p) whenever =
=
152
Lecture Notes on Complex A nalysis
I s - t o l < fJ' and s E (a , b] . However, for any such s, either rp(s ) wo, in which case g (s) = 1 (because rp(s) wo � A) or rp(s) E D' (wo , p ) and so h(rp(s)) -I- 0 and therefore g (s) = 1 because rp(s) is certainly not a limit point of zeros of h. It follows that g is continuous at to. We see then, that g is continuous at each t0 E (a, b] . Now, g only assumes at most two values, either 0 or 1 . By the Intermediate Value Theorem, it follows that g is constant. But g (a) 0 and so g (t) = 0 for all t E (a , b] Therefore g (b) 0 and so w = rp(b) E A and, in particular, h(w) = 0. This =
=
=
.
=
0
completes the proof.
Remark 8.4 In practice, the set S is usually a line segment, or a disc or part of a disc or some similarly simple geometric region. This result expresses a remarkable rigidity property enjoyed by analytic functions. For example, suppose that f is entire. Suppose that g is also entire and that g ( z ) = f ( z ) for z in, say, the very small and very remote line segment S = (10 66 , 10 66 + 10-999] . Nonetheless, by the theorem, it follows that g ( z ) = f (z ) throughout the whole complex plane.
The complex functions exp z , cos z and sin z are the only entire functions which agree with their real counterparts exp x, cos x and sin x, respectively, on the real axis. That is, if f is entire and f ( x) exp x for all x E JR, then f (z) = exp z for all z E C, and similarly for cos z and Corollary 8 . 3
=
sin z .
Proof.
Take S =
lR
and apply the theorem.
0
Remark 8 . 5 Suppose that D is a domain and suppose that f E H(D) is not identically zero on D. Let S denote the (possibly empty) set of zeros of f ;
S = { zeros of f } = { z E D : f (z ) = 0 }.
Then S can have no limit points in D--otherwise, by the Identity Theorem, f would be zero throughout D. In other words, the zeros of f are isolated: if zo E D is such that f (zo) = 0 then there is r > 0 such that f has no other zeros in the disc D(zo, r) . Definition 8.2 Suppose that f is analytic in a domain D. The point zo E D is said to be a zero of f of order m (with m � 1 ) if f (zo) = 0 and the k Taylor series expansion of f about zo has the form f (z ) = L�= m ak ( z - zo ) where a m -I- 0.
Cauchy 's Theorem
153
This is equivalent to demanding that the derivatives f(zo ) , f'(zo ) , . . . , f ( m - Il (zo ) all vanish, but f ( m l (zo ) # 0. Note also, that according to the discussion above, the Identity Theorem implies that either f vanishes throughout D or every zero of f is isolated and so has some (finite) order. (In the latter case, the Taylor series cannot be the zero series and so must have at least one non-vanishing coefficient. ) A s the next example shows, the set o f zeros may well have a limit point not belonging to the domain. Example 8.6 Let D = C \ { 0 } , the punctured plane and let f(z) sin(1 /z) for z E D. Now, sin( 1 /z) = 0 whenever 1 / z = k7r for some k E Z, i . e. , when z = 1 / (br) , k E Z \ { 0 } . Let S = { z : z 1 / (krr) , k E Z \ {0} } . We see that S � D , f vanishes on S and that S has (the single) limit point 0. However, f does not vanish on the whole of D. This does not contradict the Identity Theorem because the limit point 0 does not belong to the domain D. =
Recall that a set is said to be countable if it has a finite number of elements (or is empty) or if its elements can be listed as a sequence (that is, can be labelled by N) . For example, the sets N, Z and Q are countable, but one can show that the sets (0, 1 ) and lR are not.
Suppose that D is a domain and that f E H(D). Then either f vanishes throughout D or the set of zeros of f is countable.
Theorem 8 . 1 3
Suppose that f is not identically zero on D and let Z = { z E D : f (z) = 0 } denote the set of zeros of f in D. Let K � D be compact. We show that K n Z is either empty or has only a finite number of elements. Indeed, suppose that K n z is an infinite set. Let W I E K n z . Now let w 2 E (K n Z) \ { WI } and let w3 E (K n Z) \ { WI , w 2 } . Continuing in this way, we construct a sequence (wn ) in K n Z such that Wn # Wm for any n # m. (This construction works because (K n Z) \ { WI , w 2 , . . . , Wn } is not empty.) Now, K is compact and so (wn ) has a convergent subsequence, Wn k ---+ (, say, with ( E K . But each Wn k is a zero of f and so ( is a limit point of zeros in D. By the Identity Theorem, this is impossible and so we conclude that K n Z cannot contain infinitely-many elements. To complete the proof, we observe that by proposition 3.6, the set D has a compact exhaustion, D U:'= I Kn where each Kn is compact. B ut then Z = U := I (Kn n Z) which is the union of a sequence of finite (or empty) sets and so is countable. 0 Proof.
=
Lecture Notes on Complex Analysis
154
8.10
Preservation of Angles
Consider a path 1 and a point zo on the path. Let us say that 'Y makes an angle 0 with respect to the positive real direction at the point zo = r (t0) if r( t) =I- zo for sufficiently small t - to > 0 and if
r (t) --zo--, ----> e ·& ir (t) - zo l
.,...--'-: --
as
t l to .
•
The angle between two paths 12 and 11 , each passing through is then 82 - 81. where o1 is the angle 11 makes with the positive real direction at zo, for j 1 , Now suppose that f is analytic in some disc D(zo, r ) and is not constant there. Then we know that f has a Taylor series expansion
zo
=
2.
f(z)
00
=
L an (z - zo t . n=O
By hypothesis, f is not constant and so there must be some non-zero an with n � 1 . Suppose that a m is the first such non-zero coefficient so that the Taylor series for f has the form 00
00
f(z) = L an (z - zo) n ao + L an (z - zo t f(zo) + (z - zo) m g(z) n=m n =O where m � 1 and g(zo) = a m =/=- 0. Let r1 (t) = J ( r1 (t)) for j = 1 , Since f is not constant, the Identity Theorem implies that fj (t) =I- f(zo) for all sufficiently small t - to > 0 (where zo = r (to)). Hence f( rj (t)) - f(zo) rj(t) - f(zo) i fJ (t) - f(zo) l lf h1 (t)) - f(zo)l ( /j (t) - zo ) m g( rj (t)) - lrJ (t) - zo l lg(rJ (t)) l g(zo) as t l to, ( ei0i ) m lg(zo) I' = eim oi ei Arg g(zo) =
=
2.
---->
and so r1 makes an angle mOj + Arg g(zo) with respect to the positive real direction at f(z0). But then this means that the angle between f 2 and f1 at f(zo) i s m( 02 - 01) .
Cauchy 's Theorem
155
In other words, if the derivatives j(r) ( zo } = 0 for all r = 1, . . . , m - 1 , but f( m ) ( zo } =/:- 0, then the angle between paths intersecting at zo is multiplied by m under f. In particular, if f' ( zo) =/:- 0, then f preserves the angle between intersecting paths. ( Such maps are said to be conformal. )
Chapter 9
The Laurent Expansion
9.1
Laurent Expansion
In this chapter, we discuss a generalization of the Taylor series expansion. Rather than considering analyticity in some disc, we assume only analyticity in an annulus. This leads to a representation in terms of both positive and, possibly, negative powers of z, the Laurent expansion. The starting point is Cauchy's Integral Formula for the given function f. Here the circle of integration can be deformed, as in the Deformation Lemma, to produce , in fact, two contour integrals. Each of these is the integral of f(w)/(w - z) around a certain circle. The idea is then to expand 1 / ( w - z ) in powers of z - zo , in one case , and in powers of 1 / (z - zo ) in the other. Doing the integrals gives the coefficients and all that remains is to worry a bit about the convergence of the two series thus obtained. Now for the details. 9 . 1 Suppose that f is analytic in the annulus A A(zo ; R1 , R2 ) = { z : R1 < l z - zo l < R2 } . Let z E A, and let R1 < r1 < l z - zo l < r2 < R2 . Then
Lemma
27rif(z)
=
r
f(w)
r
f(w)
dw + }02 w - z lc, w - z
dw,
where c1 is the circle zo + r1 e 2 rrit and c2 is the circle zo + r2e 2 rr it ' 0 :::; t :::; 1 . We argue j ust as in the proof of the Deformation Lemma, lemma 8. 1 . Insert line segments between the two circular contours, as shown in Fig. 9. 1 . The idea is to put in sufficiently many cross-cuts so that each of the contours /j is so narrow that it is contained in the star-domain DJ which is itself inside the annulus A. Clearly, if r2 - R1 is small, there will need to be many such cross-cuts. We number the /j S so that 11 goes around the point z , as indicated. Proof.
1 57
158
Lecture Notes on Complex Analysis
Fi g . 9. 1 Construct cross-cuts to g ive many narrow simple closed contours ')'1 , /'2 , . . . , "Yn . The poi nt z is encircled by /'1 ·
Next, let r be a circle around z with sufficiently small radius that it is encircled by 11 , as shown in Fig. 9.2.
Fig . 9 . 2
Star- like domains D i contai nin g "Yi .
Using Cauchy's Integral Formula and arguing as in the proof of the Deformation Lemma, lemma 8 . 1 , we have
.
2m f (z) =
1 f (w ) dw j"' wf (w- )z dw. w-z r
--
=
--
159
The Laurent Expansion
(w )
, Dn are star-like and wf - z is analytic in Furthermore, since D2 , D3 , each of these domains, it follows, by Cauchy's Theorem for a star-domain, theorem 8.4, that •
1
1'2
•
•
f (w) dw = . . . -Z
W
=
1
f (w) dw = 0. -Z
"'n W
Piecing all these together, and using the fact that the contour integrals along the cross-cuts cancel out, we get
1
1
1
f (w ) d + . . . + f (w) dw + f (w ) dw 1'2 - Z w -Z -Z ) dw + r f (w) dw . = r wf (w lc2 - z }0, w - z
l'l W
W
"'n W
We conclude that
27ri f (z)
=
r 2 wf (-w )z dw - Jrc, wf (-w )z dw,
}0
D
and the proof is complete. As mentioned earlier, the next step is to expand powers. We use the formula
1/ ( w - z ) into suitable
1 = 1 + o: + o:2 + + o: k -1 + -Q: k -· · · 1 - o: 1 - o: valid for any complex number and w =/:- zo, z =/:- zo , we have
1 w-z
o: =1- 1 and any k E N.
Indeed, for any n E N,
1 1 w 1 - zo)/(w - zo)) zo) w zo (z zo) ( (z ( ( ) n 1 z - zo ( z - zo ) - 1 + ( z - zo ) n - (w - zo ) w - zo + . . . + w - zo (w - zo) n - 1 (w - z ) n 1 1 + (z - zo ) + . . . + ( z - zo ) + ( z - zo) n ..,..----.,. .. (w - zo ) (w - zo ) 2 ( w - zo) n (w - zo ) n (w - z ) Sn (zo , z , w ) , say.
(
-:----'----.----c-'----,.,..
=
)
160
Lecture Notes on Complex Analysis
Similarly, for any m E N and w =/:- zo and z =/:- zo , we have 1
1
-1
- --
w-z
(w - zo ) - ( z - zo ) ( z - zo) - ( w - zo) 1 (w - zo) . . . (w - zo)m- l (w - zo)m + + + -:--'-----:-m--7-,----...,.- + - ---:(z - zo) (z - zo) 2 (z - zo) m (z - zo) (z - w ) Sm (zo, w, z). =
=
------,----,-
Notice that the first expression involves positive powers o f z - zo, whilst the second involves negative powers of z - zo . Why should we be interested in both cases? The point is that lw - zo l < lz - zo l whenever w belongs to the inner circle C1 , whereas the reverse inequality holds when w belongs to the outer circle c2 . This means that wz- zoo < 1 in the first case -z so that Sn (zo, z , w) converges as n ---> oo . On the other hand, if w is on the outer circle c2 , then �=:z < 1 and so the second expansion, Sm (z0 , w , z) , the one with negative powers of (z - zo) , will converge as m ---> oo . Applying these considerations, together with lemma 9. 1 , leads to the Laurent expansion, as follows.
1
I
1
I
Theorem 9 . 1 (Laurent Expansion)
Suppose that f is analytic in the annulus A = A(zo; R1 , R2) . Then, for any z E A, f(z)
=
00
00
n=O
n== l
L an (z - zo t + L bn (z - zo) - n
with dw, an = -12ni- Jcr (w -f(w) zo) n + l and bn
=
1
27rZ .
for n = 0, 1 , 2, . . ,
1 (w - zo) n - 1 f (w) dw C
.
=
1
27rZ .
1
C
f(w) dw, ( W - ZQ) - n + l
for n = 1 , 2, . . . , where C is any circle encircling zo such that tr C � A . Furthermore, both of these series are absolutely convergent in the given annulus A. Let z E A = A(z0; R1 , R2 ) be given and let r1 and r2 satisfy the inequalities R1 < r 1 < l z - zo l < r2 < R2 . Let C1 and C2 be the circles zo + r1 e 2 1l"it and zo + r2 e 2 1l"it, for 0 ::::; t ::::; 1, respectively. Then using lemma Proof.
1 61
The Laurent Expansion
9 . 1 , together with the preliminary discussion above , we see that given by
f (z )
= =
=
f (z)
is
f (w) dw - � r f (w) dw 21rz 1c, w - z c2 w - z � dw + � { f (w) Sm (zo , w, z ) dw 21rz 1c, 2m 1{c2 f (w ) Sn ( zo, z, w ) + + + ao a 1 (z - zo ) · · · an - 1 (z - zo t - 1 + Pn + b 1 + · · · + bm m + Q m (z - zo ) (z - zo )
� r 21rz 1
where the as and bs are as stated in the theorem, and where the remainder terms Pn and Q m are given by
and 1 r (w - zo ) m f (w) Q m - 27ri 1c, (z - zo ) m (z - w ) dw. _
We wish to show that Pn ----> 0 as n ----> oo and that Q m ----> 0 as m ----> oo . The traces of the circles C1 and C2 are compact sets in C and therefore there is some M > 0 such that J f (w) J � M whenever w E tr C1 U tr C2 . Furthermore , since z rJ_ tr c1 u tr c2 , it follows that there is some (j > 0 such that D ( z, fJ) n (tr C1 U tr C2 ) = 0 (because z E C \ (tr C1 U tr C2 ) , an open set) . Hence Jw - zJ � fJ for every w E tr C1 U tr C2 . (One could alternatively apply proposition 3 .7 to reach this conclusion.) We can use this to estimate IPn l and IQ m J , as follows.
JPn J as n ----> oo ,
1
�
27r
---->
0,
Jz - zo J n M 21rr 2 r2n (j
Jz - zo l
since !..._ =-c < 1 . Similarly, T2 1 r m1 JQ m J � _
27r J z - zo J m
---->
0,
M
7 u
21rr 1
162
as
Lecture Notes on Complex Analysis
m � oo ,
then
since
m � oo ,
rt
l z - zo l
< 1
·
Taking the limit, say,
n � oo ,
first, and
it follows that
f(z)
=
00
00
k=O
j= l
L: a k (z - zo ) k + L bi ( z - zo)- i ,
as claimed. By an argument as in the proof of the Deformation Lemma, lemma 8. 1 , we see that the a n and bn integrals are independent of the particular circles Ct and C2 , provided that they lie in A and encircle z0• To show that the two series are absolutely convergent in A, we simply estimate the general term in each case. We have
and
By the Comparison Test, both series converge absolutely.
0
Definition 9 . 1 The (double) series constructed above for the function f, analytic in A, is called the Laurent series expansion of f in the annulus A(zo; Rb R2 ) . The series of negative powers, E:= l b n (z - zo) - n is called the principal part of f.
Our next task is to establish the uniqueness of the Laurent expansion. After all, it is not obvious that it is not possible to change some (or all) of the an s and some (or all) of the bns without affecting f.
The
0
Laurent Expansion
163
z0 + re 2 1rit , for As a preliminary observation, we note that if "Y (t) � t � 1 , is the circular contour centred at zo and with radius r, then =
11
dz
27ri ..., (z - zo) m
=
{0,
1,
m
E Z,
m
= 1.
m
-:/:- 1
This is a consequence o f corollary 7.2, since (z - zo) - m has a primitive in C \ {zo } provided m -:/:- 1 . 9.2
Uniqueness o f the Laurent Expansion
Theorem 9 . 2 (Uniqueness of Laurent Expansion) is analytic in the annulus A = A(zo ; R1 o R2 ) and that
f (z)
00
=
Suppose that f
00
L an (z - zo) n + L f3n (z - zo) - n n=O n= l
where each of the two series converges absolutely in A. Then an an and f3n = bn for all n, where the a n s and bn s are the Laurent series coefficients (as given by the integral formulae in theorem 9. 1). =
0
be given and set Rn (z) = E� n + l ak (z - zo) k and set Tn (z) E� n + l f3k (z - zo)- k , for n E N. The series for Rn converges absolutely in A, i.e . , for z with R1 < lz - zo l < R 2 . It follows that this series converges for all z with lz - zo l < R 2 . In other words, Rn is analytic in the disc D(zo , R2 ) , and, in particular, Rn is analytic in A. However, Proof.
Let E > =
Tn (z)
=
f(z) -
n
n
k=O
k= l
L ak (z - zo) k - Rn (z) - L /3k (z - zo) - k
0
and so Tn is analytic in the annulus A since this is true of the right hand side. In particular, if R 1 < r < R2 and "Y (t) = zo + re 2 1rit , � t � 1 , then both Rn and Tn are continuous on tr "Y· Furthermore , E � o la k l r k and E� 1 1 /3k l r- k both converge and so there is N E N-such that 00
L la k l r k < E
k=n+ l
00
and
L lf3k l r- k < E
k=n+ l
1 64
Lecture Notes o n Complex Analysis
whenever
n
> N. It follows that, for any z E tr "'( (so that lz - zo l
I Rn (z) l �
00
L
k=n+l
=
r) ,
la k l r k < £
and that
I Tn (z) l
�
00
L
k =n +l
whenever n > N . Finally, let m E Z be given. Choose
I .Bk l r- k <
n
c
> lml . We have
n
n k f(z) = L ak (z - zo ) + L .Bk (z - zo)- k + Rn (z) + Tn (z ) . k= l k=O Hence 1 _
27ri
1
-y
}
{
O! m , if m � 0 f(z) 1 dz + (z - zo) m+ l - .B tml • if m < 0 27ri _
But the left hand side is j ust Furthermore ,
1 Rn (z) + Tmn+(z)l dz . "I
(z - zo )
{ b�:;. i[r� � � } , according to their definitions.
1-1- 1 Rn (z) + Tn (z) dz l -< 2._ � 27rr ' 27ri 271" r m + l (z - zo) m+ l "I
for all
n
> N,
2£
=
This is an estimate for the modulus of the difference between a m and O!m or between b t m l and .Bt m l • depending on whether m � 0 or not (and assuming that n is chosen greater than both N and lml ) . Since this holds for any given £ > 0 , we conclude that O! m = a m for m = 0, 1 , 2, . . . and .Bm = b m 0 for m = 1 , 2, . . . , as required. Remark 9 . 1 Suppose that r1 < r < R < R1 and that f is analytic in the annulus A(zo ; r1 , R l ) . Then the uniqueness of the Laurent expansion implies that the coefficients in the Laurent expansion of f in the annulus A(z0; r, R) are precisely the same as those in the Laurent expansion of f in the annulus A(zo ; r1 , R1) .
165
The Laurent Expansion
What ' s
going on? A function f analytic in an annulus can be written as a sum of a series of positive powers and a series of negative powers. Both series converge absolutely in the annulus and the coefficients can be expressed in terms of the function f by means of certain contour integrals. This is the content of the Laurent Expansion Theorem . The absolute convergence of these series is part of the theorem. The uniqueness of the Laurent expansion means that whenever and however one manages to write f as a sum of absolutely convergent powers and inverse powers, then this has to be the Laurent expansion.
Examples
9.1
( 1 ) For any z =/:- 0, the power series definition o f the sine function gives 1 1 1 1 1 + + 5 1 10 - sin 2 = 2 - 7 .1 z 14 3 1. z 6 z z .z
·
·
·
which is the Laurent expansion about z = 0 of the function sin(1/z 2 ) , valid for z in the punctured plane C \ {0} . (2) f (z) = 1 / (z - a) is analytic in the annulus { z : l z - al > 0 } . The Laurent expansion of f about z = a, valid for l z - al > 0, is simply f(z) =
1
. z-a
The function f is also analytic in the annulus { z : l z l > lal 1 1 a a2 1 a f(z) = = - 1 + - + -2 + . . = - + -2 + z ( 1 - a/ z ) z z z z z
(
·
)
} . We find a2
z3
+
·
·
·
is the Laurent expansion of f about z = 0, valid for l z l > l a l . For z in { z : 0 < l z l < l al } , we find that 1 1 f ( z) = --- = - -�---..,.-:z-a a( 1 - z / a ) is the Laurent expansion of f about z = 0, valid for z with 0 < l z l < l a l (where we suppose that a =/:- 0) . This i s also valid for z = 0 . Note that these series converge absolutely for the given ranges o f z . ( 3 ) The function f(z) = z (e 1 f z - 1) i s analytic i n the punctured plane C \ { 0 } . For any z =/:- 0, we have f(z) = z
( (1
1
1
+ - + -- + z 2., z 2
+ ...) - 1 3 ., z 3
1
1 1 1 + -+ =1++, , , 2 4. z3 . . . 2. z 3. z which is the Laurent expansion of f about z = 0.
)
166
Lecture Notes on Complex Analysis
Suppose f is analytic and bounded in the punctured disc D'(zo, R) . Then there is a function F analytic in the whole disc D(zo, R) such that f F on D'(zo, R) . In other words, f can be extended so that it is defined at zo in such a way that the resulting function is analytic (in the whole disc).
Proposition 9 . 1 =
The hypotheses imply that f has a Laurent series expansion in the annulus A(zo; r, R) for any 0 < r < R. In particular, we can calculate the coefficients of the principal part of the Laurent expansion as
Proof.
j (z - zo) m - l f (z) dz zo + pe 2 1rit for 0 t 1 and p can be chosen arbitrarily in bm
=
1 -. 2 7!"2
'Y
where 1 (t) � :::::; the range 0 < p < R. By hypothesis, there is M > 0 such that D'(zo, R) . Hence, for any m 2:: 1 , =
lbm l
::=;
1 mP 1M 27rp 27!"
=
lf(z) l :::::; M for all z in
M pm .
This holds for any 0 < p < R and so it follows that b m 0 for all m 2:: 0 . The principal part o f the Laurent expansion of f vanishes and we have =
f(z)
=
00
L an (z - zo)n ,
n =O
for
z E D ' (zo, R) .
This series converges absolutely for all 0 < l z - zo l < R and so certainly for all l z - zo l < R. Hence, we may define F on the disc D(zo , R) by the power series
F(z)
00
=
l:::>n (z - zot .
n =O
The absolute convergence implies that F is analytic in D(zo, R) . Clearly F = f on the punctured disc D'(zo , R) (and F(zo) = ao). 0 What's going on? If f is analytic in a punctured disc in which it is also bounded, then, by suitable adjustment of f at the centre of this punctured disc, one finds that f is really just the restriction of a function analytic in the whole disc to the punctured disc. This is because one can estimate the coefficients occurring in the principal part of Laurent expansion and show that they must vanish. In other words, f is just a sum of positive powers, that is, it is a power series and we know that power series are analytic. We will need this result later.
Chapter 1 0
S i ngularit ies and Meromorphic Functions
10.1
Isolated Singularities
We can think of the principal part of the Laurent expansion of a function f analytic in a punctured disc as encapsulating how badly behaved f is near to the centre. There are essentially only three situations; all coefficients in the principal part vanish, only a finite number of coefficients are non-zero, or an infinite number of coefficients are non-zero. Definition 1 0 . 1 A point zo E C is said to be an isolated singularity of a function f if there is R > 0 such that f is analytic in the punctured disc D'(zo , R) but f is not analytic at zo . Examples 1 0 . 1
0 i s an isolated singularity o f the function f ( z ) � ( 1 ) The point zo This function is not defined at 0 , so is certainly not analytic there. In fact, it is not possible to assign a value at this point so that the resulting (extended) function is analytic. If this were possible, say, with f(O) defined to be a:, then the resulting function would have to be continuous at 0 and, in particular, bounded in some disc around 0. This n is evidently false, as is seen, for example, by observing that f(z) when z 1 /n. (2) Suppose f : C ----> C is defined by f(z) z for z =1- i and f( i ) 3. Then z i is an isolated singularity of f . (f is not even continuous at z i.) 0 whenever z 1 /br with (3) Let f (z) 1 / sin( 1 /z) . Now, sin(1/z) k E Z \ {0} and so f is certainly undefined at such points. However, for any given k E Z with k =1- 0, f is analytic in the punctured disc D' ( 1 jk1r , r ) provided r is sufficiently small (depending on k) . Therefore, for any k E Z with k =1- 0, the point z 1 /br is an isolated singularity =
=
=
=
=
=
=
=
=
=
=
167
=
Lecture Notes on Complex Analysis
168
of f. f is not defined for z 0 and f is not analytic in any punctured disc D'(O, r ) for any r > 0 (because any such disc will contain points of the form 1 /k1r) . Evidently z 0 is a singularity of f (in that f is not analytic there) but according to the definition, z = 0 is not an isolated singularity of f. (4) The principal value logarithm, f(z) = Log z , is analytic in the cut plane C \ { z : z + l z l = 0 } . Each point of the negative real axis { z : z + l z l = 0 } is a point of discontinuity of Log z but none of these points are isolated singularities. Note that Log z is not defined for z = 0 but is defined in the punctured plane C \ {0}. =
=
Let zo be an isolated singularity of the function f. Then, by definition, f is analytic in some punctured disc D' (zo, R) . It follows that f has a Laurent expansion valid for all z in this punctured disc: f (z)
=
00
l:::>
n=O
n n (z - zo) +
00
L )n (z - zo ) -n .
n= l
Definition 10.2
(i) If all bn 0 , then z0 is said to be a removable singularity. (By defining or redefining f at z0 to be equal to ao , we get a function analytic in n the whole disc D(zo , R) , namely, L: �= O an (z - zo ) .) (ii) Suppose that only a finite (but positive) number of the bns are non zero; say, b m -:f. 0 , but bn 0 for all n > m. Then zo is said to be a pole of f of order m. (Sometimes one uses the terms simple pole or double pole for the cases where m = 1 or m = 2, respectively. ) (iii) If an infinite number of the bns are non-zero, then zo is said to be an isolated essential singularity of f. =
=
Examples 1 0 . 2
(1) Consider the function f (z)
=
cos z - 1 z2
=
z22, + z 4, (1 - 4 ·
1
·
z2
-
z2 z4
z66,
·
+ ...) - 1
= -, + , - , + . . . 4. 6. 2.
Evidently z
=
0 is a removable isolated singularity.
169
Singularities and Meromorphic Functions
(2) Consider f ( z)
We see that z
(3) The function
=
=
zsin z = z3
z 3 + 5T zs
31
z3
-
...
1 z2 1 + = 2 3! 5T z
·
· ·
0 is a double pole.
f(z)
=
1
exp - = z
1 1 1 1 + - + 21 2 + 1 +... z .z 3. z -
3
has z = 0 as an essential singularity. (4) Let
f (z) =
(sin z) 4 4 + cos z , z
for z E
10.2
Suppose that f E H ( D'(zo , R) ) . Then z0 is a pole of if and only if the limit lim z _.z0 ( z - zo) m f (z) exists and is non
Theorem 1 0 . 1
order zero.
m
Suppose first that zo is pole of f of order m. This means that b m is the last non-zero coefficient (of the power (z - z0 ) -m ) in the principal part of the Laurent expansion of f about zo (bn = 0 for all n > m ) . Evidently, Proof.
1 70
Lecture Notes on Complex A nalysis
required. Conversely, suppose that lim z -+ zo ( z - zor f (z) = a '# 0. Let F be given by F(z) = (z - zo) m f(z) for z E D' (zo, R) . Since f and therefore F is analytic in this punctured disc, it follows that F has a Laurent expansion as
F(z)
=
00
00
n =O
n= l
L An (z - zo) n + L Bn (z - zo) - n
valid (absolutely convergent) in D'(zo, R). However, by hypothesis, we have that F(z) -+ a as z -+ zo and so F is bounded in the neighbourhood of zo, say, in D'(zo , r) . (IF(z) l is close to lal if z is sufficiently close to zo.) It follows that Bn = 0 for all n E N (as in proposition 9.1). Hence F(z) = E::'= o An (z - zo) n and 0 -:f.
a =
lim
z -�o zo
F(z)
=
Ao .
Therefore
so that
Ao A1 f(z) = ( Z - Zo ) m + ( Z - zo ) m _ 1 + · · · + A m + Am + I (z - zo) + . . . for any z E D'(zo, R) . Since Ao '# 0, we see that zo is a pole of f of
order
= a
m.
0
1 - is undefined at those points z ez - 1 for which e z 1 , i.e. , when z 2k7ri for k E Z . f is analytic everywhere else and so these are isolated singularities. Fix k E Z and set w z - 2k7ri. Then f(z) f(w + 2k7ri) 1/(ew + 2 k1ri - 1) 1/ (ew - 1 ) . Using the power series expansion of ew , we see that (ew - 1)/w -+ 1 as w -+ 0 and so we deduce that (z - 2k7ri) f (z) -+ 1 -:f. 0 as z -+ 2k7ri. It follows that for each k E Z, z 2k7ri is a simple pole of f. Proposition 1 0 . 1 Suppose zo is an isolated singularity of f and suppose that l f(z) l -+ oo as z -+ zo . Then zo is a pole of f {so f is meromorphic at zo). Conversely, if zo is a pole of f , then lf(z) l -+ oo as z -+ zo . Example 1 0 . 3
The function f(z)
=
=
=
=
=
=
=
=
Suppose l f(z) l -+ oo as z -+ zo . Then, for any given M > 0, there is R > 0 such that f E H(D' (zo, R)) and lf(z) l > M for all z E D'(zo, R) . In particular, f(z) -:f. 0 on the punctured disc D'(zo , R) . It follows that g(z) = 1/ f(z) E H(D'(zo, R)) and g satisfies l g (z) l < 1/M on D'(zo , R) .
Proof.
171
Singularities and M eromorphic FUnctions
By proposition 9. 1, we see that zo is a removable singularity of g and that g can be written as (using lim z -+ zo g ( z) 0) =
g(z)
=
A 1 (z - zo) + A2 (z - zo) 2 + . . .
for any z in D'(zo , R) . Furthermore, g is not zero in this punctured disc (because g(z)f(z) = 1) and therefore not all of the Ans are zero. That is, there is m � 1 such that Am # 0 but An 0 for all n < m . It follows that =
t
f z)
=
g(z)
=
(z - zo) m (Am + Am + l (z - zo) + . . . )
and so
1 -:----:----:-:--:- ---+ (z - zo) m f(z) A m ---+
�
z ---+ zo, i.e. , (z - zo) m f (z) # 0. By the theorem, theorem 10.1, A it follows that zo is a pole o f f o f order For the converse, suppose that zo is a pole of f o f order 2:: 1 , say. Then, by theorem 10. 1 , (z - zo) m f (z) ---+ # 0 as z ---+ zo. Hence, for z # zo (and in some neighbourhood of zo so that f(z) is defined)
as
m.
m
a
m 1 1 - 1 (z (z- zo)- zo)m f(z) 1_ I -+ I Q 1 - 0 f(z) a
as z ---+ zo and the result follows.
-
0
Suppose that zo is an isolated essential singularity of f . Then f is neither bounded near zo nor does 1/(z) l diverye t o oo as z ---+ zo . In other words, there are sequences (zn ) and ((n ) such that both Zn ---+ zo and (n ---+ zo and such that lf(zn) l ---+ oo as n ---+ oo but ( f ((n )) is bounded. Corollary 10.1
Proof. If f were bounded near zo , then zo would be a removable isolated singularity. On the other hand, if 1 / (z) l ---+ oo as z ---+ zo , then we have seen that this would imply that zo is a pole of f. By hypothesis, neither of these possibilities hold and so the result follows. 0
Lec ture Notes on Complex Analysis
172
To say that limz-.z0 (z - zo ) m f ( z ) = G of. 0 means that one can think of f ( z) as behaving rather like D oo as z -> zo . However, the converse, although true (as we have seen) , is not quite so obvious . After all , one could be forgiven for imagining that it was possible for f to have an infinite number of terms in the principal part of its Laurent expansion in some punctured disc around zo (and so zo would be an esse ntial singularity) and still be such that l f ( z ) l -> oo as z -> zo . That this cannot happen is perhaps something of a surprise. What ' s going on?
Example 10.4 any z 1- 0,
Define f(z) , for any z 1- 0, by f(z) = cos � · Then, for
f(z) = cos - = 1 - -1-2 + -1-4 - . . z 2. z 4. z and we see that zo = 0 is an essential singularity of f. Taking Zn = i/n and (n = 1 /n , for n E N, we see that Zn ----> 0 and (n ----> 0 as n ----> oo b ut f(zn ) cos (n/i) � (e n + e- n ) ----> oo as n ----> oo, whe re as the sequence (f((n ) ) = (cos n) is bounded.
1
1
1
•
=
10.3
=
Behaviour
as
lzl
- oo
Let f be entire and non-constant. Then f is a polynomial if and only if l f (z) l ----> oo as l z l ----> oo .
Theorem 1 0 . 2
Proof.
Suppose, first, that f i s a (non-constant) polynomial,
f(z)
=
a n z n + · · · + a 1 z + ao
with a n 1- 0. Then
I f tz) I I an zn + � · · + ao I ----> 0 ao an- � . . I zn ( an + -' + · + -n ) 1 z z =
=
as l z l ----> oo . Conversely, suppose that lf (() l ----> has a Taylo r expansion (about z 0)
oo
as 1 (1 ---->
oo .
=
f(z)
00
=
L an zn ,
n=O
for all z .
Since f is entire, it
Singularities and Meromorphic Func tions
For z i=
0,
173
set
g(z)
=
! (�)
00
=
-n '"' � an Z . n =O
Then g E H(C \ { 0 } ) Moreover, this series converges absolutely for lz l > 0 and so is the Laurent expansion of g ( about z 0). Now, as z ----> 0, I � I ----> oo and so , by hypothesis, I f ( � ) I ---> oo , i.e., l g ( z ) l ----> oo as z ----> 0. I t follows that z = 0 i s a pole of g and therefore .
=
for some
m
am a1 g (z) = ao + - + · · · + zm z � 1 , where a m =/= 0. Hence
f(z) = for z i=
f(O)
=
0.
g(� )
=
ao + a1 z +
· · ·
+ am zm
However, from the Taylor expansion of f above , we see that
ao and so
f(z) for all z E
C ( including z
=
=
ao + a 1z + · · · + a m z m
0).
That is,
f is a polynomial.
0
Any entire non-polynomial function f has the property that there is a sequence (zn ) such that lzn l ----> oo and lf(zn ) l ----> oo as n ----> oo and another sequence ((n ) , say, such that l (n l ----> oo as n ----> oo but the sequence (f( (n )) is bounded. Corollary 1 0 . 2
First we note that by Liouville's Theorem, theorem 8.10, since f is entire it cannot be bounded, unless it is a constant . However, it cannot be constant because it is not a polynomial. The existence of a sequence (zn ) , as above , then follows. Furthermore, again because f is not a polynomial, the existence of some sequence ((n ) , as above, follows from the previous theorem. 0 Proof.
What's going on? It is intuitively clear that if f is a polynomial then lf(z) l must diverge to oo as lzl --+ oo. It is far from obvious that the converse is true. After all, one might suspect that a non-polynomial function such as e • + e i z , or something similar, would exhibit this behaviour. Not so. (In this particular example, set z = it - t for t E lR and t > 0. Then e • + e i z e it e - t + e - t e - it -t 2e cos t which stays bounded as t --+ oo . ) =
=
1 74
10.4
Lecture Notes
on
Complex Analysis
Casorati-Weierstrass Theorem
The next result tells us that near an isolated essential singularity a function takes values arbitrarily close to any given complex number.
Theorem 10.3 (Casorati-Weierstrass Theorem) Suppose zo is an isolated essential singularity of the function f. Let r > 0, E > 0 and w E C be given. Then there is some z with l z - zo l < r such that l f (z) - w l < E. Proof. We know that there is some p > 0 such that f is analytic in the punctured disc D' (zo , p) . Let r' min {r, p} . By way of contradiction, suppose that l f (z) - w l 2: E for all z E D' (zo , r ' ) . Then it follows that g (z) = l j ( f (z) - w ) is analytic and bo unded in D' (zo , r') . This means that zo is a removable singularity of g, i.e . , we can extend g to an analytic function in the whole disc D(zo , r' ) . Let us denote this extension by G. Then =
G(z) (f (z) - w )
=
1
for z E D' (zo , r') . If G(zo ) 1- 0, this entails that f be bounded near zo . On the other hand, if G(zo ) 0 then l f (z) l must diverge as z --+ zo . The first case wo uld mean that zo were a removable singularity and the second that it was a pole of f. However, zo is neither of these, it is an essential singularity of f. We conclude that l f (z) - w l 2: E for all z E D' (zo , r') is 0 false , and the result follows. =
Remark 10. 1 In fact, it has been shown that a function takes on all values with at most one exception in any neighbo urhood of an essential singularity, but this is harder to prove (Picard's Theorem) . The example exp ( l /z) shows that there may be an exception. Evidently zo 0 is an isolated essential singularity but the exponential function is never zero . =
Chapter 1 1
Theory o f Residues
11.1
Residues
The function z n has a primitive on
Laurent expansion of a function plays a special role.
Definition 1 1 . 1 so that, for some
Suppose that R > 0,
f is meromorphic in a set S and let zo E S
oo
m f(z) L a n (z - zo) n + L bn (z - zo) - n n=O n= 1 for z E D ' (zo, R) , and where possibly depends on zo. The residue of f at zo is defined as Res (! : zo) b1 • Thus, =
m
=
Res(f : zo) b 1 =
where r (t)
=
zo + re 2 rr it , 0
Res(f : zo) Proof.
t
::;
27rZ1 . j f , '"Y
1 , for any 0 < r < R.
Suppose tha t zo is
Theorem 1 1 . 1
for any n �
::;
=
=
m.
a pole of f of order Then 1 dn - 1 ( (z - zo) n f(z) lim ) ( n - 1 ) ' z--> zo dz n _ 1 .
m.
We have
00
f(z) L ak(Z - zo) k + (z -b 1 zo) k=O =
1 75
+
· · ·
+
-,---bm---:-_ (z - z0) m
1 76
Lecture Notes on Complex Analysis
and so
(z - zo) n f(z) = (z - zo) n
00
l:::> k (z - zo) k + bt (z - zot- 1 + . . . k =O . . . + bm (Z
for any n �
m.
- zo) n - m 0
The result follows directly.
Remark 1 1 . 1 Usually, one takes n = m to find the residue . However, it is sometimes more convenient to choose a suitable n > m . Consider, for sin z example , the function f(z) = -4- . The point z = 0 is a pole of order 3. z Taking n m 3, we find =
=
Res(f : O)
=
1 -1 lim0 2 . z -->
(
d? z 3 sin z ) dZ 2 Z-4-
( )
-
d2 sin z dz 2 -z- · On the other hand, taking n 4 > 3 = m , we get
which requires calculation of
(
=
Res(f : O )
=
1 d3 z 4 sin z) lim 1 3 z --> 0 dZ 3 Z-41 1 ( - cos z) 1 z = O -6 3! •
=
-
=
-
-
which is marginally easier to calculate.
Example 1 1 . 1 The function sec z 1 / cos z has isolated singularities 0, namely, when z is of the form z (2n - 1 H for n E Z. when cos z P ut w z - k � where k 2n - 1 . Then =
=
=
=
=
sec z
= sec ( w + k � )
=
{
cos w cos k 2,.
-1
sin w sin k �
1
. . w sm k 2,. } - sm (- 1) n
=
sin w
Since w j sin w ----> 1 as w ----> 0, we deduce that (z - (2n - 1) � ) sec z ----> (- 1 ) n as z ----> (2n - 1 ) � - It follows that these are all simple poles, with residues ( - 1 ) n , respectively.
Theory of Residues
11. 2
Let
1 77
Winding Number (Index)
1(
t) =
zo
+ r e 2 rrikt for t E [0, 1 ] and k E N. Then we find that
1 ____!!3__ =
1
'Y
__
27ri
z - zo
k,
the number of times the circular path 1 "goes around" the point zo . By deformation, we might expect that this holds even if 1 is changed slightly so that it is no longer circular but still goes round zo k times. In fact, we can use the integral formula to tell us how many times 1 does encircle the point zo . This leads to the notion of winding number for any contour (not necessarily circular) , as we discuss next.
For any closed contour 1 and any point zo rJ_ t r 1 ,
Theorem 1 1 . 2
1
__
27ri
Proof.
Suppose first that
1
:
10r a < t < b . Then h' (t) = _
_
continuo us. Hence
�t ( ( ( t) r
zo
'Y
z - zo
E z.
[a , b] ----> C is a smooth path. Let
h(t) = c
1 ____!!3__
1t a
r' ( s ) ds ( 'Y ( S ) - ZQ )
r' ( t ) on ( a , b) , s1nce t he ( 'Y ( t ) - zo )
) exp( - h ( t ) ) )
·
m ·
t egrand 1s ·
= r ' ( t ) e - h(t) - ( r ( t ) - zo ) h ' (t) e- h (t) = 0.
We deduce that ( r ( t ) - z0 ) e- h(t) is constant on [a , b] . For a general contour 1 : [a , b] ----> C , we may write 1 = /1 + /2 + + 'Ym < t m = b and smooth sub-paths rj : [tj - 1 , tj] ----> C, for a = to < h < 1 :::; j :::; m. The function h is now defined by ·
h(S) -
_
{
·
·
·
·
·
s -y ' s Jra ( 'Y ( s )(-)zo ) d h(tj - d + � s
8 '
tj - 1
-y ' ( s ( ( s ) -)zo ) ds , 'Y
for a :::; s :::; h , for tj - 1 < s :::; t1· and 2 :::; j :::;
m.
1 78
Lecture Notes on Complex Analysis
Then h is continuous on [a , b] and, as above, (r(t) - zo) e - h(t) is constant on each interval (tj _ 1 , tj) and therefore on the whole of [to , t m ] = [a , b] . Hence , equating values at the end-points, we get
(r(b) - zo ) e- h(b) = (r (a) - zo ) e - h(a) . It follows that e - h (b) = e - h ( a ) , since r (a) r(b) (because 'Y is closed) and zo rJ_ tr ,. Hence h (b) = h(a) + 21rki for some k E Z. But h (a) = 0 , and dz h(b) = and so z zo '"(
1
=
--
1
dz = 27r k z" '"( z - zo
for some k E Z,
as
--
0
required.
If tr 'Y is a circle centred on zo , then we know that k is j ust the number of times 'Y goes aro und the point zo . This suggests the terminology of "winding number" or "index" of a (closed) conto ur with respect to a given point not on the conto ur.
Definition 1 1 . 2 integer
For any closed conto ur 'Y and any point zo rJ_ tr '' the
1 � is called the winding number � 27rz z - zo 'Y
(or index) of 'Y with
respect to zo , and is denoted by Ind(r : zo) .
Examples 1 1 . 2 ( 1 ) For 'Y(t) = re 2 rrit , 0 � t � 1 , we find that Ind(r : O) 1 . (2) If r(t) = re-6rr it , 0 � t � 1 , then Ind(r : O) = -3. (3) For 'Y (t) = e 4 rrit , 0 � t � 2, Ind(r : O) = 4 , Ind(r : 3i) = 0. ( 4) Deformation techniques can be used to determine winding numbers. For example , let 'Y be the simple closed conto ur whose trace is the diamond with vertices ±i and ± 1 . The function 1 /z is analytic in the star-domain D = { z : z = rei0 , r > 0 , - :i: 7r < e < � 1r } and so it follows that =
J
[ l , i]
dz
-;
=
1 dz '{J
-;
where cp is the quarter-circle cp(t) = eit , 0 � t � � 1r. Applying this idea to the other sides of the square (in the suitably rotated star-domains,
Theory of Residues
179
ein: / 2 D, ein: D and eiJn: / 2 ) and adding the results, we see that
1 dz = 1 dz 'Y
where '1/J is the circle '1/J ( t ) should.
11.3
=
z
"' z
eit , 0 � t � 27r. Hence Ind(-y : O) = 1 , as it
Cauchy's Residue Theorem
The next theorem is yet another fundamental result.
Let f be a mero morphic function with a finite number of poles, (1 , . . . , (m , say, in the star domain D and let ')' be a closed contour with tr ')' � D \ { (1 , . . . , (m } (i. e., the contour does not pass through any of the poles). Then
Theorem 1 1 .3 ( Cauchy's Residue Theorem)
1f 'Y
Proof.
S uppose that
=
2rri
't, Res ( f : (k) Ind( T (k ) ·
k= 1
(k is a pole of order mk.
that
Then there is
Rk > 0 such
00
L a�k l (z - (k t + Pk (z) n =O k for z E D ' ((k , Rk) , where Pk(z) = z=:! 1 b� ) (z - (k)- n . The function Pk k is analytic in
=
m
g(z) f(z) - L Pk (z). k= 1 Then g E H(D). Note that g ((j ) is defined to be =
g ((j )
Since
m
=
( ! - Pj ) ((j ) - L Pk((j )
k #-j D is star-like , it follows that J'Y g = 0.
=
a bi )
Hence
m
- L Pk((j ) · k #-j
180
But
Lecture Notes on Complex Analysis
1'Y 1'Y �1 � Pk
b k ) (z - (k ) - n d z
=
n=
=
=
b �k )
21ri b �k ) Ind ( - y: (k )
!:__ 1'Y __! Z - (k
and the result follows. Example 1 1 . 3
0
We will show that
1-y Z Z - 1 cos z
(
)2
d
z
=
27ri (1 - sin 1 - cos 1 ) ,
where 'Y is the circle 'Y(t) = 3 e 2 11"it , 0 � t � 1 . To see this, first note that the singularities of the integrand f (z) cos zjz(z - 1 ) 2 are at z 0 and z = 1 . Clearly, Ind('Y : O) = 1 = Ind('Y : 1 ) . Therefore =
=
1
·1
27f Z -y
-
Now,
f
=
Res(f : O) + Res( ! : 1).
z cos z =1 z ->O z (z - 1 ) 2
Res(! : 0) = lim and Res (! : 1 )
=
d
.
z -> 1 Z
= lim
z -> 1
=
(
cos z Z (Z - 1 ) 2 cos z z2
hm (z - 1 ) 2 d
( - sin z
_
z - sin 1 - cos 1 .
)
)
Example 1 1 .4 Suppose that f is analytic in C except for poles at the points ± 1 where it has residues Res (! : - 1 ) 5, 2 and Res(! : 1 ) respectively. Let 'Y : [0, 1] � C b e the contour
'Y (t)
=
{
-1 + e 1 2 i11"t '
for
1
-r
f
=
2
=
0�t<
1 2
for � � t � 1 .
1 + e i11" ( l - St) ,
Then 'Y winds 3 times around z in the negative sense. Therefore
27rz �
=
=
- 1 and winds around z
X 3
+5
X
( -2)
=
-4 .
=
1 twice, but
181
Theory of Residues
The next example indicates a way of performing real integrals involving trigonometric functions-courtesy of the relationship eit = cos t + i sin t. Example 1 1 .5
dt [ 2" = lo J5 + cos t 7r.
To show how such integrals might be evaluated, notice first that an integral from 0 to 27r suggests an integral around a circle. Let us write the integrand in terms of eit . We have cos t = � (eit + c i t ) . Let 'Y (t) = eit , 0 � t � 21r. Then cos t i"( (t) so the integral H 'Y(t) + 1 h (t) ) and 'Y'(t) = ieit becomes =
1
1 1 'Y' (t) 1 [ 2" - dz dt = 1 1 i (t) i"( VS + � ('Y (t) + 'Y(t) - ) -r v's + Hz + z ) z lo = 21ri L Res (f : (k ) IC:k l < 1
where f =
1
{ vRu + 21 (z + z - 1 ) } iz
. and the sum 1s over those poles (k of f
which lie inside the unit circle (because Ind('Y : ( ) = 0 for any pole ( of f outside this circle) . Notice that there are no poles o n the circle tr 'Y because J5 + cos t is never zero (for t real) . To find the poles of f , we write f ( z) =
1
--=----:-- -
i ( z vts + � (z2 + 1 ) ) 2
i (z2 + 2 VS z + 1 )
2
i(z + J5 - 2) (z + J5 + 2)
and so we see that f has poles at z = - J5 ± 2. Now, Ind ('Y : -J5 - 2) = 0 because -J5 - 2 is outside "( and Ind(T - J5 + 2) = 1 . Therefore,
1 2" a
R
dt
v u + cos t
= 21ri Res(! : -v's + 2)
(z + J5 + 2) lim i z -+ - .J5+2 (z + J5 - 2) (z + J5 + 2)
= 27ri
�
= 27ri
�� 4
= 7r.
i
Lecture Notes on Complex Analysis
182
Suppose that f is analytic in { z : Im z 2: 0 } \ { z 1 , . . . , zn } with poles at the points z 1 , . . . , Zn in the upper half-plane (i. e. , Im zk > 0, 1 � k � n). Suppose further that zf(z) ---) 0 whenever l zl ---) oo with Im z 2: 0 . Then n R limoo f (x) dx 2rri L Res ( ! : Zk ) . Theorem 1 1 .4
1
R-+
-R
=
k= l
Proof. Let c > 0 b e given. B y hypothesis, there i s R so that l zf(z) l < c for all l zl > R with Im z 2: 0. Let E be the semicircular path with centre 0 and radius p and let r [-p, p] + E. Choose p so large that r encloses all the poles z 1 , . . . , Zn. Then Ind ( r : zk ) = 1 for all 1 � k � n and so, by the Residue Theorem, =
1f r
2rri
=
t Res (f : z ) . k
k= l
Now,
hf
and we claim that
hf
....-+
=
0 as
l
f+
- p ,p]
p
---)
hf
To see this, suppose
oo.
p > R.
for any z E tr E, l zf (z) l < c , that is, p if(z) i < c , since lz l = p for z Hence if(z) i < sjp for all z E tr E. By the Basic Estimate, we get
I lr.r f I
and it follows that
hf
....-+
�
0 as
� L ( E)
p
p
---)
1
=
� rrp
p
oo, as
f The result now follows because . [- p,p]
=
=
Then, E
tr E.
c 7r
claimed.
1P f (x) dx. -p
0
Note that in the theorem above, there are no poles on the real axis. Example 1 1 .6
We evaluate
1 00 (x2 + 1 ) (x2dx+ 4) (x2 + 9) . - oo
1 . Evidently, f satisfies the (z 2 + 1 ) (z 2 + 4) (z 2 + 9) hypotheses of the theorem. The poles of f are at ±i, ±2i and ±3i and are
To do this, let f(z)
=
Theory of Residues
all simple poles. By the theorem,
1
00
_ 00
dx = (x 2 + l ) (x 2 + 4 ) (x 2 + 9) p�� = 27ri
1
P
-p
poles
183
dx (x 2 + l ) (x 2 + 4) (x 2 + 9) (
L
in { Im z
>
Res(! : () 0}
= 21ri ( Res( ! : i) + Res(! : 2i) + Res (! : 3i ) ) . All we now need do is to calculate the residues. For example, Res(f : i) = lim (z - i)f (z ) =
.
1
2z X 3 The other two residues can be similarly calculated. z-H
X
8
Chapter 1 2
The Argument Principle
1 2. 1
Zeros and Poles
We shall see that the number of zeros and poles of a meromorphic function f is determined by the behaviour of the quotient f.
Suppose that f is analytic at zo and that zo is a zero of f of order m. Then f is meromorphic at Zo and Res ( r zo ) m. In fact, the point zo is a simple pole of f.
Theorem 1 2 . 1
=
Proof.
By hypothesis, we can write
f (z)
00
=
an (Z - zo) n L n =m
where am =f. 0 and this power series converges absolutely for all z in some disc D(z0 , R) . Since am =f. 0, it follows that f is not identically zero in D(zo , R) and so there is r > 0 such that f(z) =1- 0 for all z in the punctured disc D' (zo , r) ( otherwise f would vanish, by the Identity Theorem) . It follows that 11 f is analytic in D' ( z0 , r) and so, therefore, is f' I f . Hence, z0 is an isolated singularity of f' I f. Write
f(z) = (z - zo ) m (am + am+ I (z - zo) + . . . ) . 'P (Z)
for z E D(z0 , R) . Differentiating, we get
f'(z) = m(z - zo) m - 1 cp(z) + (z - zo) m cp ' (z) . 185
186
Lecture Notes on Complex Analysis
Now, cp ( zo) = a m =f. 0 and so, by continuity, there is p > 0 such that cp ( z) =f. 0 for all z E D ( z0, p ) . Hence, for any z E D' ( z0 , p ) , we may write m
f' ( z )
f( z )
( z - zo )
cp' ( z )
+
cp ( z )
·
Furthermore, cp' I'P is analytic in the disc D ( z0, p) and so has a Taylor series expansion there. Therefore
f' ( z) f ( z)
=
� n � A n ( z - zo ) ( z - zo ) + n=O m
for suitable coefficients (A n ) , where the series converges absolutely in D( zo , p) . This is the Laurent expansion of f ' If at zo and the result follows. 0
There is a corresponding result for poles.
Suppose that zo is a pole of f of order k . Then simple pole at zo with Res ( { zo) = -k .
Theorem 1 2.2
Proof.
f has a
By hypothesis, f has the Laurent expansion
where bk =f. 0, valid in some punctured disc
f ( z)
=
D ' ( z0, R) .
Hence we may write
'1/J( z)
( z - zo) k in D( zo , R))
where '1/J is analytic at zo (in fact , and '1/J ( zo ) = b k =f. 0 . By continuity, it follows that '1/J is non-zero in some neighbourhood of z0 and so there is some p > 0 such that '1/J is analytic and does not vanish in the disc D( z0 , p) . From the equality above, we see that f does not vanish in the punctured disc D' ( z0, p) . Hence
f' ( z) f ( z)
-k
( z - zo )
+
'1/J ' ( z) '1/J(z)
for z E D ' ( z0 , p) . The last term '1/J' N is analytic in this disc (so has a Taylor expansion) and we deduce that z0 is a simple pole of f' If and that Res( ! ' If : zo ) -k, as claimed. 0 =
187
The Argument Principle
12.2
Argument Principle
Applying these results, we obtain the following theorem.
Suppose that D is a star domain and 'Y is a contour in D such that Ind ('Y : ( ) = 0 or 1 for any ( ¢:. tr 'Y. Suppose, further, that f is meromorphic in D, has a finite num ber of poles in D and is such that none of its poles nor zeros lie on tr "(. Then 1 !' = N-y - P-y 21l"i -y f Theorem 1 2 . 3 (Argument Principle)
1
where N-y is the number of zeros of f inside "( (counting multiplicity) and P-y is the number of poles of f inside "( (counting multiplicity). Proof. This is an immediate consequence of the previous two theorems together with the Residue Theorem, theorem 1 1 .3. 0
It is called the Argument Principle for the following rea son. The expression 2 ; i J'Y f ' / f looks as though it should be j ust 2 ;i log f ] 'Y but we have seen that the logarithm must be handled with care. To make some sense of this, suppose that we break the contour 'Y into a number of ( possibly very small ) subcontours, 'Yk · The integral is then the sum of the integrals along these subcontours. Since f does not vanish on tr "f, we can imagine that each subcontour is contained in some disc also on which f does not vanish. (This is a consequence of the continuity of f.) We can also imagine that these discs are so small ( if necessary) that the values f(z) taken by f on each such disc lie inside some disc in C which does not contain 0. This means that there is a branch of the logarithm defined on the values f(z) , as z varies on a given subcontour, 'Yk · Now the integral really is the logarithm Remark 1 2. 1
1 2 1!"�.
1
'Yk
!' f
=
log f(zk ) - log f (zk - 1 )
= log l f (zk ) l + i arg f (zk ) - log lf(zk- d l - i arg f (zk 1 ) = log lf(zk ) l - log lf(zk - d l + i l:!. k arg f , -
where log denotes the branch constructed above, Zk- 1 and Zk are the ends of "fk and l:!. k arg f denotes the variation of the argument of f ( z) as z moves along 'Yk ·
Lecture Notes on Complex Analysis
188
The idea is to do this for each of the subcontours, but being careful to ensure that the choices of branch of logarithm (or argument) match up at the values of f at the ends of the "fk S · Integrating along "Y means adding up these subintegrals and we see that the log lf(z) l terms cancel out leaving j ust the sum
il\ 1 arg f + il\ 2 arg f + . . . , which is the total variation of the argument of f(z) as z moves around the contour "f. (The point is that we have to keep making possibly different choices of the argument as we go along, depending on the winding behaviour of f(z) around zero.) Given "Y : [a, b] --> C, let r be the contour r(t) Then
Remark 1 2 . 2
for a
�
t � b.
=
f( "Y (t)),
b 1 fJ' = 1 f' (J"Y((t))(t)"Y') (t) dt "Y -; b - J r' (t) dt a
-
= =
a
r(t)
l d:
21ri Ind(r : o) .
Hence, by the Argument Principle, we see that the number of zeros minus the number of poles inside "Y is determined by the winding number of the contour r f 0 "( about the origin;
=
N-; - P-;
=
Ind(r : 0) .
In particular, if f E H (D), then the number of zeros of f inside the closed contour "Y is precisely Ind (f o "(= 0) . Now for fixed w0 E C, any solution to f(z) w0 is a zero of the function f(z ) - w0 , and vice versa. By arguing as above (and using (f - wo ) ' !') , =
=
The Argument Principle
we see that
189
1 (! wo ) ' = 1 f' ..., (! - wo) ...,bf -' wo f ('y (t))'y ' (t) = 1 f(-y(t) dt ) Wo -
a
-
=
-
{
dw lr w - wo 2rri Ind (r : wa )
which is to say that the number of solutions to f(z) = w 0 inside 'Y is equal to the winding number Ind ( f o -y: wa ) . 1 2. 3
Rouche's Theorem
This relationship between the number of zeros of a function and certain winding numbers suggests that the comparison of two functions might be attacked by comparing winding numbers. Suppose that -y1 and 'Y2 are two given closed contours, parameterized by [a, b] and neither passing through the origin. If 'Y2 (t) is "close to" 'YI (t) for every t E [a, b] , then we would expect that they have the same winding number around 0. We can think of t as "time" and the pair of points -y 1 (t) and -y2 (t) moving as a composite system around 0 as if j oined by a spring. An alternative picture is an earth moon system. Here the earth and the moon have the same winding number around the sun, even though they have quite different traj ectories. The issue is what is meant by "close to" in this context? To be sure that the two points -y 1 ( t) and 'Y2 ( t) encircle the origin the same number of times, we must ensure that one of them, say 'Y2 (t) , cannot "duck under" the origin whilst the other, -y 1 (t) , goes "over the top" . This is ensured if -y2 (t) always lies in the disc D('y 1 (t) , l"f 1 (t) 1 ) , with its centre at 'YI (t) and with radius l "f 1 (t) l , since in this case -y2 (t) and -y 1 (t) are always "to the same side of 0" . The requirement that -y2 (t) belong to D ('y 1 (t) , 1"f 1 (t) l ) is to demand that l"f 1 (t) - -y2 (t) l < i'YI (t) l . This discussion leads to the following proposition. Proposition 1 2. 1
by [a , b] , such that
Suppose -y 1 and -y2 are closed contours, parameterized
Lecture Notes on Complex Analysis
190
for all a � t � b. Then First we note that neither of -y1 nor -y2 pass through 0 (otherwise the inequality above would fail) . For t E [a, b] , set -y (t ) -y2 (t) h1 (t) . Then -y ' 'Y2 - 'Yi 'Y 'Y2 'Y l and 'Y is a closed contour satisfying Proof.
=
I I - -y (t ) l < 1 for each t E [a , b] . It follows that tr -y indeed, by theorem 8.2) , Ind(-y: O) However, Ind( - y: O)
C
1 = 2m
.
1 jb
D(I , 1) and so, by theorem 8.4 (or,
1 dzz = 0 . '"Y
-
j
dz b -y ' (t) dt I 1 2rri '"Y z 2rri a -y(t) 1 'Y2 (t) 'Y� (t) dt dt - 1 = 2rri a 'Y2 (t) 2rri a 'YI (t) = Ind ('Y2 : 0) - Ind('Y1 : 0) , =
__
and we conclude that Ind('YI : 0)
=
=
__
jb
0
lnd('Y2 : 0) .
Suppose that f and g are ana lytic in the star-domain D and that 'Y is a contour in D, as in Theorem 1 2. 3. Suppose, further, that I f(() - g (() l < lf(() l , for all ( E tr -y . Then f and g have the same number of zeros inside 'Y (counted according to multiplicity).
Theorem 1 2 .4 (Rouche's Theorem)
Let 'YI f o 'Y and 'Y2 g o 'Y so that I'YI - 'Y2 I < I'YI I · Then, by the Argument Principle, theorem 12.3, and proposition 12. 1 , we have
Proof.
=
=
0
as required.
We shall use Rouche's Theorem to show that z 5 + 14z + 2 has precisely 4 zeros inside the annulus { z : � < lz l < 2 } . To show this, set f(z) z 5 and g (z) z 5 + 14z +2 . Then l z l 2 implies that lf(z) l 32 and l f (z) - g (z) l = I I 4z + 2 1 � 1 4 l z l + 2 30. Hence
Example 1 2 . 1
=
=
=
=
=
191
The Argument Principle
lf( z) - g (z) l < lf(z) l on the circle lzl 2.
By Rouche's Theorem, f and g have the same number of zeros inside the circle, that is, in { z : l z l < 2 } . This number is 5. Therefore z 5 + 14z + 2 has 5 zeros in { z : l z l < 2 } . =
Now put
f(z) 14z and leave g as before. For lzl lf(z) l 14 l z l 14 x � 21 =
=
=
=
� ' we find
=
and
l f(z) - g (z) l l z5 + 2 1 � l z l 5 + 2 (�) 5 + 2 < 10. Therefore lf(z) - g (z) l < I J (z) i on i z l = � and so neither f nor g can vanish for such z and 14z and z 5 + 14z + 2 have the same number of zeros in { z : i z l < � } , namely 1. It follows that four of the five zeros of the polynomial z 5 + 14z + 2 lie in the annulus { z : � < lzl < 2 } . =
=
As another example, reconsider the Fundamental Theorem o f Algebra.
For any n E N and any complex numbers ao , a t , . . . , an - 1 , the polynomial g(z) z n + an - 1 Z n - 1 + · · · + a 1 z + ao has precisely n zeros in the complex plane (including multiplicity). zn . Let r > l ao ! + · · · + !an - 1 ! and also r > 1. Then, Proof. Set f(z) for any z with i z l r, we have n- 1 lf(z) - g (z) l = L a k z k
Theorem 1 2 . 5 (Fundamental Theorem of Algebra)
=
=
=
I
I k=O
�
=
n- 1
L: ia k i i z k l k= O n- 1 L ia k i r k k=O
n- 1 �r =
n- 1
L ia k l k =O
,
because r k � rn - 1 , for 0
�
k � n - 1,
l f(z) / .
By Rouche's Theorem, it follows that f and g have the same number of zeros inside the circle 'Y(t) re 2 rrit , 0 � t � 1, namely n. 0 =
192
Lecture Notes on Complex A nalysis
The following result is a corollary to Rouche's Theorem.
Suppose that f is analytic in the star-domain D and 'Y is a contour in D as in Theorem 12. 3. Suppose, further, that f - w0 does not vanish on tr "f. Then there is o > 0 such that the equations w = f ( z) and Wo f ( z) have the same number of roots inside 'Y whenever w E D( Wo ' o) .
Theorem 1 2 . 6
=
Proof. Since f - wo is continuous and never zero on tr 'Y and tr 'Y is compact, it follows that there is some o > 0 such that l f (z) - wo l ;:::: o for all z E tr "(. ( 1 / (/ - wo ) is continuous and bounded.) Let F(z) = f (z) - wo and G(z) f (z) - w and suppose lw - w0 1 < o. Then =
I F(z) - G(z) !
=
!w - wo !
I F(z) l
for all z E tr "(. By Rouche's Theorem, Theorem 12 . 4, it follows that F and G have the same number of zeros inside 'Y and the result follows. 0 This has the following interesting corollary. Theorem 1 2 . 7 Suppose that f is analytic in a domain D and zo E D is such that f'(z0 ) =/: 0. Then there is some r > 0 such that f : D(zo , r) -+ C is one-one .
f has the Taylor expansion f (z) L:�=O a n (z - Zo ) n about zo , valid i n some disc D(zo , R) . Since f'(zo ) =/: 0, i t follows that a 1 =/: 0 . Rewrite this expansion as Proof.
=
f (z)
=
ao + a 1 (z - zo) + (z - zo)g (z)
where g (z) = a 2 (z - zo ) + a3 (z - z0) 2 + . . . and note that f ( zo) ao . Since g (z) -+ 0 as z -+ zo , there is p > 0 such that lg(z) i < ia 1 1 whenever z E D (zo , p) . But since the equality f(z) = ao entails (z- zo) (a 1 +g (z) ) 0 , i t follows that f ( z ) - a o can have n o zeros i n the disc D ( zo, p ) apart from zo which is a zero of order one. By Theorem 12. 6, there is o > 0 such that if w E D (a0 , o), then f (z) - w also has exactly one zero inside the circle iz - zo l = p . Now let 0 < r < p be sufficiently small that l f(z) - f(zo) l < o whenever iz - zo i < r. Suppose that Z 1 E D(zo , r) . Then w f(zi) E D(ao , o) and so f(z) - f(z 1 ) has j ust one zero inside the circle iz - zo ! p, which must 0 therefore be z 1 . In other words, f : D(zo , r) -+ C is one-one. =
=
=
=
193
The Argument Principle
1 2.4
Open Mapping Theorem
We show that non-constant analytic functions map open sets into open sets. (The image of a constant function is j ust a single point.)
Suppose f is analytic . and not constant in a domain D Then f (D) is an open set. In particular, if G � D is open, then f (G) is open.
Theorem 12.8 (Open Mapping Theorem)
Proof. Let wo E f(D). Then there is some zo E D such that f (z o) wa . Since f is not constant, the point z0 is an isolated zero of f - w0 which means that there is r > 0 such that D (z0 , r) � D and f(z) - wa has no zeros in the punctured disc D' ( z0 , r) . In particular, f - w0 does not vanish on the circle l z - zo l �r. Let 'Y be the contour given by 'Y(t) = �re 2 ,.it , for 0 � t � 1 . B y theorem 12. 6 , we conclude that there is some o > 0 such that f (z) - w certainly has zeros inside 'Y whenever w E D(wo , o) . B ut this simply means that D (wo , o) � f (D) which shows that f (D) is open. Suppose G � D is open. Then G is a union of open discs in D. By the Identity Theorem, f is not constant in any of these discs and so by the first part, the image of any such disc under f is open. But then f (G) is a union of open sets and so is open. 0 =
=
Chapter 1 3
Maximum Mo dulus Principle
13.1
Mean Value Property
The first result of this chapter is a certain mean value property enjoyed by analytic functions and is a direct consequence of Cauchy's Integral Formula. Theorem 13.1
for any 0 < r < R,
Suppose that f is analytic in the disc D (z0, R) . Then,
f (zo) = Proof.
'Y(t) = gives
r"' f(zo + re i8 ) d() .
1
2rr Jo
For given 0 < r < R, let 'Y be the circle around z0 with radius r; z0 + reit , for 0 :::; t :::; 2rr. Cauchy's Integral Formula, theorem 8 . 5 ,
f(zo)
=
� 2rrz
1 wf -(w)zo dw 'Y
{ 2 rr J ('Y (t) ) 'Y' (t) dt 2rri }0 'Y(t) - zo 1 . r 2 rr f ('Y (t) ) 1
-
=
2rrz }0 1
re•t
riei t dt
r 2 rr f (zo + reit ) dt
2rr Jo
0
and the proof is complete. 195
Lecture Notes on Complex Analysis
196
Remark 1 3 . 1
Writing zo
u (xo , Yo ) + i v (xo , Yo ) =
=
Xo + iyo and f = u + iv, we obtain
{ 2 7r { u (xo + r cos 0, y0 + r sin 0) 2 71" lo + iv(xo + r cos O, Yo + r sin O) } dO. 1
Equating real and imaginary parts gives 1 { 2 7r u (xo , Yo ) u(xo + r cos 0, y0 + r sin 0) dO 2 71" lo 1 { 2 7r v(xo , yo ) = v (xo + r cos O, y0 + r sin O) dO . 2 71" lo For the next result, we recall that a continuous real-valued function on a closed interval [a, b] is bounded and attains its supremum. =
Lemma 1 3 . 1
Suppose that cp : [a, b] --+ lR is continuous and that
1b cp(s) ds = M (b - a)
where M = max{ cp(s) : a :::; s :::; b } . Then cp(s) = M for oll a :::; s :::; b. Proof.
For s E [a, b] , let g (s) = M - cp(s) . Then, by hypothesis,
1b g (s) ds = M(b - a) - 1b cp(s) ds = 0.
However, g : [a, b] --+ lR is continuous and non-negative and so must vanish, 0 i.e. , cp = M on [a, b] . 13.2
Maximum Modulus Principle
This next major result says that analytic functions do not have maxima, that is, their moduli do not.
Let D be a domain and suppose that f E H (D) . Suppose, further, that there is M > 0 such that l f (z) l :::; M for oll z E D. Then either f is constant on the domain D, or else 1 /(z) l < M for all z E D. In other words, 1/ 1 cannot attain a maximum on D, unless f is constant.
Theorem 1 3 . 2 (Maximum Modulus Principle)
Proof. Suppose that M > 0 and that 1/ (z) l :::; M for all z E D , and suppose that there is some point zo = xo + i yo E D such that 1/(zo ) l = M.
Maximum Modulus Principle
197
In particular , l f (zo) l =f. 0 and so f(z)/f (zo) E H ( D) and l f (z)/f (zo) l � 1 on the domain D. Now, z0 E D and so there is some R > 0 such that D(zo , R) � D. Let 0 < r < R. By theorem 13. 1 , f (zo) _..!:._ { 2 " f (zo + re it ) 1= dt. = f (zo) f(zo) 27r Jo
Writing f (x+iy) / f(zo) = u (x, y)+iv (x, y) , and equating real and imaginary parts, we obtain
1
1 2 7r 1=u (xo + r cos t, yo + r sin t) dt 27r 0 and
1
1 2" 0=v (xo + r cos t, yo + r sin t) dt. 27r 0 Let cp(t) = u (x0 + r cos t, y0 + r sin t) . Then cp : [0, 27r] ---> and satisfies cp (t) = Re �
lR
is continuous
( f(zo + reit) ) f (zo ) e i t)
l f(z;�� I • using the inequality Re w � lwl , l f(zo + reit) l
� 1.
M
By lemma 13. 1 , together with the equality ( * ) , it follows that cp(t) = 1 for all 0 � t $ 27r. Hence 1 = lcp(t) l 2 = u (xo + r cos t, yo + r sin t) 2 u (xo + r cos t, Yo + r sin t) 2 + v (x0 + r cos t, Yo + r sin t) 2 f(zo + reit) 2 = f ( zo ) �
I
1
From this, w e deduce that v(xo + r cos t, y 0 + r sin t ) = 0 and therefore f(zo + reit) - 1 + z.0 ' f(zo) _
Lecture Notes on Complex Analysis
198
giving f (z0 + re i t) f (zo) for all 0 � t � 2rr and 0 < r < R. In other words, we have shown that f (z) = f (zo) for all z E D(z0 , R) . By the Identity Theorem, theorem 8 . 12, it follows that f (z) f (z0) for all z in D. Thus, if I J I has a maximum in D, then f must be constant in D. If f is non-constant, l f l does not achieve a maximum in D. 0 =
=
From this, it readily follows that (the modulus of) non-constant analytic functions have no local maxima.
Suppose that f is analytic on a domain D. Then 1!1 has no local maxima in D, unless f is constant on D.
Corollary 1 3 . 1
Proof. Suppose that zo E D is a local maximum for l fl , that is , there is some disc D(zo , R) in D such that l f(z) l ::; l f (zo) l for all z E D(zo , R) . If l f (zo) l = 0, then f vanishes on D (zo , R) . On the other hand, suppose that lf (zo) l > 0. Then, by the Maximum Modulus Principle applied to f on the domain D(zo , R) , we find that f is constant on D(zo , R) . In any event, f is constant on the domain D, by the Identity Theorem. 0
An alternative version of this can be given using Rouches Theorem via Theorem 12.8, as follows .
Suppose f is analytic and non-constant in a domain D . Then for any zo E D, there is z E D such that lf(z) l > l f(zo ) l .
Theorem 1 3 . 3
Proof. Since f(zo) E f (D) and f (D) is open (by Theorem 12.8) , there is some p > 0 such that D (f(zo) , p) � f (D) . Suppose that f (z0) Re i9 where R = l f (zo) l and 0 E R Then for any 0 < r < p, f (z0) + re i9 E D(f(zo) , p) � f(D). Hence there is z E D such that f (z) = f(z0) + re i9 . But then =
l f(z) l as required.
=
1 Re i9 + re i9 l
=
R + r = l f (zo ) l + r > lf (zo ) l , 0
This discussion suggests that if I J I is to achieve a maximum, then this should occur on the boundary of a domain-assuming that the function f is defined there and sufficiently well-behaved. A formulation of this is contained in the next theorem. For this, we note that the closure of a set is the union of the set together with its boundary.
Let D be a bounded domain and suppose that f : D ---+ C is continuous and that f is analytic in D . Then either f is constant on D
Theorem 1 3 .4
Maximum Modulus Principle
199
or l fl attains its maximum on the boundary of D but not in D. In fact, for any z E D (and assuming that f is not constant), l f(z) l < sup l f(w) l w ED
= m � lf(() l = max lf(() l . ( E 8D
(ED
First we note that the boundedness of D implies that of D. This set is also closed and therefore is compact. The continuity of f on D implies that of lfl which therefore is bounded on D and achieves its supremum. That is, there is some ( E D such that Proof.
lf (() l = sup lf (z) l = M, say. zED
By the Maximum Modulus Principle, theorem 1 3 . 2 , if f is non-constant then l f(z) l < M for all z E D. It follows that ( E D \ D = aD, the boundary of the domain D. 0 Remark 1 3. 2
A slight rephrasing of this theorem is as follows . Suppose
that D is a domain with D bounded and suppose that f : D ---> C is continuous and that f is analytic in D. If lf(z) l � M for all ( E an , then either l f(z) l < M for all z E D, or else f is constant on D. For unbounded regions the above result may be false. For example , let D be the infinite horizontal strip
Example 1 3 . 1
7r
7r
D = { z : - - < Im z < - } , 2 2 and let f ( z) = exp ( exp z) , z E C. Then f is entire and so is certainly continuous on D. We claim that f is bounded on the boundary of D. To see this , let ( E an , so that ( = X ± i � for some X E R We have
f(() = exp ( exp ( x ± irr/2)) exp ( e"' e±i,. / 2 ) = exp (±ie"' ) = cos e"' ± i sin e"' . =
It follows that lf (() l = 1 for every ( E aD, the boundary of D.
200
Lecture Notes on Complex Analysis
Is l f(z) l < 1 for all z E D? The answer is no. For example, suppose that z = x E lR s;;; D. Then we find that f (z) = f (x) = exp (exp x ) = ee x . Clearly l f (x) l = eex ---> oo as x never mind being less than 1 . 13.3
---> oo
and so 1 / 1 is not even bounded on D
Minimum Modulus Principle
Maxima of I f I correspond to minima of 1 ] 1 , assuming that we do not need to worry about f being zero. This observation leads to the following Minimum Modulus Principle. Theorem 1 3 . 5 (Minimum Modulus Principle) Suppose f is analytic and non-constant on a domain D and that lf(z) l � m > 0, for all z E D . Then l f (z) l > m for all z E D .
If, in addition, D is bounded and f is defined and continuous and non zero on D, then 1/ (z) l > inf 1/(w) l = min 1 /(() 1 wED
(, E 8D
for all z E D. Proof.
Apply the Maximum Modulus Principle to g =
1 y-
0
As motivation for the next theorem, we observe that if w = a + ib, then = ea ei b , so that I e w I = ea. This idea of looking at exponentials leads to maximum and minimum principles for the real and imaginary parts of an analytic function. ew
Let D be a domain and suppose that f = u + iv E H (D) is non-constant. Suppose M is some real number such that u � M on D. Then u < M on D . Similarly, if there is m E lR such that m � u on D, then m < u on D . A similar pair of statements hold for v . Furthermore, if D is bounded and f i s defined and continuous o n D, then the suprema and infima of u and v over D are attained on the boundary of the domain D .
Theorem 1 3 . 6
Proof.
Suppose that u( x , y)
� M
for z = x + iy E D. Set g(z) = exp (f(z)) eu e iv , for z E D . =
201
Maximum Modulus Principle
Then g is non-constant on D ( otherwise f would be ) and lg (z) l = eu ( x, y ) , so that u � M implies that eu � eM. ( The real exponential function is monotonic increasing. ) Hence l g (z) l � eM on D. By the Maximum Modulus Principle, theorem 13.2 , it follows that and therefore u ( x , y) < M for all x + iy E D. Next, suppose that m � u ( x, y) for all z
=
h(z)
=
exp ( - f(z))
=
x
+ iy E D. P ut
e - u e - iv , for z E D.
Then, h is non-constant on D and , for any z E D ,
so
that
on D , by the Maximum Modulus Principle. It follows that m < u ( x , y) for any x + iy E D , as required. The analogous results for v are obtained by considering the functions exp ( ::r= if(z)) . Finally, we note that g and h are defined and continuous on D if f is. Furthermore , if f is non-constant, neither are g nor h. By the Maximum Modulus Principle, the suprema of the moduli of these functions is attained on the boundary of D ( and not on D itself) . This amounts to saying that the maximum and the minimum of u over D are both attained on the boundary of D ( and not in D) . A similar argument applied to the two functions exp ( ::r= if) leads to the similar statements for v rather than u . 0 1 3.4
Functions on the Unit Disc
Theorem 1 3.7 ( Schwarz's Lemma )
open unit disc D (O, 1) and satisfies f(O) Then lf(z) l � M l z l fo r all lz l < 1 .
Proof.
Let
f(z)
=
=
Suppose that f is analytic in the 0 and l f(z) l � M for all lz l < 1 .
ao + a 1 z + a 2 z 2 + . . .
be the Taylor series expansion of f(z) about ao = f(O) = 0, by hypothesis.
z0
=
0, for z E D (O, 1 ) . Then
202
Lecture Notes on Complex Analysis
=
Put g(z ) a 1 + a 2 z + . . . , so that f(z ) z g(z). The function analytic in D(O, 1) and g(z) f(z)/z for 0 < l z l < 1 . Let 0 < r < 1 and suppose that lzl r. Then
=
i g(z) i
=
= f( z) lf(z) I I= r l �
M
r By the Maximum Modulus Principle, it follows that =
Z
g
is
.
M lg(z) l � r
for all lzl � r. Letting r ----> 1 we see that therefore
i f(z) i for all
z E D(O, 1 ) , as required.
lg(z) l
= lzl lg(z) l �
M
� M,
for all
l zl
< 1 , and
izi 0
Remark 1 3. 3 In fact, the inequality lg(z) l � M for lzl < 1 implies that i g(z) i < M for all lzl < 1 or else g is a constant, g(z) o: and lo:l M. =
But then
for all 0 <
= =
=
lf(z) l i z g(z) i < M i zl lzl < 1 , or else f(z) o: z, with lo:l M . ,
=
We can use Schwarz 's Lemma to classify those mappings of the open unit disc D(O, 1 ) onto itself which are analytic, one-one and with analytic inverse. First we consider such maps which also preserve the origin. Theorem 1 3 .8
Suppose
f : D(O, 1)
---->
D(O, 1 ) is analytic and satisfies:
f : D(O, 1) ----> D(O, 1) is one-one and onto, f-1 : D(O, 1 ) ----> D(O, 1) is analytic, (iii) f(O) 0. Then f(z) = o:z for some o: E C with lo:l 1 , for all z E D(O, 1 ) . Proof. B y hypothesis, f is analytic, f(O) 0 and lf(z) l < 1 for all z in the disc D(O, 1 ) . By Schwarz 's Lemma, it follows that lf(z ) l :s; lzl for all z E D(O, 1 ) . Exactly the same reasoning applied to the inverse function f- 1 implies that lf- 1 (z) l :s; lzl for all z E D(O, 1 ) . Hence l zl l f - 1 ( J(z)) i � l f(z) l � lzl (i) (ii)
=
=
=
=
203
Maximum Modulus Principle
and we see that l f(z) l l zl for all z E D(O, 1 ) . Let h (z) = f(z)jz for z in the punctured disc D' (O , 1 ) . Then h is analytic in this punctured disc and obeys l h(z) l 1 there. We deduce that h is constant on D'(O, 1 ) , i.e. , h(z) = a: for some a: E C with l o:l = 1 . Thus f(z) o:z for all z E D'(O, 1 ) . This equality persists even for z = 0 since f(O) 0 and the proof is complete. 0 =
=
=
=
Before considering the general case ( i.e. , removing the assumption that f(O) 0) , we need one more observation. z-a Theorem 1 3 .9 For any a E D(O, 1 ) , z � 9a (z) = --- is a one-one 1 - az mapping of D (O, 1) onto itself. =
Proof.
Let z E D(O , 1 ) and set w 9a (z) . Then 1 - l wl 2 1 - w w (z - a) (z - a) = 1 _ ( 1 - a Z) ( 1 - a z) (1 - l a l 2 ) (1 - lz l 2 ) (1 - a z) (1 - a z) (1 - la 1 2 ) ( 1 - l z l 2 ) > 0 1 1 - a z l2 =
=
=
and so we see that 9a : D (O , 1 ) --+ D (O, 1 ) . Furthermore, 9 - a i s the inverse of 9a and s o we deduce that one-one and onto D (O, 1) .
9a
is both 0
Suppose that f : D(O, 1 ) --+ D (O, 1 ) is analytic, one-one, onto and such that f - 1 is analytic. Then z a f(z) = a: �a 1 for some a E D(O, 1 ) and some a: E C with lo:l = 1 . Theorem 1 3 . 1 0
J
(
Since f is one-one and onto D(O, 1 ) , there is a unique a E D(O, 1) such that f(a) 0. Let 9-a be the transformation z � (z + a ) / ( 1 + a z) , as above, and let r.p be the composition r.p : z � /( 9 - a (z)) . This is a composition of analytic one-one mappings, each mapping the disc D (O, 1 ) onto itself and each with an analytic inverse. The same is therefore true of r.p. Furthermore, by construction, r.p(O) 0. It follows that r.p(z) = a: z for some a: E C with lo:l 1 . Hence r.p(ga (z)) = o: ga (z) , i.e . , f(z) = o: ga (z) for all z E D (O , 1 ) . 0 Proof.
=
=
=
Lecture Notes on Complex A nalysis
204
13.5
Hadamard's Theorem and the Three Lines Lemma
The following is a maximum modulus result on an unbounded domain. Theorem 1 3 . 1 1 (Hadamard) Suppose that f is analytic in the (verti cal} strip S = { z : 0 < Re z < 1 } and continuous on its closure S. Suppose,
further, that if(z)i � K for all z E S and that if(z) i � Then lf (z) l � 1 for all z E S.
1 for all z E 8S.
E N, let 9n (z) = f (z) exp ( z 2,:;- 1 ) . Then, for each fixed z E S , 9n (z) � f(z) , as n � oo , since exp (�( z 2 1 ) ) � 1, as n � oo . Moreover, with z = x + iy, Proof.
For
n
-
I exp (z 2 : 1) I = I exp ( x2 - �2 - 1 + 2i:y ) I x2 - Y2 - 1 ) = exp ( � exp ( - � ) , since x2 1 for z E n
�
S.
Fix z E S and n E N. Choose yo > llm z l so large that K e- y� / n < let R be the rectangle
R = { z : 0 < Re z <
1,
1
and
-yo < Im z < Yo } .
Then z belongs to the interior of R. We apply the Maximum Modulus Principle to the function obtain
9n on R to
l9n (z) l � (,mE a8xR l9n (() l � 1 . Therefore l f(z) l
=
limn -+oo
l9n (z) l � 1 for any z E S.
0
Suppose that f is as in the theorem, but satisfies the bounds l f (z) l � K on S, lf(z) l � Mo when Re z = 0, and lf (z) l � M1 when Re z = 1 . Then lf (z) l � M�- Re z M�e z for all Z E S. Corollary 1 3 . 2 (Three Lines Lemma)
In other words, l f(x + iy) l � M�- x Mf for all 0 � x � so
1 and y E lR, and
Maximum Modulus Principle
where Mx Proof.
=
205
sup y l f (x + iy) l .
f(z) M�- 1 M! z , for z E S (principal values) . Now, I M� - 1 1 = I M�e z - 1 M� lm z I = I exp((Re z - 1 ) Log Mo ) exp(i lm z Log Mo ) I = exp ( (Re z - 1 ) Log Mo ) _ - M0Re z - 1 Set h(z)
=
•
{
So
I M� - 1 1 = M�e z - 1
=
R I M-1 z l = M-1 e z
=
Similarly,
Hence l hl
:S
1 if Re z = 0 M0 ' 1'
{ M- 1 , 1,
1
if Re z = 1 .
if Re z = O if Re z = l .
1 on aS, and, by the theorem, l h(x + iy) l
:S
1 on S. That is,
on S, or on S, as required. This, together with the hypotheses, means that
l f(x + iy) /
:S
MJ-x Mf
for all 0 :S x :S 1 and y E R Since the right hand side is for the left hand side, as y varies in IR, we conclude that sup l f(x + iy) / y E IR as
claimed.
:S
an
upper bound
MJ -x Mf , 0
Chapter 14
Mobius Transformat ions
14.1
S p e c ial Transformations
We can think of a complex function f : C -+ C as a transformation of the complex plane into itself. This introduces a strong geometrical flavour to the discussion. We first consider the four basic transformations of translation, rotation, magnification and so-called inversion. A translation is any transformation of the form z �----> w = z + a, with a E C. A rotation is any map of the form z �----> w = ei'P z, with cp E JR. A magnification is a map of the form z �----> w = rz with r > 0. Note that if r > 1 , then this is a magnification in the conventional sense, but if r < 1 then it is a contraction rather than a magnification. Illustrations of these three mappings are given in Figs. 14. 1 , 14.2 and 14.3, below.
w=z+a
Fig. 14.1
Translation:
z ,__. w
207
=
z
+ a, a E IC .
208
Lecture Notes on Complex Analysis
Fig. 14.2
Fig. 14.3
Rotat ion: z
>--+ w =
Magnification: z
e i
>--+ w
=
rz, r >
0.
Notice that under any of these three special transformations, any given region retains its general shape but may be moved around and magnified (or shrunk) . Next , we discuss inversion , which is somewhat less transparent.
209
Mobius Transformations
14.2
Inversion
Inversion is the map z f-> w = � , for z =f. 0. To see what happens under an z inversion , write z r ei 9 . Then =
1 1 -t w = - = -e 9 z r
Thus, points near zero are transformed far away, points in the upper half plane are mapped into the lower half-plane and vice versa. 1 1 z . Indeed , we can wnte z f-> w = - as z f-.-> w = - = -2 .
z
z
lz l
Let us see what happens to circles and straight lines under an inversion. To say that z lies on a circle, we mean that if z x + iy, then x and y satisfy an equation of the form =
for real numbers a, b, c, d, with a =f. 0. Explicit inclusion of the coefficient a means that we can set a 0 to get the equation of a straight line. That is, equation ( * ) determines either a circle (if a =f. 0) or a straight line (if a = 0) in C . This circle/line passes through the origin depending on whether d 0 or not. Actually, if a =f. 0, then by completing the square, we see that ( *) can be written as =
=
This really is a circle (with strictly positive radius) provided b2 + c2 Let us write � = w = u + iv. Then z
1
z = - =
w
1
u+w
--.-
=
u - iv u 2 + v2
=
. x + zy.
It follows that
u u 2 + v2
X = --;:---:::
and y
Now , if x and y satisfy (* ) , then
a (u 2 + v 2 ) bu + 2 2 2 u +v u + v2
=
-
v . u 2 + v2 --::---:::
>
4ad .
Lecture Notes o n Complex Analysis
210
and so
i.e. , ( u , v ) lies on a straight line or a circle (depending on whether d 0 or not) . It follows that the family of circles and straight lines is mapped into itself under the inversion z �--+ w = ! (the points z = 0 and w 0 z excepted) . =
=
(* )
(** )
a ,;:. 0, d ,;:. 0
circle not through (0, 0)
circle not through (0, 0)
a = 0, d 1:- 0
line not through (0, 0)
circle through (0, 0)
a ,;:. 0, d = 0
circle through (0, 0)
a = 0, d = 0
line through ( 0, 0 )
line not through ( 0 , 0 ) line through ( 0, 0 )
So the family of circles and straight lines is mapped into itself under all four of the special maps considered so far, and therefore also under any composition of such maps. 14.3
Mobius Transformations
Definition 14. 1
A Mobius transformation is a map of the form z �--+
T(z ) =
az + b ez + d
-
where a, b, e, d E C satisfy ad - be 1:- 0. Mobius transformations are also called bilinear transformations or frac tional transformations. The condition ad - be 1:- 0 ensures that the map is defined (except, of course, at z = -d/e if e '#- 0 ) and is not a constant. If a 1:- 0 and c 1:- 0, then T(z ) a(acz + bc )/c(acz + ad ) . If be = ad then this reduces to a constant, namely aje. Also, one can check that the condition ad - be 1:- 0 means that T(z ) = T(w ) if and only if z = w. If w = T(z) (az + b)/ (ez + d ) , then "solving for z" , we calculate that =
=
Mobius Transformations
211
z = ( -dw + b)/ (cw - a ) and (-d) ( -a) - be = ad - be =F 0. Hence -dw + b S(w) cw - a =
is also a Mobius transformation and is the inverse of T. Furthermore, if TI and T2 are Mobius transformations, then so is their composition T2 TI (which maps z into T2 (TI (z) ) ) . It follows that the set of Mobius transfor mations forms a group ( under composition ) with the group identity being the identity transformation I ( z ) z. =
I t turns out that any any Mobius transformation T(z)
=
az + b can be cz + d
expressed in terms of the special transformations considered earlier. To see this , suppose first that c =F 0. Then we write
az + b cz + d
(a/c ) (cz + d ) - ( ad/c ) + b cz + d a (ad / c ) - b cz + d c a be - ad = + 2 � c (z + (d/c)) "
be - ad =F 0. Then c2 T(z)
Let A =
=
a A + ( z + (d/ c) ) c ·
Consider the following maps:
d +c 1 WI 1--+ W2 = WI w2 �---+ WJ IAI ei arg >-w2 a WJ �--+ W = WJ + Z
1--+
WI
=
Z
translation inversion magnification / rotation
=
translation.
C
Then
w i.e., z
�---+
=
a a wa + - = AW2 + C
C
=
A a A a +-= +W I C Z + (d/c) C
-
T(z) = w is given by the composition Z
f--->
trans
WI
f---> i
nv
W2
f--->
mag/rot
W3
f--->
trans
W.
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Lecture Notes o n Complex Analysis
Now consider the case c = 0, which is a little simpler. First note that now ad =f. be demands that ad =f. 0. In particular , d =f. 0. Therefore T(z) = Consider
az + b = � z+ �. d d d magnification/rotation translation
giving T(z) = w via z
�---+
mag/rot
w1
�---+
trans
w.
We have seen that any Mobius transformation can be expressed as a suitable composition of the special mappings, namely, translation, rotation, magnification and inversion. It follows that the family of circles and straight lines is mapped into itself under any Mobius transformation. Let us consider the inversion z 1 / z in more detail . Consider points z E C with Re z constant, that is points of the form z = a + iy, y E IR , for fixed a E R Such zs form a line parallel to the imaginary axis. Suppose that a =f. 0, so that the line does not pass through the origin. 1 1 u - iv Now, if - = w = u + iv, then z = . and so z = a + iy = u 2 + v 2 z u+w giving v u a = --� and y = 2 2 u +v u u That is, w is such that u and v satisfy a = 2 , or (u 2 + v 2 ) - - = 0 u + v2 a or 2 u+ v2 = 1 2 . 4a 2a
......
--
( 2_ )
__
This is the equation of a circle, centred at ( 2� , 0) and with radius � . So w lies on this circle. To find the range of values taken by w, we use 1 1 a - iy . =-= w = u + zv = z a + iy a 2 + y 2
---
to get and
...::.,..
__
213
Mobius Transformations
z varies on the line, i.e., as y varies in JR., we see that u takes all values between 0 and 1/a, including 1/a but not including 0. Thus ( u - � ) takes 2 on all values between ± 2� , not including - 2� and v takes on all values between ± 2� (including both values ± 21<> themselves) . It follows that, for any a =f. 0, the image of the line { z : Re z a } under the inversion z f---> is the set { - � I � } \ {0}. This leads to the 21 1 2 following theorem. As
w � =
w : lw
=
w
Inversion z f---> one-one onto the disc { �I
Theorem 14. 1
=
1
- maps the half-plane { z : Re z > 1 } z � } , and vice versa.
w : lw - < We can use the circles, varying a > 1 , above, or prove this directly as follows. Setting z = x + iy, w 1/ z, we have 2-z < l w - � 1 = 1 � - � � < � 1 -2z I 2 21 - z l < l z l 12 - z l 2 < l z l 2 (2 x) 2 + Y 2 < x2 + Y 2 - 4x + x2 + Y 2 < x2 + Y 2 4( 1 - x) < 0 =
Proof.
as
=
1
{==:}
{==:}
{==:} {==:}
_
{==:} 4 {==:}
{==:}
x > l.
Hence inversion maps the half-plane { z : Re z > 1 } into the disc D ( � , �) and it also maps the disc D (� , �) into the half-plane { z : Re z > 1 } . But inversion is its own inverse, so it follows that it maps { z : Re z > 1 } onto D ( � , �) and D ( � , �) onto { z : Re z > 1 } . 0 We can use this fact to help construct Mobius transformations mapping various given half-planes into discs and vice versa by manoouvering via the "standard" half-plane/disc pair, as above.
w : lw - <
Example 14. 1 We shall find a Mobius transformation which maps the half-plane { z : Im z > 2 } onto the disc { 21 3 }.
As can be seen from Fig. 1 4 . 4 , the idea is to manipulate the original half-plane into the standard half-plane { w2 : Re w2 > 1 } using rotations and translations (in fact, one of each) . Inversion now transforms this stan dard half-plane into the standard disc. Further rotations, translations and magnifications now bring the image into the required position.
Lecture Notes on Complex Analysis
214
-i1rf2z
=
-iz
0
1
Fig. 14.4 Manreuvring via the standard half-plane/disc pair.
215
Mobius Transformations
The overall transformation is built from a number of simple ones :
z �---+ WI
e - i'Tr / 2 z = -iz , 1 W 2 I-+ W3 = - ' W3 W2 W4 I-+ W 5 = 6 W4 ,
=
I-+
W4
=
W3 - 2I ,
Combining these, we get
w = w5 + 2 = 6w4 + 2 = 6 (w3 - �) + 2 6 6 = -1 6w3 - 1 = - - 1 W2 WI - 1 6 + iz + 1 6 _1= -iz - 1 -iz - 1 iz + 7 -iz - 1 =
14.4
--
Mobius Transformations in the Extended Complex Plane
The Mobius transformation T : z �---+ (az + b)/ (cz + d) is not defined when z = -d/c (when c =f. 0 ) . Now, we can extend the inversion mapping z �---+ 1 / z on C \ {0} to a mapping from Coo to Coo by the assignments 0 1---+ oo and oo ...... 0 . We mimic this for any Mobius transformation as follows. First recall that ad - be =f. 0, so c and d cannot both be zero. If c = 0 ( so that a =f. 0 ) , set
Tz = If c =f. 0, set
Tz
=
{
{ �� + b) jd,
:o: z� C, .
(az + b) / (cz + d) , for z E C and z =f. -d/c, oo , z = -d/c ajc, z = 00 .
In this way, any Mobius transformation T : z ,..... (az + b)j(cz + d) can be extended to a mapping on C00 • Moreover, one checks that the mapping z ...... ( dz - b)/ ( -cz + a) is the inverse to T on Coo , so that the collection of Mobius transformations on Coo also forms a group under composition.
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Lecture Notes o n Complex Analysis
L et z1 , z2 and Z3 be any three distinct points in Coo . Then there is a Mobius transformation mapping the triple (z1 , z 2 , z3 ) to (0, 1 , oo) .
Proposition 14. 1
Proof.
Suppose first that z1 , z2 and z3 all belong to C. Set
Tz
=
(z - zi ) (z2 - z3) (z2 - z i ) (z - z3)
·
Then we see that T is a Mobius transformation with the required properties . Now, the cases when one of z1 , z2 or z3 is equal to oo are handled as follows. We define T by if Z l if Z2 if Z3 = =
00 ,
=
00 , 00 .
D
Again, one sees that T has the required properties.
Suppose that z1, z2 and z3 are any distinct points in Coo and that S is a Mobius transformation such that S zJ z1 for j 1 , 2 , 3 . Then Sz z for all z E C00 . az + b Proof. Suppose first that (z1 , z 2, z3) (0, 1 , oo) . Writing Sz = cz + d ' we calculate that
Theorem 1 4 . 2
=
=
=
=
O f---4 0
===}
O = b/d
===}
b=O
and 1
f---4
1 and oo f---4 oo
===}
aj(c + d)
=
1
and c = 0
===}
a = d.
It follows that S z = z for all z. Now consider the general case o f any distinct points z1 , z2 and z3 . Let T be a Mobius transformation which maps (z1 , z 2 , z3 ) to (0, 1 , oo). Then the composition T o S o r- 1 is a Mobius transformation mapping (0, 1 , oo) to (0, 1 , oo) and so according to the above argument it is equal to the identity transformation. It follows that S = r- 1 o T, that is, S is also the identity D transformation, S z z for all z . =
Mobius Transformations
217
If S and T a re Mobius transformations obeying Szj Tzj for any three distinc t points z1 , z2 and z3 in C00 , then S = T. Corollary 1 4 . 1
=
1 , 2 , 3 , is fixed under the map Proof. We simply note that each Zj , j S o r- 1 so that S o T- 1 z = z for all z, by the theorem. It follows that Sz = Tz for all z . D =
Chapter 1 5
Harmonic Functions
15.1
Harmonic Functions
We recall that if f u + iv is analytic in a domain D then it obeys the Cauchy-Riemann equations there; Ux = Vy and uy = -Vx · Indeed, we have seen that =
f' = Ux + ivx
=
Vy - iu y .
Moreover, we know that f has derivatives of any order and so u and v also possess partial derivatives of any order. It follows that Vx y = Vy x and so we find that Uxx = (ux ) x = (vy )x = (vx) y = -Uyy , that is, u satisfies Laplace's equation
Uxx + Uyy = 0 in the domain D. Definition 15.1 We say that the function cp(x, y) is harmonic in the domain D in IR2 if all partial derivatives of cp up to second order exist and are continuous in D and if cp satisfies Laplace's equation 'Pxx + 'P yy = 0 in D.
We can summarize the remarks above by saying that the real part of a function analytic in a domain D is harmonic there. Note also that if cp is harmonic in D , then the mixed partial derivatives 'Px y and 'Py x are equal in D. It is natural to ask whether every harmonic function is the real part of some analytic function. Let cp(x, y) = log y'x2 + y 2 for (x, y) =f. (0, 0) . Then one is harmonic in this region. However, cp(x, y) = Re Log z
Example 1 5 . 1
checks that
cp
2 19
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Lecture Notes on Complex Analysis
where z = x + iy and it follows that cp is not the real part of any function analytic in the punctured plane D = C \ {0} . In fact, if f E H(D) and if cp = Re f in D, then f and Log z have the same real part in the cut-plane C \ { z : z + l z l 0 } and so differ by a constant there; f(z) = Log z + a for some constant a E C for all z E C \ { z : z + l z l 0 } . But this is not possible because the left hand side has a limit on the negative real axis (with 0 excluded) whereas the right hand side does not. =
=
15.2
Local Existence of a Harmonic Conjugate
Notice that the domain in the example above is not star-like, and it is the presence of holes which allows for such a counterexample. For star-like domains, every harmonic function is indeed the real part of some suitable analytic function, as we now show. Definition 1 5 . 2 Let cp be harmonic in a domain D. A function '1/J (x, y) is said to be a harmonic conj ugate for cp in D if the function f(x + iy) = cp(x, y) + i'ljJ (x, y) is analytic in D.
The analyticity of f implies that '1/J is also harmonic in D. Furthermore, if w(x, y) is also a harmonic conj ugate for cp in D, then the functions cp + i'ljJ and cp + iw are analytic in D and have the same real part and so they differ by a constant. In other words, if cp possesses a harmonic conj ugate in D, then it is unique to within a constant. We can now prove the main result here.
Suppose that u is harmonic in a star-domain D . Then u possesses a harmonic conjugate there, that is, there is some f analytic in D such that u (x, y) = Re f(z) for all z = x + iy E D .
Theorem 1 5 . 1
Proof. For any z = x + iy E D, set g (z) since u is harmonic in D , it follows that
=
u x (x, y) - iu y (x, y). Now,
so the real and imaginary parts of g satisfy the Cauchy-Riemann equations in D. Furthermore, all partial derivatives are continuous in D (because u is harmonic) and so we deduce that g is analytic in D . By hypothesis, D i s a star-domain and s o g possesses a primitive i n D , that is, there i s G E H(D) such that G' g i n D. =
221
Harmonic FUnctions
Let G = U + iV and set f = u + iV. We shall show that f E H(D) . Indeed, general theory tells us that
G'
=
Ux + iVx
=
Vy - iUy
but we know that
G'
=
g
=
U x - iu y
so that U x = Vy and u y = -Vx . In other words, the real and imaginary parts of f obey the Cauchy-Riemann equations in D. Moreover, the partial derivatives of u and V are continuous in D (because u is harmonic and G D is analytic) and so f E H(D) and the proof is complete. Corollary 1 5 . 1 Let u be harmonic in a domain D . For any z0 E D, there is some r > 0 such that u has a harmonic conjugate in the disc D (z0 , r) . Proof. For given z0 E D, there is some r > 0 such that D (z0 , r) Since D (z0 , r) is star-like, u has a harmonic conj ugate there.
15.3
<:;;:
D.
D
Maximum and Minimum Principle
Suppose that u is harmonic and is non-constant in a domain D . Then u has neither local maxima nor local minima in D .
Corollary 1 5 . 2
Proof. Suppose that u does have a local maximum at some point , say at zo = Xo + iyo E D . Then there is r > 0 and f E H (D (zo , r) ) such that u = Re f in D (z0 , r) . Let g = e f . Then g E H(D(z0 , r)) . Moreover, for any given z = x + iy E D (z0 , r) ,
l g(z) l
=
eu( x ,y )
�
e u( x o,yo )
=
l g (zo) l
and so g is constant in D (z0 , r) by the Maximum Modulus Principle. In particular, e f ( z) = g(z) = g (zo) = ef( zo ) and so (f(z) - f(z0)) /27ri E Z for all z E D (zo , r) . Since f is continuous, this means that f is also constant in the disc D (z0 , r) and so f is constant on D, by the Identity Theorem. But then u Re f is constant on D , a contradiction. We conclude that u has no local maxima. By replacing u by - u in the above, we see that u also has no local D minima. =
222
Lecture Notes on Complex Analysis
Suppose that u i s harmonic and non-constant i n IR2 • Then u is neither bounded from above nor from below.
Corollary 1 5 . 3
First we note that, by the theorem, there is some function f analytic in
Chapter 1 6
Local P rop erties of Analyt ic Funct ions
16.1
Local Uniform Convergence
We begin with a definition. A sequence of functions Un) converges uniformly to f on the set A if and only if for any c: > 0 there is N E N such that n > N implies that l fn (z) - f(z) l < c: for any z E A. ( The important point is that the same N works no matter which z E A is selected. ) For any given domain D, we say that the sequence Un) of complex valued functions on D converges locally uniformly to the function f on D if and only if for each point z0 E D there is some r > 0 such that D ( zo , r) � D and (Jn ) converges uniformly to f on D(zo, r) .
Definition 1 6 . 1
We show next that local uniform convergence is equivalent to uniform convergence on compact sets. Theorem 1 6 . 1
The sequence Un ) converges locally uniformly to f on D
if and only if Un) converges uniformly to f on any compact set K � D .
Suppose that the sequence Un) converges to f uniformly on any compact set K in D and let z0 E D. Then there is some R > 0 such that D(z0 , R ) � D. In particular, D(z0 , R/2) C D and Un) converges uniformly to f on the compact set D (z0 , R/2) . In particular, Un ) converges uniformly to f on the disc D(z0 , R/2) and so it follows that Un) converges locally uniformly to f on D. Conversely, suppose that Un) converges to f locally uniformly on D and let K be a given compact subset of D. For each point z E K, there is some r2 > 0 such that D(z, r2 ) � D and such that fn -7 f uniformly on the disc D (z, r2 ) . The collection { D (z, r2 ) : z E K } is an open cover of K Proof.
223
224
Lecture Notes on Complex Analysis
and so has a finite subcover; for some z1 , . . . , Z m E K . Let c > 0 be given. Then for each j = 1 , . . . , m , there is some Nj E N such that n > N1 implies that
lfn ( z) - f ( z) l < c
for any z E D ( zJ , rzJ ) . Setting N = max { Nj : 1 $ j $ n > N implies that
m
,
} we see that
lfn ( z) - f ( z) l < c for any z E K . That is , ( fn) converges to f uniformly on K . An alternative proof o f this, without using open covers, can be given as follows. Suppose that fn --+ f locally uniformly on D but that ( fn) does not converge uniformly on some compact subset K of D. Then there is some c o > 0 such that for any N E N there is some n > N and some point , ( , say, in K ( and depending on N ) such that
lfn ( () - f ( ( ) l We construct a sequence (wj ) in K and some Zn1 , say, in K such that Let w1 = Zn1 - Then setting N and some Zn2 E K such that Let w2 = such that
Zn2 • Now let N
=
=
as
� co ·
follows. For N
=
1,
there is n 1 >
1
n 1 , we may say that there is some n 2 > n 1
n 2 . Then there is some n 3 > n 2 and Zn3 E K
Let w3 = Zn3 • Continuing in this way, we obtain a sequence (wj ) in K ( and integers n 1 < n 2 < . . . in N ) such that for all j E N. Since the sequence ( w1 ) lies in K and K is compact, there is a convergent subsequence, WJk --+ ( , say, in K as k --+ oo .
225
Local Properties of Analytic FUnctions
But since fn ---7 f locally uniformly, there is some r > 0 such that fn ---7 f uniformly in D((, r) . In particular, there is k0 E N such that if k > ko then l fni k (z) - f(z) l < co for all z E D (( , r) . However, for sufficiently large k , wik E D ((, r ) and so
for sufficiently large k. This contradicts ( * ) and we conclude that fn uniformly on K , as required.
---7
f
D
Suppose that fn ---7 f locally uniformly on a domain D and that each fn is continuous. Then f : D ---7 C is continuous.
Proposition 1 6 . 1
For any given z0 E D there is some R > 0 such that fn ---7 f uniformly in the disc D(z0 , R) . Let € > 0 be given. Then there is some N such that l fN (() - f(() l < � € for all ( E D(zo , R) . Since fN is continuous, there is p > 0 such that l f(w) - f(zo) l < � € whenever w E D(zo p ) . Set r = min { R, p } . Then for any w E D(z0, r) , we have Proof.
,
l f(w) - f(zo) l
�
l f(w) - fN (w) l + l fN (w) - fN (zo) l + l fN (zo) - f(zo) l
< l3 c + l3 c + l3 c
=€.
D
It follows that f is continuous at z0 and so the proof is complete.
Suppose that Un) and (gn ) are sequences of contin uous functions such that fn ---7 f and 9n ---7 g locally uniformly on the domain D . Then
Proposition 16.2
(i ) ( ii) (iii ) ( iv)
a.fn ---7 a. f locally uniformly on D, for any a. E C . fn + 9n ---7 f + g locally uniformly o n D . fn9n ---7 f g locally u niformly o n D . If, furthermore, each 9n =1- 0 and g =1- 0 on D , then uniformly on D.
b. 9n
---7
l 9
locally
Proof. It is enough to show uniform convergence on any given compact set K in D and it is straightforward to show that this is true of a.fn to a.f and fn + 9n to f + g. Let us show that fn9n ---7 fg uniformly on K . The uniform convergence on K implies that both f and g are continuous on K and therefore bounded there. Hence there is M > 0 such that l f(z) l � M and lg(z) l � M, for all
226
Lecture Notes on Complex Analysis
z E K . Also, the uniform convergence of fn to f on K implies that l fn (z) l for all sufficiently large
n
l fn (z)gn (z) - f(z)g(z) l
l fn (z) - /(z) l + l f(z) l �1+M �
and for any z E K . But then, for given c > 0 , �
�
l fn (z) l l gn (z) - g (z) l + l fn (z) - f(z) l lg(z) l (M + 1) c + M c
for all sufficiently large n and any z E K . It follows that fngn ---7 fg uniformly on K and hence locally uniformly on D. To show that fnlgn ---7 f l g locally uniformly, we need only show that 1 I gn ---7 1 I g locally uniformly ( and then apply the argument above ) . So suppose that gn =/:- 0 and g =/:- 0 on D. Let K C D be compact. As above, we know that g is continuous on K . It follows that lg l attains its lower bound on K and so there is m > 0 such that lg(z) l � m, for all z E K . Furthermore, lg(z) - gn (z) l < � m for sufficiently large n and all z E K and so m � lg ( z) l � l g(z) - gn (z) l + l gn (z) l < I¥ + lgn (z) l ,
)
giving lgn (z l >
� m for large n and any z E K .
1 gn_1(z)
_
_
1 g (z)
_
Hence
gn(z) I < 2 lg (z) -2 gn (z) l ' I = I ggn(z )(z)- g(z) m
for sufficiently large n and any z E K . The result now follows because D gn ---7 g uniformly on K . 16.2
Hurwitz's Theorem
Locally uniformly convergent sequences of analytic functions are very well behaved, as we see next.
Suppose that D is a domain and that (in) is a sequence in H (D) which converges locally uniformly to f on D . Then f E H(D) and moreover, for each k E N, the sequence of derivatives converges locally uniformly to the kth derivative on D.
Theorem 1 6 . 2
j( k)
(!Ak) )
Proof. First we shall show that f is analytic in D. Let zo E D . Then there is r > 0 such that D(z0 , r) <:;;; D . Let T be any triangle in D(zo , r) .
227
Local Properties of Analytic FUnctions
By Cauchy's Theorem,
r
laT
fn = 0
for every n E N. Now, Un ) converges to f uniformly on the compact set aT and so f is continuous on aT. Let c: > 0 be given. Then there is N such that l fn (z) - f(z ) l c: for any z E aT , whenever n > N. Hence
<
I faT f - faT fn I I faT (f - fn ) I =
� c: L (aT)
by the Basic Estimate, theorem 7. 1 . It follows that
r
JaT
f
= n-.oo lim r Ja
T
fn = 0.
By Morera's Theorem, theorem 8.8, we conclude that f is analytic on D. To show that the derivatives of fn converge to those of the function f locally uniformly, we use Cauchy's Integral Formulae. Let zo E D and k suppose R > 0 is such that D (z0 , 2 R) <:;;: D. We will show that f� ) ---7 f( k ) uniformly on D(zo , R) . Let k E N and z E D(z0 , R) and let c: > 0. Then, by Cauchy's Integral Formula
I f ( k l (z) - fn( k ) (z) I =
- fn(w) I � 1 f(w) dw I (w - z) k+l 2m
"/
where we take 1 to be the circle 1(t) = z0 + � Re21rit, 0 � t � 1 . Now, for all sufficiently large n, l f(w) - fn (w) l c: for all w in the compact set tq. Moreover, for all such w, lw - z l 2:: ! R. Hence, by the Basic Estimate, theorem 7.1 , we see that the right hand side above is bounded by
<
- fn (w) 1 _1 1 f(w) 2_ ( � R) k+l dw I < (w - z ) k+ 1 27ri
7
2 1r
c:
provided n is sufficiently large. The result follows.
2 7r 23 R , D
Suppose that Un ) is a sequence of functions analytic on the domain D such that fn ---7 f locally uniformly on D. Suppose, further, that fn =/= 0 in D. Then either f is identically zero on D, or f =/= 0 on D .
Theorem 1 6 . 3 (Hurwitz's Theorem)
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Lecture Notes on Complex Analysis
We know that f E H (D) . Suppose that f is not identically zero in D. Let zo E D. Then there is p > 0 such that D(z0 , p) � D . Since f is not identically zero, there is 0 < r < p such that f(z) =f:. 0 for all z with l z - zo l = r. (Otherwise, zo would be a limit point of zeros of f.) Since the circle C = { z : l z - zo l = r } i s compact and f i s continuous , there is ( E C such that 1 /(z) l � l/ (() 1 for all z E C. By construction, f is not zero on C and so m = l/(() 1 > 0 , i.e. , 1/ (z) l � m > 0 on C. Let n be so large that 1 / (z ) - fn (z ) l < m for all z E C. (This is possible, since fn ---> f uniformly on compact sets. ) Then Proof.
1/(z) l
�
m > lf(z) - fn (z) l
on the circle C. By Rouche's Theorem, theorem 1 2 . 4 , it follows that f and fn have the same number of zeros inside the circle C, that is , none. In particular, /(zo) =f:. 0 which is to say that f has no zeros in D. 0
Suppose fn E H(D) and that fn ---> f locally uniformly on the domain D . Suppose that zo E D is a zero of f of order m, then for any R > 0 there is a disc D(z0 , r) , with r < R, and N E N such that for all n > N, fn has exactly m zeros in D(zo, r) .
Theorem 16.4
Let R > 0 b e given. Since fn ---> f locally uniformly, there is some disc D (z0, p) such that fn ---> f uniformly on D(zo , p). Then, arguing as above, we deduce that there is 0 < r < mi n {p, R} and a > 0 such that 1/ (z) l � a for all z E C = { l z - zo l = r} . Moreover , we may assume that zo is the only zero of f inside the circle C (because the zeros of f are isolated) . The uniform convergence on compact sets implies that there is N E N such that if n > N then Proof.
lf(z) - fn ( z) l < a for all z E C. Hence
1/1
�a>
I f - fn l
on C and so, by Rouche's Theorem, f and fn have the same number of 0 zeros inside the circle C, namely m.
Local Properties of Analytic FUnctions
16.3
229
Vitali's Theorem
Definition 1 6 . 2 A sequence Un) of functions is said to be locally uni formly bounded on a domain D if for each z0 E D there is some r > 0 with D(zo , r) <:;;; D and some M > 0 such that l fn (z) l < M for all n and all z E D(z0 , r) . Note that both r and M may depend on z0•
The result of interest in this connection is Vitali's Theorem ( which we will not prove here ) , as follows.
Let Un) be a sequence offunctions analytic in a domain D. Suppose that Un) is locally uniformly bounded in D and that there is some set A in D such that A has a limit point in D and such that Un(z)) converges for all z E A. Then there is a function f such that fn ---7 f locally uniformly in D (so that, in particular, f E H (D)).
Theorem 1 6 . 5 (Vitali's Theorem)
App endix A
S ome Result s from Real Analysis
We collect here some of the basic results from real analysis that we have needed. They all depend crucially on the Completeness Property of R We begin with some definitions. A.l
Completeness of IR
Definition A.3 A non-empty subset S of IR is said to be bounded from above if there is some M E IR such that a � M for all a E S. Any such number M is called an upper bound for the set S. The non-empty subset S of IR is said to be bounded from below if there is some m E IR such that m � a for all a E S. Any such number m is called a lower bound for the set S. A subset of IR is said to be bounded if it is bounded both from above and from below. Suppose S is a non-empty subset of IR which is bounded from above. The number M is the least upper bound of S ( lub S ) if
( i ) a � M for all a E S ( i.e . , M is an upper bound for S ) . ( ii ) If M' is any upper bound for S, then M � M ' . I f S is a non-empty subset o f IR which i s bounded from below, then the number m is the greatest lower bound of S ( glb S ) if
( i ) m � a for all a E S ( i .e . , m is a lower bound for S ) . ( ii ) If m' is any lower bound for S, then m' � m. Note that the least upper bound and the greatest lower bound of a set S need not themselves belong to S. They may or they may not. The least 231
232
Lecture Notes on Complex A nalysis
upper bound is also called the supremum (sup) and the greatest lower bound is also called the infimum (inf) . Evidently, if M is an upper bound for S, then so is any number greater than M. It is also clear that M is an upper bound for any non-empty subset of S. In particular, sup S is an upper bound for any such subset of S. Note that if M = lub S, then there is some sequence ( xn ) in S such that Xn ---7 M, as n ---7 oo . (Indeed, for any n E N, the number M - � fails to be an upper bound for S and so there is some Xn E S such that Xn > M - � · Hence Xn obeys M - � < Xn ::; M which demands Xn ---7 M.) Analogous remarks apply t o lower bounds and glb S. sup[O, 1] = 1 = sup(O, 1) and inf[O, 1] = 0 = inf(O, 1 ) . Example A.l Note that (0, 1 ) has neither a maximum element nor a minimum element. The essential property which distinguishes IR from Q is the following. The Completeness Property of IR Any non-empty subset of IR which is bounded from above possesses a least upper bound.
A consequence of this property, for example, is that any positive real num ber possesses a square root. In particular, thanks to this we can be confident that y'2 exists as a real number . (It is given by sup{ x : x2 < 2 } .)
If (an ) is an increasing sequence of real numbers and bounded from above, then it converges.
Proposition A.3 is
By hypothesis, { an : n E N } is bounded from above. Let K = lub { an : n E N } . We claim that an ---7 K as n ---7 oo . Le t c: > 0 b e given. Since K i s an upper bound for { an : n E N } , it follows that a n ::; K for all n. On the other hand, K - c: < K and K is the least upper bound of { an : n E N } and so K - c: is not an upper bound for { a n : n E N } . This means that there is some aj , say, with aj > K - c: . But the sequence (an ) i s increasing and so an � ai for all n > j . Hence an > K - c: for all n > j . We have shown that Proof.
K - c: < an
::;
K < K + c:
for all n > j and so the proof is complete.
D
Any sequence (bn ) in IR which is decreasing and bounded from below must converge. Corollary A . l
Some Results from Real A nalysis
Proof.
Just apply the above result to the sequence an = -bn .
233
D
In fact, bn converges to the greatest lower bound of { bn : n E N } . A.2
Balzano-Weierstrass Theorem
Theorem A.6 (Bolzano-Weierstrass Theorem) Any bounded sequence of real numbers possesses a convergent subsequence. Proof.
Suppose that M and m are upper and lower bounds for ( an ) ,
We construct a certain bounded decreasing sequence and use the fact that this converges to its greatest lower bound and so drags a suitable subse quence of ( an ) along with it. To construct the first element of the auxiliary decreasing sequence, let M1 = lub { an : n E N } . Then M1 - 1 is not an upper bound for { an : n E N } and so there must be some n 1 , say, in N such that
Next, we construct M2 as follows. Let M2 = lub { an : n > n 1 } so that M2 :-::; M1 . Moreover, M2 - � is not an upper bound for { an : n > n1 } and so there is some n 2 > n1 such that
M2 - � < an 2 :<:; M2 . Continuing in this way, we construct a sequence (Mj )jEN and a sequence
(nj )jEN such that Mj +l :<:; Mj , ni + 1 > nj , and
for all j E N. Now, m :<:; ani :<:; MJ and so (Mj ) is a decreasing sequence which is bounded from below. It follows that (Mj ) converges, say Mi ---7 JL, as j ---7 oo . However, by our very construction,
and so ani
---7 JL,
as j
---7 oo ,
Mj - I < ani :<:; Mj
and the proof is complete.
D
234
Lecture Notes on Complex Analysis
Remark A . l Note that if a n E [a , b] for all n , then the limit of any convergent subsequence also belongs to the interval [a, b] . c
Recall that a sequence (an) in lR is a Cauchy sequence if for any given > 0 there is N E N such that l an - a m i < c whenever both n, m > N.
Theorem A. 7
Every Cauchy sequence in lR converges.
First , we show that any Cauchy sequence (an ) in lR must be bounded. Indeed, we know that there is some N E N such that both n > N and m > N imply that Proof.
In particular , for any j >
N,
l ai l � l aj - a N+l l + l a N+l l < 1 + l a NH I It follows that if M
=
·
1 + max{ l a 1 l , l a 2 l , . . . , l a N+l l } , then
for all k E N, i.e. , (an ) is bounded. To show that (an ) converges , we note that by the Bolzano-Weierstrass Theorem , (an ) has some convergent subsequence, say an k � a, as k � oo . We show that an � a. Let c > 0 be given. Then there is k0 E N such that k > k0 implies that
l an k - a l < � €
Since (an ) is a Cauchy sequence , there is m > No imply that
N0 such that both n > No and
I an - a m I < � €
Let
•
•
N = max { ko , No } . Then
I an - al � I an - an k I + l a n k - a l < � € + � € = €
whenever n >
N. Thus an � a as k �
oo as
required.
D
Remark A. 2 Note that a convergent sequence is necessarily a Cauchy sequence. Indeed , if an � a, then the inequality
shows that (an ) is a Cauchy sequence.
235
Some Results from Real Analysis
A.3
Comparison Test for Convergence of Series
The above result enables us to conclude that various series converge even though we may not know their sum. Theorem A.8 (Comparison Test) Suppose that ao , a1 , . . . and b0 , b1 , . . . are sequences in IR such that 0 ::; an ::; bn for all n = 0 , 1 , 2 , . . . .
If the series L::= o bn converges, then so does the series L::= o an .
Proof. Let Sn = L::�=O a k and Tn = L:: �=O b k be the partial sums. Then we see that for n > m
n
n
k = m+l
k = m+l
If L:: := o bn is convergent , then (Tn ) is a Cauchy sequence and so therefore D is ( Sn ) . Hence L::= o an converges.
A.4
Dirichlet's Test
Theorem A.9 (Dirichlet's Test) Suppose a0 , a 1 , is a sequence in IR such that the partial sums Sn = L:: �=O a k are bounded and suppose � 0 is a decreasing positive sequence in IR such that Yo � Yt � Y 2 � Yn 1 0, as n ---7 oo . Then L::= o an Yn is convergent. •
·
·
•
•
·
The proof uses a rearrangement trick ( called "summation by parts" ) . Let Tn = L:: �= O a kYk · Then, for n > m ( by straightforward verification ) , Proof.
Tn - Tm =
n
n
k = m+l
k = m+ l
L a kYk = Sn Yn + l - SmYm+ l + L Sk ( Yk - Yk+ t ) .
Let c: > 0 be given. By hypothesis , there is M > 0 such that I Sn l ::; M , for all n and since Yn 1 0 , as n ---7 oo , there is N E N such that 0 ::; Yn < t: / 2M,
236
for all
Lecture Notes on Comple,; Analysis
n
> N. Hence, if n , m > N,
ITn - Tm l
:S
I Sn Yn +l l + I SmYm+l l +
n
L I Sk l ( Yk - Yk+ I )
k = m+l n :S M Yn +l + M Ym+l + L M(yk - Yk+ l ) k = m+ l = M Yn + l + M Ym+l + M( Ym+ l - Yn + l ) = 2M Ym+l < e.
It follows that (Tn) is a Cauchy sequence in R and therefore convergent. A.5
E �= O an Yn is D
Alternating Series Test
The Alternating Series Test is an immediate consequence ,
as
follows.
Theorem A . l O (Alternating Series Test) Suppose x0 � x1 � x2 � · · · � 0 is a decreasing positive sequence in R such that Xn 1 0 as n � oo .
Then the alternating series xo -x1 +x2 -x3+ · · = E�=0( - l ) n xn converges. n Proof. Set an = ( - l ) and Yn = Xn · Then the partial sums Sn = n L:: ;=O an = � ( 1 + ( - l) ) are bounded so we can apply Dirichlet 's Test to
obtain convergence , as stated. A.6
D
Continuous Functions on [a, b] Attain their Bounds
Next , we look at properties of continuous functions.
Suppose that the function f : [a, b] � R is continuous. f is bounded on [a, b] and achieves its maximum and minimum on Then [a , b] .
Theorem A. l l
We argue by contradiction. Suppose that f is continuous on [a, b] but is not bounded from above. This means that for any given M whatsoever , there will be some x E [a, b] such that f (x) > M. In particular , for each n E N there is some an , say, with an E [a, b] such that f(a n ) > n. The sequence (an)nEN lies in the bounded interval [a , b] and so , by the Bolzano-Weierstrass Theorem , it has a convergent subsequence (an k )kEN , Proof.
Some Results from Real Analysis
237
say; a nk ---+ a as k ---+ oo. Since a � an k � b for all k, it follows that a � a � b. By hypothesis, f is continuous at a and so ank ---+ a implies that f(a nk ) ---+ f ( a ) . But any convergent sequence is bounded (mimic the argument above for Cauchy sequences) . This is a contradiction, so we conclude that f must be bounded from above. To show that f is also bounded from below , consider g = -f. Then g is continuous because f is. The argument j ust presented, applied to g , shows that g is bounded from above. But this j ust means that f is bounded from below , as required. We must now show that there is a: , f3 E [a, b] such that f ( x ) � f ( a ) and f ( x ) � f(/3) for all x E [a , b] . We know that f is bounded so let M = sup{ f ( x ) : x E [a, b] } . Then f ( x) � M for all x E [a , b] . We show that there is some a E [a , b] such that f ( a ) = M. To see this , suppose there is no such a: . Then f(x) < M and so, in particular , M - f is continuous and strictly positive on [a , b] . It follows that h 1/(M - f) is continuous and positive on [a, b] and so is bounded , by the first part. Therefore there is some constant K such that 0 < h � K on [a , b] , that is, =
0<
1 5:_ K . M-f
Hence f 5:_ M - 1 /K which says that M - 1/K is an upper bound for f on [a, b] . But then this contradicts the fact that M is the least upper bound of f on [a, b] . We conclude that f must achieve this bound, i.e. , there is a E [a , b] such that f ( a ) = M = sup { f ( x ) : x E [a , b] } . In a similar way, if f does not achieve its greatest lower bound, m , then f - m is continuous and strictly positive on [a , b] . Hence there is L such that
0<
1
--
on [a , b] . Hence m + 1 / L 5:_ f and m + 1/ L is a lower bound for f on [a, b] . This contradicts the fact that m is the greatest lower bound for f on [a, b] and we can conclude that f does achieve its greatest lower bound , that is , there is f3 E [a , b] such that f (/3) = m . D Remark A . 3
Note that neither a nor f3 need be unique.
238
A.7
Lecture Notes on Complex Analysis
Intermediate Value Theorem
Any real-valued function f continuous on the interval [a , b] assumes all values between f(a) and f (b) . In other words, if ( lies between the values f(a) and f(b) , then there is some s with a :<:; s :-::; b such that f(s) = ( . Theorem A . 1 2 (Intermediate Value Theorem)
Proof. Suppose f is continuous on [a, b] and let ( be any value between f(a) and f (b) . If ( = f(a) , take s = a and if ( = f(b) take s = b. Suppose that f(a) < f (b) and let f(a) < ( < f(b) . Let A be the set A = { x E [a, b] : f(x) < ( } . Then a E A and so A is a non-empty subset of the bounded interval [a, b] . Hence A is bounded and so has a least upper bound, s , say. We shall show that f(s) = ( . Since s = lub A, there is some sequence (an ) in A such that an --+ s. But A � [a , b] and so a :<:; an :-::; b and it follows that a :-::; s :<:; b. Furthermore , by the continuity of f at s , it follows that f(an ) --+ f(s ) . However , a n E A and so f(an) < ( for each n and it follows that f(s) :<:; ( . Since , in addition, ( < f(b) , we see that s =1- b and so we must have a :<:; s < b. Let (tn ) be any sequence in (s , b) such that tn --+ s. Since tn E [a , b] and tn > s , it must be the case that tn tf_ A, that is, f(tn ) 2": ( . Now , f is continuous at s and so f(tn ) --+ f (s) which implies that f( s) 2": ( . We deduce that f (s) = ( , as required.
Now suppose that f (a) > ( > f(b) . Set g(x) = - f(x) . Then we have that g(a) < -( < g(b) and applying the above result to g , we can say that there is s E [a , b] such that g(s) = - ( , that is f(s) = ( and the proof is complete. D
A.8
Rolle's Theorem
For differentiable functions , more can be said.
Suppose that f is continuous on the closed interval [a , b] and is differentiable in the open interval (a, b) . Suppose further that f(a) = f(b) . Then there is some � E (a , b) such that f'( O = 0 . (Note that � need not be unique.)
Theorem A. 1 3 (Rolle's Theorem)
Since f is continuous on [a, b] , it follows that f is bounded and attains its bounds, by Theorem A . 1 1 . Let m = inf{ f (x) : x E [a , b] } and Proof.
Some Results from Real Analysis
239
let M = sup { f(x) : x E [a , b] } , so that
m ::; f(x) ::; M ,
for all x E [a, b] .
If m = M , then f is constant on [a, b] and this means that f ' (x) = 0 for all x E (a, b). In this case, any � E (a, b) will do. Suppose now that m =f. M , so that m < M. Since f(a) = f(b) at least one of m or M must be different from this common value f(a) = f(b) . Suppose that M =f. f(a) ( = f(b) ) . As noted above, by theorem A. l l , there is some � E [a, b] such that f(�) M. Now , M =/:- f(a) and M =/:- f(b) and so � =f. a and � =f. b. It follows that � belongs to the open interval (a , b) . We shall show that f ' (�) = 0. To see this , we note that f(x) ::; M = f(�) for any x E [a , b] and so (putting x = � + h) it follows that f(� + h) - f(�) ::; 0 provided I hi is small enough to ensure that � + h E [a , b] . Hence =
f(� + h) - !(�) < - 0 for h > 0 and small h and
�
f(� + h - ! (�)
� 0 for h < 0 and small.
But ( * ) approaches J'(�) as h ! 0 which implies that J'(�) ::; 0. On the other hand, ( * * ) approaches !' (�) as h i 0 and so !' (�) � 0. Putting these two results together, we see that it must be the case that f ' (�) = 0 , as required. It remains to consider the case when M = f(a). This must require that m < f(a) ( = f(b)). We proceed now just as before to deduce that there is some � E (a , b) such that f(�) = m and so ( * ) and ( * * ) hold but with the inequalities reversed. However, the conclusion is the same, namely that
f ' � ) = O.
A.9
D
Mean Value Theorem
Suppose that f is continuous on the closed interval [a , b] and differentiable on the open interval (a , b) . Then there is some � E (a, b) such tha t
Theorem A. 1 4 (Mean Value Theorem)
f(b) - f(a) J ' (�) = . b-a
240
Lecture Notes on Complex Analysis
Let y = f(x) = mx + c be the straight line passing through the pair of points (a, f (a)) and (b, J(b)). Then the slope m is equal to the ratio Proof.
(J(b) - f(a) ) / (b - a). Let g(x) f(x) - f(x) . Evidently, g i s continuous o n [a , b] and dif ferentiable on (a, b) (because f is) . Furthermore , since f(a) = f(a) and f(b) f(b) , by construction, we find that g(a) = 0 = g(b) . By Rolle's Theorem, theorem A.13 , applied to g , there is some � E (a , b) such that g'(�) = 0. However , g' (x) = f'(x) - m for any x E (a, b) and so =
=
f ' (�)
=
m=
f(b) - f (a) b-a 0
and the proof is complete.
We know that a function which is constant on an open interval is dif ferentiable and that its derivative is zero. The converse is true. Corollary A, 2
and that f' (x)
=
Suppose that f is differentiable on the open interval (a, b) 0 for all x E (a , b) . Then f is constant on (a, b) .
Let a < j3 be any pair of points in (a, b) . Applying the Mean Value Theorem to f on [a, /3] , we may say that there is some � E (a, /3) such that Proof.
J(/3) - f(a) . /3 - a However, f' vanishes on (a, b) and so f'(�) = 0 and therefore f(a) f'(�)
The result follows.
=
=
J (/3) .
0
B ibliography
The following list includes books that have been useful for the preparation of these notes as well as some which may be of interest for further study. Ash , R. B. Complex Variables. New York , London, Academic Press , 1 97 1 . Bak, J. and D . Newman. Complex A nalysis. 2nd ed ition , Springer-Verlag, 1 997. Beardon, A . F. Complex A nalysis: the argument principle in analysis and topology. Chichester , Wiley, 1979. Conway, J. B. Functions of One Complex Variable . 2nd ed ition, Springer-Verlag, New York , 1978. Duncan, J. Elements of Complex A nalysis. London , J. Wiley, 1968. Grove, E. and G. Ladas . Introduction to Comp lex Variables. Houghton Mifflin , Boston, 1974. Hairer , E. and G . Wanner. A nalysis by Its History. Springer-Verlag, New York 1996. Marsden, J. E. and M. J. Hoffman. Basic Complex A nalysis . 3rd ed ition , W. H. Freeman, 1999. Rud in, W. Real and Complex A nalysis . 3rd edition, New York, London, McGraw Hill, 1987. Stewart, I . and D. Tall. Complex A nalysis. Cambridge University Press , 1993. Tall, D. Functions of a Complex Variable. Routledge and Kegan Paul, London , Henley & Boston , 1977. For real analysis, any of the following books make a very good start . Abbott , S . Understanding Analysis . Springer-Verlag, 200 1 . Bartle , R . G . and D . R . Sherbert . Introduction t o real A nalysis . J . Wiley & Sons Inc . , 1982 . Rud in, W. Principles of Mathematical A nalysis. International Series in Pure and Applied Mathematics, McGraw-Hill, New York, 1964 . For a history of the various characters in mathematics, The Mac Tutor History of Mathemati cs archive ,
http : / /www - groups . dc s . st-and . ac . uk/ -hi story/
see :
which is located at the url
( created by John J . O 'Connor and Edmund F. Robertson ) 24 1
Index
11" ,
compact exhaustion, 46 compact set, 41 Comparison Test, 235 completeness of C, 22 completeness of IR, 232 complex conj ugate, 1 complex number , 1 absolute value, 3 formal definition , 12 imaginary part , 1 modulus, 3 real part , 1 complex plane, 2 complex power , 100 conformal, 155 continuous function , 59 contour , 1 1 2 length, 1 1 3 contour integration, 1 1 6 convex domain , 57 cosine function , 79 countable set , 1 53
87
accumulation point, s ee limit point Alternating Series Test, 236 analytic function , 70 angle between two paths, 154 annulus, 1 57 Argand diagram, 3 argument , 8, 93 principal value, 9, 93 Argument Principle, 187 Basic Estimate, 120 bilinear transformation, 210 Bolzano-Weierstrass Theorem, 233 boundary, 38 Cantor's Theorem, 41 Casorati-Weierstrass Theorem, 1 74 Cauchy 's Formula for Derivatives, 142 Cauchy 's Inequality, 146 Cauchy's Integral Formula, 138 Cauchy 's Residue Theorem, 179 Cauchy 's Theorem for a star-domain , 133 for a triangle, 127 Cauchy-Riemann equations, 67 chain rule, 63 closed disc , 3 1 closed set , 32 closure , 36 cluster point , s ee limit point
De Moivre's formula, 10 Deformation Lemma, 1 36 derived series, 74 d iameter, 40 differentiable complex function, 61 Dirichlet's Test , 74 , 235 disc of convergence , 74 domain, 56 243
244
Lecture Notes on Complez Analysis
e,
81 irrationality of, 83 entire function, 70 exponential function, 79 extended complex plane , 1 5 field , 1 2 fractional transformation , 210 Fundamental Theorem of Algebra, 147, 191 of Calculus, 1 2 1 greatest lower bound { i n
R) , 23 1
Hadamard , 204 harmonic conjugate, 220 harmonic function , 219 holomorphic function, 70 Hurwitz's Theorem , 228 hyperbolic functions, 80 Identity Theorem , 150 for power series, 77 imaginary axis , 3 independence of parametrization, 1 1 8 index , 1 78 infimum {in R) , 232 infinity, 15 interior point , 29 Intermed iate Value Theorem, 238 inversion , 209 isolated singularity, 167 isolated zero, 152 Laplace's equation, 219 Laurent expansion , 160 principal part , 162 uniqueness, 163 least upper bound (in R) , 231 lim sup, 25 limit of sequence, 1 8 limit o f series, 2 3 limit point , 34 line segment , 49 Liouville's Theorem, 147 local uniform convergence, 223
locally uniformly bounded , 229 logarithm , 98 branch , 104 of analytic function , 132 principal branch, 104 principal value , 98 Mobius transformation , 210 magnification, 207 Maximum Modulus Principle, 196 for harmonic functions, 221 mean value property, 195 Mean Value Theorem (in R) , 239 meromorphic function, 169 Minimum Modulus Principle , 200 for harmonic functions , 221 Morera's Theorem, 145 nth_Root Test , 26 nth_root of analytic function , 132 nested sequence of sets , 40 open d isc , 29 Open Mapping Theorem, 1 93 open set , 30 order, 14 path, 50 closed , 1 1 1 final point , 50 impression , 50 initial point, 50 length, 1 1 3 piecewise smooth, 1 1 2 reverse, 1 1 2 simple, 1 1 1 smooth, 1 12 trace, 50 trace is compact , 5 1 track, 50 with infinite length , 114 Picard 's Theorem, 174 pole, 168 polygon, 49 polynomial , 147 , 172 power series, 73
245
Index
primitive , 123 existence in star-domain, 130 punctured disc, 34 quotient rule, 63 radius of convergence, · 74 Ratio Test, 26 real axis, 3 residue , 1 75 Riemann sphere , 14 Rolle's Theorem, 238 rotation, 207 Rouche's Theorem, 190 Schwarz's Lemma, 201 sequence, 17 Cauchy, 21 Cauchy ( in IR) , 234 convergent , 18 series, 23 absolutely convergent , 24 conditionally convergent, 24 convergent , 23 d ivergent, 23 partial sum, 23 set bounded , 40 closed, 32 compact , 4 1 connected , 52 disconnected , 52 open, 30
pathwise connected , 51 polygonally connected , 51 stepwise connected , 51 sine function, 79 singularity essential, 168 isolated , 167 pole, 168 removable, 168 star-centre , 56 star-domain, 56 star-like, 56 subpath, 1 1 2 subsequence , 17 summation by parts, 235 supremum ( in JR) , 232 Taylor series, 139 Three Lines Lemma, 205 translation , 207 triangle, 127 triangle inequality, 4 trigonometric functions, 79 add ition formulae, 84 uniform convergence, 223 Vitali 's Theorem, 229 winding number, 1 78 zero of order
m,
1 52
