Lectures on the Theory of Functions of a Complex Variable 0882755315, 9780882755311

Table of contents :
Introduction
Contents
Chapter 1. The Real and Complex Number Fields
Chapter 2. Elementary Point Set Topology
Chapter 3. Continuous Functions
Chapter 4. Differentiation of Complex Valued Functions of a Complex Variable
Chapter 5. Complex Integration
Chapter 6. Entire and Meromorphic Functions
Chapter 7. Conformal Mapping
Chapter 8. Analytic Continuation and Riemann Surfaces
Chapter 9. Algebraic Functions and Their Integrals (Resumé of Results)

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VAN NOSTRAND MATHEMATICAL STUDIES Editors Paul R. Halmos, The University of Michigan

Frederick Wu Gehrine:. The LJni,,~r~it\,

nf 1\/li"h;,Y"'i .

Paul R. Halmes-LECTURES ON BOOLEAN ALGEBRAS

Shmuel Agmon-LECTURES ON ELLIPTIC BOUNDARY VALUE PROBLEMS Noel J. Hicks- NOTES ON DIFFERENTIAL GEOMETRY Leopoldo Nachbin-TOPOLOGY AND ORDER

Sterling K. Berberian-NOTES ON SPECTRAL THEORY Roger C. Lyndon-NOTES ON LOGIC Robert R. Phelps.- LECTURES ON CHOQUET'S THEOREM

L. E. Sigler-EXERCISES IN SET THEORY George W. Mackey-LECTURES ON THE THEORY OF

FUNCTIONS OF A COMPLEX VARIABLE Lars V. Ahlfors-LECTURES ON QUASICONFORMAL MAPPINGS

J. Peter May-SIMPLICIAL OBJECTS IN ALGEBRAIC TOPOLOGY

W. H. J. Fuchs-TOPICS IN THE THEORY OF FUNCTIONS OF ONE COMPLEX VARIABLE •

Lennart Carleson-SELECTED PROBLEMS ON EXCEPTIONAL SETS Leopoldo _N achbin- ELEMENTS OF APPROXIMATION THE·ORY

S.S. Chern-COMPLEX MANIFOLDS WITHOUT POTENTIAL THEORY Additional titles will be listed and announced as issued .

.

•

Lectures on

THE THEORY OF FUNCTIONS OF A COMPLEX VARIABLE by GEORGE W. MACKEY Harvard University

D. VAN NOSTRAND COMPANY, INC. PRINCETON, NEW JERSEY TORONTO

LONDON

MELBOURNE

VAN NOSTRAND REGIONAL OFFICES: New York, Chicago, San Francisco

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D. VAN NOSTRAND AvsTRA.LIA PTY. LTD., Melbourne Copyright© 1967, by D. VAN NOSTRAND COMPANY, INC.

Published simultaneously in Canada by D. VAN NOSTRAND CoMPANY (Canada), LTD.

No reproduction in any form of this book, in whole or ih part (except for brief quotaiion in critical articles or reviews), may be made without written authorization from the publisher.

PRINTED IN THE UNITED STATES OF AMERICA

INTRODUCTION Mathematics 213, the standard introduction at Harvard to the theory of functions of a complex variable, has been given over the years by a number of people from a number of different points of view. These notes-written down at the time for the be,1efit of the studentsconstitute a fairly accurate transcription of the lectures -given by me in this course during the academic year 1959-60. For better or worse I chose to de-emphasize the intuitive geometric aspects of the subject and to present it as a deductive system starting with axioms for the real and complex numbers. While I assumed a fair amount of "mathematical maturity" on the part of the students I made a point of assuming no previous formal knowledge of mathematics. I defined all terms-even those of calculus and elementary algebra-and proved all theorems except those which I decided to leave to the students as exercises. In principle (but not in practice of course) the notes can be read with understanding by someone with-no previous knowledge of mathematics. Of course, in order to avoid excessive pedantry I left many simple arguments to the imagination of the student-expecially after the first few chapters. The first three chapters are preparatory and deal in turn with the real and complex number systems, elementary metric space topology and the basic properties of continuous functions. The theorems on metric space topology are applied later not only to obtain needed results about the topology of the complex plane but also to various spaces of functions-above all in the chapter on conformal mapping. The chapter on continuous functi.ons concludes with proofs of the fundamental theorems of the differential and integral calculus for real functions of one real variable. The chapter on complex differentiation contains the basic elementary theorems about power series in one complex variable including their term by term differentiability. It also contains a: proof that 111

iv

Theory of Functions of a Complex Variable

complex differentiability implies the Cauchy-Riemann equations. It concluded with a detailed study of the power series

1 + z + z 2 /2! + z 3 /3! + ... leading up to the definition and properties of the elementary transcendental functions sin x and cos x for real and complex values of the argument. Chapter V, on complex integration, is the heart of the course. It begins with the definition and properties of complex line integrals and proceeds more or less directly to a proof of Cauchy's integral theorem for star shaped regions. With this p~werfµl tool available, the basic theorems of the theory are developed in short order, the existence of power series expansions, the unique determination of an analytic function by its values on an infinite set with a limit point, the principle of the maximum, Liouville's theorem, etc., etc. The chapter also contains treatments of the Artin winding number form of the Cauchy theorem, Rouche's theorem, the implicit function theorem, and the evaluation of definite integrals by means of residues. The remaining chapters reproduce the lectures given during the second semester and are of a more advanced and specialized character. Chapter VI deals with entire and meromorphic functions in the plane, Chapter VII with conformal mapping, and Chapter VIII with analytic continuation and Riemann surfaces. More detailed outlines of these chapters will be found in the summaries with which they begin. Chapter IX eptitled "algebraic functions and their integrals" is only an outline of the chapter that would have been written had the course lasted a few weeks longer. · In conclusion, I should like to emphasize to the reader that what follows is an informal set of lecture notes and not a finished book. Except for the summaries at the beginnings of the later chapters and the correction of a few mistakes, they are just as written down while the course was being given. There has been no revision and no intensive effort to find and correct mistakes.

ex,

George W. Mackey

CONTENTS CHAPTER

I II

1

The Real and Complex Number Fields Elementary Point Set Topology

25

, .....

III

Continuous Functions

IV

Differentiation of Complex Valued Functions

41

•. . •. . . , . . . . , , •, . , •

of a Complex Variable ••.. , . . . . . , , , • V

57

79

Complex Integration . . . . . . • , , , . . . .

131

VI

Entire and Meromorphic Functions

VII

Conformal Mapping .• , • , .•...

175

Analytic Continuation and Riemann Surfaces . ,

209

VIII

IX

0

OI

• 0

0

0

0

0

e

Algebraic Functions and Their Integrals (Resume of Results) .•.•.••..•..

255

CHAPTER I THE REAL AND COMPLEX NUMBER FIELDS

Let 1 be a set with at least two elements and let there be given two "operations" + and · in 1. That is, we suppose that for each pair x and y of elements of 1 we are given two other elements which we denote by x + y and x · y respectively. We shall say that 1 is a field with respect to these two operations provided that the following conditions are satisfied. I. x+ y = y+ x and x · y = y · x for all x and y in 1. II. x+(y+z) = (x+y)+_z and (x·y) · z = x· (y·z) for all x, y, z in 1. III.x·(y+z) = x·y+x·z forall x,y,and z in 1. IV. There exists an element O in 1 such that x+ 0 = x for all XE 1. V. There exists an element 1 I, 0 in 1 such that x· 1 = x for all XE 1. VI. For each XE 1 there exists y E 1 such that x+ y = 0. VII. For each x E 1 with x I, 0 there exists y E 1 with x·y = 1.

Remarks: There can be at most one element O having property IV. Indeed, if x+ 0' = x for all x in 1 then letting x = 0 we have 0+0' = 0 But O+O' = O'+O by I and 0'+0 = O' by IV. Hence 0 = 0 ~ Similarly if x · 1' = x for all x then 1 = 1 · 1' = 1 '· 1 = 1 so that the 1 of V is also unique. THEOREM

1.1.

If

x+ y = x+ z then y = z.

Proof: Choose w so that w+x w+ (x+y) The symbol l

=

= 0. Then

w+ (x+z)

ls an abbreviation for "is in" or "is a member of."

1

2

Theory of Functions of a Complex Variable

so

(w+x) + y

so

O+ y

so

y

(w+x) + z O+z z

COROLLARY. The y of VI is uniquely determined by x. O will Definition: The unique element y such that x+ y be denoted by the symbol - x. Thus x + (- x) = (- x) + x = 0. THEOREM 1.2. If x and y are elements of S: then there exists one and only one element z such that x+ z = y.

Proof: If x+z = y and x+z' = y then x+z = x+z' so that z = z' by Theorem 1.1. On the other hand x + x) +y) = (x+(-x))+y = O+y = y so that we may take (-x)+ y as z.

«-

Definition: The unique z such that x+ z = y will be denoted by y - x. COROLLARY. THEOREM 1.3.

y - x = y+(-x) = (-x)+ y. 0 · x = 0 for all

Proof: O · x+O · x Theorem 1. 1, 0 · x = 0.

Xf

s=.

(O+O) · x = O· x = O· x+O. Hence, by

THEOREM 1.4. For all x

f

S:,

-x = (-1) x.

Proof: x+(-l)x= l·x+(-l)x = (1+(-l))x = O·x = 0. But x+(-x) = 0. Hence x+(-l)x = x+(-x).Hence (-l)x=(-x). TuEOREM 1.5. -(- x) = x.

Proof: (- x) + -(- x) = 0 by definition. Hence (- x) + x (-x) +-(-x ). Hence x = -(-x) by Theorem 1. 1. COROLLARY. THEOREM 16.

(-1)(-1) = 1. If

x

-1-·o and

xy = xz then y = z.

Proof: Choose w so that wx = 1. Then w(yx) = w(xz) (wx)y = (wx) z so 1 · y = 1 · z so y = z. COROLLARY. The y of VII is uniquely determined.

so

0

3

The Real and Complex Number Fields

Definition: The unique y such that xy = 1 will be denoted by £ 1 Thus x£ 1 = £ 1 x = 1. THEOREM 1.7.

r

(£ 1 1 = x.

r

r

Proof: (£ 1 1 • x- 1 = 1 = x£ 1 . Hence x = (£ 1 1 by Theorem 1.6. THEOREM 1.8. If x and YE j= and y -/, 0 then there exists one and only one element z in j= such that zy = x.

Proof: We may take z = xy- 1 since (xy- 1 )y = x(y- 1 y) = x, 1 = x. If zy = x then zy· = z'y so that z =z' by Theorem 1.6. Definition: We shall denote the unique z such that zy = x by x/ y and call it the quotient of x by y or the result of dividing Of course, £ 1 = 1/ x.

x by y.

Remarks: Since z · 0 = 0 for all z the equation z · 0 = x has no solution for z unless x = 0 and then any z will do. Thus the expression x/ 0 has no meaning for any value of x. Division by O is not possible. THEOREM 1.9. If x, y, z and w € j= and y f. 0, w f. 0 then (x/y)(z/w) = ;:, , (x/y)/(z/w) = (x/y)(w/z) and XW± yz x/ y ± z/ w = · yw .

Proof: (x/y)(z/w)(yw) = ((x/y) y)(z/w) w = x · z = Hence (x/y)(z/w) = ((x/y)(w/z))(z/w)

;!

;! (yw).

by Theorem 1.6.

= (x/y)(wz/zw) = x/y · 1 = x/y.

Hence (x/y)(w/z) = (x/y)/(z/w). Finally (x/y + z/w) yw = (x/y)(yw) + (z/w) yw = ((x/y) y) w + ((z/w) w) y

= xw + zy. Hence xwy~Zl

= x/y + z/w.

THEOREM 1.10. If y f. 0 then -(x/y) = -x/y = x/-y.

4

Theory of Functions of a Complex Variable

= (-l)(x/y) = (-l)(xy- 1) = [(-l)x]y- 1 (-x)(y- 1 ) = -x/y. On the other hand (x/-y)(-y) = x. Hence (x/-y)y(-1) = x. Hence (x/-y)y = -x. Hence -x/y = x/-y. Proof: -(x/y)

- zy COROLLARY. xI y - z I W = xwyw

z"f

y r_1_ -0, w r_j_o.

An element x € j= is said to be a positive integer if it is one of the elements 1, 1+ 1, 1+ 1+ 1, ... , that is if it is in the smallest subset of j= which contains 1 and contains x+ 1 whenever it contains x. We denote the set of all positive integers in 1 by g+_ Consider the set of two elements e and z where we define e e e z e + e e z = z e e+ z Definition:

z + e z+ z

z z

e

=

z

e z

z z

It is easy to verify that these operations satisfy I through VII and hence convert the set of e and z into a field. In this field e = 1 and z = 0. Since 1+ 1 = 0, 0 is a positive integer. There are many fields in which O is a positive integer. They are called modular fields and are important in algebra. We shall not, however, be concerned with them. Definition: A field in which O is not a positive integer is

said to have characteristic O (some authors say characteristic THEOREM

1.11.

S

If

x ( g+ and y

€

g+ then

x+y

€

oo ).

g+_

be the set of all members x of g+ such that € g+_ Then 1 € S by definition. On the other hand, if x € S then (x+ 1) + y = (x+ y) + 1 and x+ y € g+ by definition of S. Hence (x+y)+ 1 € g+_ Thus x+l € S. But g+ is the smallest set which contains 1 i:md contains x+ 1 whenever it contains x. Since S has just been shown to have this property, it follows that g+ C S. Proof: Let

x+ y

€

g+ whenever y

5

The Real and Complex Number Fields COROLLARY. then - x is not in g+.

1

If

is of characteristic zero and x

THEOREM 1.12. If x

all y xy

€

€

g+ then

xy

€

g+,

g+,

Proof: Let S be the set of all x f g+ such that xy € g+ for g+, Then clearly 1 f S. Let x f S and y € g+, Then

g+,

€

g+ and y

€

€

Hence xy+ y

Thus, since 1 f

S

€

g+.

and x f

Hence (x+ 1) y

S

S

argument of the last theorem that THEOREM 1.13. If x

either x-y

€

g+ or

y-x

€

€

implies x+ 1 f

€

;;;i

S.

g+.

Hence x+ 1 f

S,

it follows by the

g+.

g+ and y

€

g+ and

x

f

then

y,

g+.

Proof: Exercise I. Hint: First use induction to prove as a lemma that if z € g+ and z f 1, then z-1 f g+, Then use induction to prove the statement of the theorem using the lemma on y-x (or x-y). COROLLARY. If

are in

1

is of characteristic zero and x and y

g+ then one and only one of the following holds: x-yfg+ yX

Definition: If

1

X €

g+

= y

is of characteristic zero and x

f

g+

then

- x is a negative integer. We denote the set of all negative integers

1

in

by

g-.

We say that an element of

1

is an integer if it is a

positive integer, a negative integer or zero. We denote the set of all integers by g_

x

€

THEOREM 1.14. Let 1 be of characteristic 0. Then if € g we have also x+y € g, x-y € g, xy € g,

g and y

Proof: If x f rem 1.11. lt x f

g-,.

g+

and y

and y

€

g-

€

g+

then x+ y

€

g+

by Theo-

then x+y = x-l-y) and hence by

6

Theory of Functions of

a

Complex Variable

Theorem 1. 13 is either O in g+ or in g-. Similarly x + y E g if x € g- and y E g+_ If either x or y = 0 then x+ y = x or y and is in g_ If x E g- and y f g- then x+ y = -((- x) + (-y)) and (- x) + (-y) E g+ by Theorem 1. 11. Thus in all possible cases x + y E g if x € g and y E g_ Since x-y = x+(-y) and -y E g whenever y E g we conclude that x- y € g whenever x · and y are in g_ The fact that xy

E

g may be proved by an obvious case by case anal-

ysis analogous to that just given for x+ y. The details are left to the reader.

Definition: The rational elements of 1 are the elements of the form x/y where x E g, y E g and y /, 0. We denote the set of all rational elements of 1 by '.R. x € '.R and y E '.R then x+ Y, x-y and If in addition y -f- 0 then x/y € '.R.

THEOREM 1.15. If

xy are in

'.R.

Proof: Let. X= r/ s y = ti u where r, s, t, u E g, s -f- 0 and u I, 0. By Theorem 1. 9 x+ y = ru+ st and ru+ st and su are SU

in g by Theorem 1.4. Thus x+ y x/y are in

'.R

E'.Jl.

The proof that x-y, xy and

are similar and will be left to the reader.

COROLLARY. '.R is itself a field under the operations inherited from 1. !tis the smallest subset of 1 which is itself a field under these operations.

Definition: Two fields 11 and 12 are said to be isomorphic if there exists a function cp where domain is 11 and whose range is 12 such that ( a) cp(x) = cp(y) only if x = y (b) (c)

cp(x+y) = cp(x)+cp(y) for all x and y in 11 cp(xy) = cp(x) cp(y) for all x and y in 11 .

Suppose that 11 and 12 are both of characteristic zero. Then we may define cp by induction for all x E g+ by setting ¢( 1) = 1, cp(x+ 1) = cp(x) + 1. Setting. cp(O) = 0 and ¢(- x) = -cp(x) we define cp for all x E g_ Finally we define cp(x/y) = cp(x)/cp(y). Formal Definitions of "function," "domain" and "range" will be found in the first paragraph of Chapter III.

7

The Real and Complex Number Fields It is straightforward to show that ¢ is thus unambiguously defined

and sets up an isomorphism between

9lj 1

and

9lj 2

•

It follows in

particular that (to within isomorphism) there is just one field which is of characteristic zero and has no proper subfields. We call it the rational field or the field of rational numbers.

Definition: Let 1 be of characteristic zero. We shall call an element of 9l positive if it may be put in the form x/y where x f g+ and y f g+ . We shall denote the set of all positive elements in

9l

by

P9{ .

1.16. The sum of two positive elements is positive and the product of two positive elements is positive. If x f 9l then one and only one of the following holds: x is positive, x = 0, -x is positive. (In the last case we shall say that x is negative.) THEOREM

. g+ .xw+ zy Proof: If x, y, z and w are m then x/y+ z/w = yw and both xw+ zy and zw are in g+ by previous theorems. Similarly (x/y)(z/w) = E.. and xz and yw are in g+. Thus the statements in the first se!t!nce are verified. If x = r/ s where r and s f g and s I= 0 then we may suppose that s f g+ since x also = -r/- s. If r f g+ then x f P9{ . If r f g- then -r f g+ and - x . = -r/ s f P9{ . If r = 0 then x = 0 . If x and - x are both in P9{ then x+ - x = 0 f P9{. Hence O = r/ s where r f g+ and- .s f g+. Hence r = s · 0 = 0 and this is a contradiction. Similarly x cannot be both O and positive or both zero and negative,

Definition: If x and y f 9l then x > y, if x-y is positive and x < y if x-y is negative.

1

Definition: Let 1 be an arbitrary field. By an ordering for we shall me an a subset P of 1 having the following properties. ( a) If x f P and y f P then x + y f P and xy f P (b) If x f 1 then one and only one of the following holds: (i) (ii)

X f

=0

X

(iii) -x

P

f

p

8

Theory of Functions of a Complex Variable

p:J{ is an ordering for :Jl.

COROLLARY.

Definition: An ordered field is a field together with a particu-

lar ordering ments of :f.

P

for it. We call the members of

P

the positive ele-

Remarks: Some fields have several different orderings, some have ex-

actly one and some have none at all. THEOREM

1.17. In any ordered field

Proof: If 1

f P,

then -1

f

P.

g+ C

P.

Hence (-1)(-1)

and this is a contradiction. Hence 1 f P. Hence, if x P. Hence P :::J g+ since g+ is the smallest subset of tains 1 and contains x+ 1 whenever THEOREM

H contains

= 1f P P, 1 + x f P which conf

x.

1.18. An ordered field is necessarily of character-

istic 0. Proof: If O any ordering P.

g+ then O

f

f

P.

But by definition O

I P

for

A modular field has no orderings.

COROLLARY.

:f

Definition: If

is an ordered field and

P is the set of po-

sitive elements we say that x > y if x- y f P and we say that x < y if -(x-y) f P. If x } y so that either x = y or x < y we write x::; y. Similarly we write x THEOREM

x+z

> y+z

1.19.

If

:f

2:

y or x

= y.

is an ordered field then x > y implies

for all z X

X X

X

> > > >

y and y >z implies X > Z y and z>O implies xz > yz y and z < 0 implies xz y implies -x < -y

x> y

> 0 implies x1 < 2 y


y then x-y r P. Hence ((x+ z) - (y+ z)) x-y € P. Hence x+z > y+z. If x > y and y > z then x-y € P and y-z € P. Hence (x-y)+(y-z) € P. Hence x-z € P. Hence x > z. If x > y and z > 0 then x- y € P and z € P. Hence (x-y) z € P. Hence xz-yz € P. Hence xz > yz. If x > y and z < 0 then x-y € P and -z € P. Hence -zx+ zy € P. Hence zy > zx or zx < zy. If x > y then x-y f P so -y- (-x) € p so -y > -x so -x < -y.

P and y f P. If .! i P then - ~ € P. Hence -1 = y(-1/y) € P and this is a c:ntradiction. Hence 1/y € P. Similarly 1/x € P. Hence (x-y) ~ f P. 1 Hence x· 1 . .! - y . .! . _yl f P. Hence -y - j f P. Hence 1 1 X y X y>x. If x > y > 0 then x-y

f

i·

Definition: If x

f

j'. where j'. is an ordered field then

Ix\ lxl THEOREM

1. 20.

If

x and y are elements of an ordered

field then lx+yl S lxl + IYI

t 1 xy,

= 1x if x 2: 0 = -X if X < 0

and \xy\

\x\ \y\.

Proof: We have in all cases \x\ x or -x. Hence lxyl = \x\ = t 2 x, \y\ = t3 y where t 1 = ± 1, t 2 = ± 1, t 3 =

± 1. If t2 = t 3 then \xi IY\ = 1. Moreover xy = \x\ \YI 2: Thus lxy\ = Ix\ \y\. If t 2 t -xy 2: 0. Hence lxy\ = -xy

= t 2 t 3 xy =

xy

since 1 · 1

= (-1)(-1)

0 so lxy\ = xy so t 1 = 1. t 3 then lxl IY\ = -xy. Hence and t 1

= -1. Hence

lxy\

=

-xy

=

\x\ IYI· Thus the second statement is proved. Now consider the first. If \xi = -x then \x\ + x 2: 0 so -x S \x\, also x S lxl. Hence \x\ ± x 2: 0 and IYI ± y 2: 0. Hence if t 1 = ± 1 and t 2 = ± 1 then Ixi + t 1 x + IY\ + t2 y 2: 0.

Hence -(t 1 x + t 2 y)

S \xi + \y\.

10

Theory of Functions of a Complex Variable

Hence

-x + y -x - y X - y x+y +y

are all

< lxl + IYI

-x - y. Hence Ix+ YI :::; Ix! + !YI, Definition: Let A be a subset of the ordered field J. If 3 b

But Ix+ YI

=

X

or

such that x < b for all x E A then b is said to be an upper bound for A, (b need not be in A) and A is said to be bounded above (by b). Similarly if 3a such that x ?: a for all x E A we say that a is a lower bound for A and that A is bounded below. If A is bounded above and below it is said to be bounded. Definition: Let A be bounded above. If the set of all upper bounds for A has a least element b0 , that is, if there exists an upper bound b0 such that b 0 .:S b for all upper bounds b then b 0 is called the least upper bound of A. We use the definite article

since a least upper bound, if it exists, is clearly unique. Similarly we define the greatest lower bound of A if it exists as the element a 0 such that a 0 is a lower bound for A and a 0 bound a of A. THEOREM 1.21.

Let

J. be an

?:

a for every lower

ordered field.

If

every set

bounded above has a least upper bound then every set bounded below has a greatest lower bound. Conversely, if every set bounded below has a greatest lower bound, then every set bounded above has a least upper bound. Proof: Suppose that every set bounded above has a leastupper

bound. Let A be boundefi below. Then 3 a such that x ?: a for all x E A. Hence -x < -a for all x E A. Let K be the set of all x such that -x E A. Then -a is an upper bound for A-. Since Ais bounded above, A- has a least upper bound b0 . We show now that -b 0 is the greatest lower bound of A. Indeed, since The symbol

3

is an abbreviation for ~11there existssis'o

-x

:S b0

11

The Real and Complex Number Fields for all x

f

2:

A, we have x

-b 0 for all x

A, so that -b 0 is a < x for

f

lower bound for A. if a is any lower bound for A then a all x

A so that -a

f

2:

-x for all x

f

A

so that -a is an upper

bound for A-. Since b0 is the least upper bound of A it follows 2: b 0 . Hence a :; -b 0 . Thus -b 0 is the greatest lower bound of A as was to be proved. The proof of the converse is strict-

that -a

ly analogous.

Definition: If an ordered field has any one and hence all of the following properties,. it is said to be complete. (a) Every subset bounded above has a least upper bound. (b) Every subset bounded below has a greatest lower bound. (c) Every subset bounded above has a least upper bound and every subset bounded below has a greatest lower bound. An example of an ordered field which is not complete is the rational field. The set of all rational x such that x bounded but has no least upper bound.

> 0 and x 2 < 2 is

We shall not give a detailed

proof of this fact but shall simply make the following remarks: (a) A least upper bound b, if it existed, would have to be such that

b2 = 2.

(b) A slight development of the theory of oddness and eveness suffices to show that r 2 = 2 is impossible for rational r. THEOREM 1.22.

In a complete ordered field g has no upper

bounds. Proof: If g has an upper bound it has a least upper bound. Call this least upper bound b 0 . Since n :S b 0 for all n

f

g it fol-

lows that n - I :S b0 - I for all n f g_ But every m f g is of the form (m+ 1)-1 where m+ 1 f g_ Hence m _'.S b0 - I for all m f g_ Hence b0 - I 2: b0 . Hence -1 g has no upper bound. COROLLARY.

::3 n

f

If

> 0. This contradiction proves that

x is in a complete ordered field,

g+ such that n > x.

then

12

Theory of Functions of a Comp! ex Variable

We note that if a is a lower bound for g then -a is an upper bound for g, Hence, Theorem 1.22 also holds for lower bounds. To see that there can exist ordered fields in which g is bounded above (and below), consider the set j= all "rational functions" a 0 +ax+· 1

00

+axn n

b0 + b 1 X + "• + bm Xm where the

aj

and bk are in the rational field. These functions are

of course, defined only for such x that the denominator does not vanish. j= is a field under the familiar rules for adding and multiplying functions. Moreover, j= may be ordered by taking a 0 + a 1 x + , · · + anxn to be positive whenever the last non vanishing coefficient an is a positive rational number and defining a positive rational function to be the quotient of two positive polynomials. In this ordered field g is the set of all those constant polynomials for which the constant is an integer. On the other hand, since x- a 0 = -a0 + 1, x is positive for all x 0 , it follows that 1, x is an upper bound for g,

Definition: An ordered field in which g is not bounded above (and hence not bounded below either) is said to be Archimedean order-

ed. Examples of Archimedean ordered fields: (i) The rational field. (ii) Any subfield of a complete ordered field. 1.23. If j= is an ordered field, then every nonempty subset of g+ has a least element. THEOREM

Proof: Let A be a non-empty subset of

g+. If

1

E

A then

1 is the required least element of A. If 1 / A then there exist members of g+ which are less than all members of A. Let S be the nonempty set of all

z

€

g+

such that z


a 0 for all b f A. THEOREM

1.24. Let x and y be two elements in an Archi-

medean ordered field with x < r < y.

:f

with x < y. Then 3 a rational element r

Proof: We may suppose that x > 0, y > 0. Indeed, if y > 0 and x < 0, then we may take r = 0 and if y < 0 and x < 0 we may shift attention to -y and -x. If -y < s < -x then x < -s < y. Given O < x < y dean that there exists n

f

it follows from the fact that :f is Archimeg+ with n > ~ and hence with

; < y- x. For the same reason there exists m f g+ with m )> nx and hence m/n > x. Let m0 be the least member of g+ such that !!1n..9 > x. Then (m 0 _- l)/n < x. Hence mo < x+ l. < x+y-x = y. n n Hence x < < y as was to be proved. Definition: Let x be any element in the ordered field :f. By

:o

L X we shall mean the set of all rational members which are less than x. If :f is Archimedean ordered, then it follows from Theorem 1. 24 that whenever L X = L y we have also x = y. THEOREM 1.25. Let :f be a complete ordered field. Let A be a subset of RJ, Then A is of the form Lx for some x f :f if and only if A has the following properties:

(i) (ii)

r < s and s f A implies r f A A is not empty and is not all of R:f

(iii) Whenever

r

f

A

there exists

s

f

A

with s > r.

Proof: It is obvious that L x has properties (i), (ii) and (iii). Let A have these properties. Let y be in R:f and not in A. Then if a f A, we cannot have a > y since then y f A by (i). Hence

14

Theory of Functions of a Complex Variable

y ::: a. Hence A is bounded above. Since

'.J

is complete A has a unique least upper bound b. We shall show that A = Lb. Let x f Lb.

By Theorem 1.24 3 r f R'.J with x < r < b. Since b is the least upper bound for A 3 s € A wil:h x < r < s < b. Hence by (i) x f A. Thus Lb c; A. Suppose now 3 x f A with x ~ Lb. Then x f A and x ~ b . Since b is an upper bound for A, x :S. b . Hence x = b. But this means that A has a greatest element which contradicts (iii). Thus Lb = A . . Now let

'.J 1

and

'.J 2

be two complete Archimedean ordered

fields. Let ¢ set up the unique isomorphism between R'.J 1 and Rcr . Let x be any element of '.J 1 . Then L has properties (i), J2

X

(ii) and (iii) of Theorem 1. 25. Since ¢ is an isomorphism ¢(L) also has these properties. Hence by Theorem 1. 25 ¢(L X ) = L X ,

for some x' in '.J 2 which by Theorem 1. 24 is unique. We set 0(x) x '. It is not hard to show that 0 is an order preserving isomorphism of '.J 1 on '.J 2 . We leave the details to the reader as an exercise. It follows that to within isomorphism there is just one complete ordered field. We call it the field of real numbers. Definition: Let x1, x2, x3, ., . be an infinite sequence of members of the ordered field '.J. We shall say that xn converges to zero or has zero as a limit and write xn --> 0 provided that given any

positive plies

E there exists a corresponding n 0 such that n

\xnl
0

2: n 0 im-

1.26. If xn --> 0 and y n -> 0, then xn + y n -> 0 and yn is bounded, then xnyn-> 0.

If xn-> 0

Proof: Ixn + yn I -< Ixn 11 + IYn so that n 2: n 0 implies \ xn I < E/2

\.

Given

E > O choose n 0

2: m0 implies y n < E /2. Then if n > n O + m 0 we have lxnl < E/ 2, \Ynl < E/2 and hence Ixn + yn I < E/2 + E/2 = E. Thus xn + yn -> 0. If y n is bounded there exists M > 0 such that \y n I < M for all n. If x n -> 0 and E > 0 is given, then E / M is also and choose m0 so that n

15

The Real and Complex Number Fields

positive. Hence :3 n 0 such that n

2:

n 0 implies

Jxnl


0. To prove the middle statement of the theorem, we need only show that if y n

n

2:

->

0, then y n is bounded. But by definition :3 n 0 such that

jyn I

n 0 implies that

_::; 1. Thus if we leave out y 1 , y2 · · · y n _ 1 0

we certainly get a bounded set. On the other hand, an obvious induction showsthatthesetconsisting of yl' y 2 , ... , yn is bounded no matter what M is. Since the union of two bounded sets is obviously bounded the proof of the theorem is complete. Definition: Let x 1 , x 2 ,... be an infinite sequence of members

1.

of the ordered field

If there exists a member a of

1

such that

xn - a converges to zero, we say that xn converges to a or has a

as a limit. We write xn THEOREM

->

a.

1.27. If x n -. a

and

x n -. b

then

a= b.

Proof: x - a -> 0 and x - b -> 0 hence b - x -> 0 . Hence n n n b x - a+ b - x -> 0. Hence b - a -> 0. If b - a f- 0 then I aJ > 0.

H:nce Ib - aJ < Jb proves the theorem. THEOREM

and xn y n

->

2-9J .

Jb - aJ

Hence

2

< 0. This contradiction

1.28. If xn-> a and Yn-> b then xn+Yn ->a+b

ab.

Jxn+yn-a-bJ

Proof:

=

Jxn- a+yn-bJ

which approaches O by Theorem 1. 26.

Jxnn Y -ayn +ayn -ab!

.:S

Ixn Yn - abJ =

-< IYn I Ixn -al+ Jal IYn -bJ . yn - b is bounded. Hence y n is bounded. Thus

Jal Jyn -bl

->

0

->

Since yn -b-> 0, jyn JJxn - aJ +

by Theorem 1.26.

THEOREM

1/xn

Jxn-aJ + Jyn-bJ

1.29. If

1/ a. Proof: (Exercise).

X

n

->

a and no x n = 0 and a

f- 0,

then

16

Theory of Functions of a Complex Variable THEOREM

1.30.

1

If

is the real number field and x

then there exists a sequence r 1 , r 2 , r 3 , •.. such that each ti.onal and rn --> x.

'J

€

1

zs

ra-

Proof: We need only consider the case in which x > 0 since -x if rn --> x. For each n € g+ 3 m with m > nx. Let m -1 be the least such member of g+_ Then mn /n > x and--%-- < x. m m 1 Hence O < x < ti. Let rn = nn . Then, since lx-rnl


-if -

1,

Ix-rn I

0 3 n >

-->

1/ E

and hence

THEOREM

ordered field

1

0. indeed, since

1

is Archimedean given any E >

1/n < E

. Thus rn

-->

x.

1.31. Let xn be a sequence of elements in the

such that xn

-->

any E > 0 3 n 0 such that if \xn - xml < E .

a

for some

n > n0

a

€

1.

m > n0

and

Then given then

Proof: \xn -xm \ = Ixn -a+a-x·m \ -< Ixn -al+ \a-xm I = Jxn-al + lxm-al. Choose n 0 so that n > n 0 implies lxn -al < s/2. Then if n

> n 0 and m > n 0 we have lxn -xml < s/2+ E/2 =

s. Definition: Let {xn l be any sequence of elements in the 1. Suppose that for any E > 0 3 n 0 such that

ordered field

n > m 0 , m > n 0 implies lxn - xml < E. We then say that thesequence is a Cauchy sequence or a fundamental sequence. THEOREM

1.32. Let

1

be

a

complete ordered field. Let

be any Cauchy sequence of elements of such that

xn

-->

< 1. Hence -1
m0 • Hence -1 + x

Then there exists

a.

Proof: Let Sn be the set containing n > m0 , lxn - xml

1.

mo

+l

x

< xm < 1 +

m

x

xn, xn+l' ... . For m,

- x mo

mo

+1

< 1 for all

+l . Hence

17

The Real and Complex Number Fields

bounded. Hence Sn is bounded for all n. Let Yn be the least upper bound of Sn. Then y1

2:

y2

2:

y3 ... and y1 ,y2 , ... isclearly

bounded below. Let a be the greatest lower bound of y1 , y2 , ...• Then given E > 0 choose n 0 so large that n plies \xn - xJ

> 'n 0 , m > n 0 im-

< E/2.

Let n' be any integer > n 0 • Then for all m > n 0 we have

\xm- xn,\ < E/2 Hence x n ,_ E/2

< xm q< xn, + E/2. Hence for all x,-E

=

JJ 01 [f(x)-g(x)j 2 dx

0 in this case, fn is said to converge in the mean to

g. The example just described converges to zero in the mean but not uniformly. Example 3. Let Rn denote the set of Example 1 and let

max [x 1 -y1 [, [x2 -y2 [ ·•· [xn-Yn [ be denoted by p'(x, y). We may define convergence with respect to the distance p' by saying that xi

-->

x (with respect to p ') if p '(xi, x)

-->

0. Unlike the case of

function spaces, it turns out that p and p' convergence are equivalent; that is, if p (xi, x)

-->

0, then p '(xi, x)

-->

0, and conversely.

A great deal of what we shall wish to say about the convergence of sequences of real and complex numbers applies as well to sequence of n-tuples and to sequences of functions. In addition it applies to a great many other kinds of convergence which we cannot

27

Elementary Point Set Topology

discuss here. We can develop the theory of all of these cases at the same time by introducing the concept of a metric space.

Definition: Let M be any set. Suppose that we are given for each pair p, q of points of M a non-negative real number

p (p, q) . We shall call the function p a metric for M if it satisfies the following conditions: (a)

p (p, q)

p (q, p) for all p and q in M

(b)

p(p, q)

0 if and only· if p

(c)

p(p, q) < p (p, r) + p (r, q) for all p, q, r in M

= q

A set M together with a metric for M we shall call a metric space. We shall call p (p, q) the distance between the points p and q. Wehavealreadyseenthat lz 1 -z2 1 = p(zi, z2 ) isametricfor the complex numbers and that lx 1 -x2 1 = p(x 1, x 2 ) is a metric for the real numbers .

Definition: Let M be a metric space with metric p. If p 1 , p 2 , • • • is a sequence of points of M and p is any point of M such that p (pn, p) --> 0 as a sequence of real numbers we say that pn converges to p or has p as a limit and we write pn THEOREM

2.1. Let p 1' p 2 ,... be a sequence of points in

the metric space M and suppose that pn p

=

p.

-->

-->

p and pn

-->

q then

q.

Proof: p (p, q) .:S p(p, pn) + p(pn' q). Since p(p, pn) and p (p, qn)

-->

0 it follows that p (p, q)

S yn

where y n

->

This contradicts the definition of convergence unless p (p, q) Hence p

=

0

-->

0 .

=

0.

q.

THEOREM

2.2.

If Pn->

p in the metric space M then for

each E > 0 3 n 0 such that n > n 0 , m > n 0 implies p (pn, pm)

< E.

28

Theory of Functions of a Complex Variable Proof: Left to the reader. Definition: If {pn l is a sequence of elements in the metric

space M and if for each n 0 implies p (pn, pm)

E

> 0 3 n 0 such that n > n 0 and m >

< E, then {pn l is said to be a Cauchy se-

quence. Definition: A metric space in which every Cauchy sequence

has a limit is said to be a complete metric space. We have seen that the real and complex numbers are complete as metric spaces. Let a and b be real numbers. The set of all real numbers x with a ,:S x ,:S b is said to be the closed interval a, b and is

often denoted by the symbol [a, b]. The set of all real numbers x with a


la- cl for all n. This is the prop-

erty which makes the notion of closedness important for intervals. We generalize it as follows: Definition: Let F be a subset of the metric space M. We

say that F is closed if every point p of M which is a limit of a sequence of members of F is itself a member of F. Definition: Let A· be a subset of the metric space M. Th~n

by the closure

A of A we shall mean the set of all

p

that qn-+ p for some sequence {qnl of members of A.

f

M such

29

Elementary Point Set Topology

2.3. For any A, A SA

THEOREM

and A= A.

Proof: Since q, q, q, ... --> q, it follows that A ~ A , Now let pn--> p where each Pn

if. Then for each Pn 3~€Aso

f

< ¾ . Then

that cf...m --> p n . Choose m(n) so that p (P,n , qm( n)) p(qm(nJ p) Thus

p

if .

f

S p(qm(nJ pn) + p(pn, p) < ¾ + p(pn, p)-> 0

A = A.

Thus

COROLLARY:

is the smallest closed set containing A.

if

Definition: If A and B then x

f

A

if and only if x

f

A and x

f

B

and

f

A U B if and only if x

f

A and x

f

B or both.

x

S:

If if

nB

are subsets of the same set M,

X €

is any family of subsets then x

A for at least one A

for all A

€

s:.

€

nci:

X €

A€J

1.

U er A if and only

f

A €J

A if and only if

X €

A

THEOREM 2.4. If A and B are closed then A U B is closed. If j'. is any family of closed sets, then Oci: A is closed. A€J

Proof: Let qn-> p where qn exists a subsequence q

nl

; q

n2

€

A U B.

Then either there

, ... all of whose members are in A

or there exists a subsequence q

nl

, q , . . . all of whose members are n2

in B . Since we still have qn. -> p we have p

f

A or p

€

B.

1

Hence p

f

A U B. Thus A U B is closed.

Now let q -> p where q n

q

n

€

A for all n. Hence p

for all A

€

j'.. Hence p

€

€

n

€

Oci:

A€J

A.

Then for each A

A since A is closed. Hence p

nci: A.

A€J

Hence

ncr A

A€J

is closed.

Definition: The subset S of M is said to be dense if

S=

M.

€

1,

€

A

30

Theory of Functions of a Complex Variable Let x be a point in the open interval (a, b). Then

E

> 0 such that for all

y with \y - x\

< E

3 an

y is also in (a, b).

This is the property which makes the notion of openness for intervals important. We generalize it as follows.

Definition: Let G be a subset of the metric space M. If for each x

f


0 such that

G 3 E

y

G for all y with

f

we say that G is an open subset of

E

THEOREM

M.

2.5. If A is a subset of the metric space M,

then A is open if and only if M - A is closed.

f

Proof: Suppose that A is open and let xn -> x where each x n M -A. If x f A, then 3 E > 0 such that p(y, x) < E · :J · y f A ..

Hence p(x, x ) n

> -

for all n. Hence x -/, x. Hence x

f

n

M-A.

f

Hence M-A is closed. Conversely, suppose that M-A is closed and let x

A. Suppose that for each n

f

with p(y, x) p (xn, x)

<

· M - A. Hence p (y, x)

open. If

1

x. This is a contradiction since x


x where xn E A for all n. Then {·xn l is a Cauchy sequence. Since A is complete 3 y E A such that xn-> y. Since xn-> y and xn-> x it follows that x = y. Hence x

E

A. Hence A is closed.

The strict converse of Theorem 2.9 is not true. For example, the whole space is al ways closed but of course need not be complete. On the other hand, the following conditional converse is true. THEOREM

2.10.

If

A is a closed subset of a complete met-

ric space, then A as a metric space is itself complete. Proof: Let xn be a Cauchy sequence of elements of Since M is complete there exists y M is closed, y

f

E

A .

M such t'hat xn-> y. Since

A. Hence A is complete.

32

Theory of Functions of a Complex Variable

Roughly speaking, we can say that a closed subset of a metric space M is as complete as the lack of completeness of M allows it to be.

It should be noted that closure is a relative property depending on the nature of the large space in which the subset A is embedded. Completeness, on the other hand, is an absolute property. It has nothing to do with anything except the points of A.

Definition: A subset A of a metric space is said to be bounded if the set of all real numbers of the form p (x, y) where x

€

A and y

f

A is a bounded set of real numbers. In other words,

A is bounded if

3 b > 0 such that p (x, y)

.S

b for all x, y

f

A.

The least upper bound of the set of all p (x, y) is called the diameter of A. THEOREM

2.11. A set of real numbers is bounded in the met-

ric space sense if and only if it is bounded in the order sense. Proof: Left to reader. THEOREM

2.12. Let A be a subset of the complex field;

then if A has any one of the following properties, it has them all. ( a) A is bounded with respect to the metric. (b) There exists a positive real number R such that

Izl .S

R

for all

z

€

A.

(c) There exist real numbers a, b, x + iy

€

c

and d such that

A implies a < x < b and

c

_s

y

_s d .

Proof: Exercise I. THEOREM

2.13. Let A be any bounded set of complex num-

> 0 there exist an integer m and subsets Sn each of diameter < E such that A C S1U S2 U ...

bers. Then given any S1, ~, ... ,

u sn

E

33

Elementary Point Set Topology Proof: Exercise II.

While Theorem 2.13 may be formulated in an arbitrary metric space, it is not true except for rather special cases. For example, let M be any set and let p (x, y) = 1 whenever x -1- y. Then M is certainly bounded. On the other hand, a set of diameter

½ cannot

contain more than one point. Hence Theorem 2. 13 cannot be true if we take M to be any infinite set. The property announced in Theorem 2.13 is the most important property of bounded sets of complex numbers. Since S is not true of bounded sets generally, it is convenient to give it a special name.

Definition: A subset A of the metric space M is said to be totally bounded if given any E > 0 3

subsets S1, S 2 , ... , Sn of

A (in a positive integer) such that S.

has diameter

J

A

f


0 such that

Choose n 0 > 2/E

and so that n >

< E/2 . Then if n > n 0 and z

n 0 implies p(xn, y) have p(z, y)

0.

x 2, ... is a Cauchy sequence.

Xi,

Since A is complete there exists y

are both in

€

C n we

< 1/n + E/2 < E/2 + S/2 =

E..

t) and this contradicts the fact that en

cannot be covered by a finite subset of :J. This contradiction completes the proof.

Remark: The classical Heine-Borel theorem is the special case in which A is a closed and bounded subset of Euclidean n space.

Definition: We shall say that A has the Heine-Borel property if whenever :J is a family of open sets such that A S t)~:J 0, then A S aY:J, 0 THEOREM

for some finite subfamily :J' of :J.

2.19. A set

A S

M has the Heine-Borel prop-

erty if and only if for every family :J of closed subsets of M such that

A

n(

that

A

n ( F~:f ,F)

F~:f F)

Proof: If A FY:J (M-F),

= 0

there is a finite subfamily :J' such

= 0.

n( F~:f

F)

= 0 then A SM -

and conversely. Thus to say that A

0 is the same thing as to say that M - F with F

€

(F~:f F) =

n ( FQ:f F) = :J

form an

37

Elementary Point Set Topology open covering of A. Similarly, to say that A is to say that the M - F with F

f

S:'

n ( p~j=, F) = 0

form a finite subcovering.

The truth of the theorem should now be an obvious consequence of that of the preceding theorem. THEOREM

2.20. lf A has the Heine-Borel property then it

is complete and totally bounded. Proof: Let A have the Heine-Borel property. For each 71 > 0 and each x in A

p(y, x) < x

f

be the set of all y with x,71 71. Given E > 0 let S: be the set of all Sx,E/ 3 with

A. Clearly A ~

Borel property A f:

let

S

rNs= 0.

01

U

02

Hence, since A has the Heine... U

0n

where each

the diameter of S x, E/3 is not more than } +

i

0j

f

S:.

But

= ~ E and hence

is less than E . Thus A can be covered by a finite number of sets whose diameters are less than E. Thus A is totally bounded. Now let x 1 , x 2 , .. . be any Cauchy sequence of members of A. Let F n be the closure of

{xnl U {xn+ 1 1 U .... Then A

n Fn is not empty for any n. F2 n ... is not empty. Let x

n F1 n F2

Hence, by Theorem 2.19, A

••.

n Fi n

be any element of this set. Given

E > 0, choose n 0 so large that m > n 0 and n > m0 imply that p(xn, xnJ < E/2. m

> m0 ,

p(xn, xnJ

F , there exists m such that no+ 1 E/2. Thus n > n 0 implies p(xn, x)


x 0 then f(xn)

> 0 .3

plies p(f (x), f(x 0))

o> 0


f(x 0 )

such that p(x, x 0 )


x then (£1 + f2 )(1ti) = f1 (xn) + fixn), But f1 (xn) ---> f 1(x) and f 2 (xn) ---> f 2 (x) since f 1 and f 2 are continous. By an earlier theorem (Chapter I)

f 1(xn) + f 2(xn)---> f 1(x) + f 2(x) =

43

Continuous Functions

U 1 +f2 )(x). Hence (f 1 +f2 )(xn)--,(f 1 +f2 )(x). Hence f 1 + f 2 is continuous. The rest of the argument proceeds along analogous lines. COROLLARY., If a 0 , a 1 , a 2 , ... ,

an are real numbers then

the unique function f such that f(x) = a 0 + a 1 x + ... + an~ for all real x is continuous. If a 0 , a 1 , ... , an are complex numbers then the unique function f such that f(z) = a 0 + a 1 z + a 2 z 2 + ... + a n zn for all complex z is continuous. Definition: If f is a function with domain S 1 and range S S 2 and if A is any subset of S2 then F 1 (A) is the set of all X € sl such that f(x) f A. THEOREM

3.4. If f is

a

function whose domain is the met-

ric space M1 and whose range is contained in the metric space M2 then f is continuous if and only if r 1(0) is open whenever 0 is an open set in M2 . Proof: Exercise II. Note: If f is continuous and

0

is an open set in the domain of f

then f(0), the set of all f(x) for x c 0, is not necessarily open. For example, if f(x) = x 2 and 0 is the interval - 1

--->

0.

0 , and

F (f (x 0 + hn)) - F (f (x 0 )) f (x 0 + hn) - f(x 0 )

F(f(x 0 ) + Hn)- F(f(x 0 ))

-.. F '(f(x ))

Hn and the theorem is proved in this case.

If f(x 0 + hn) - f(x 0 ) = 0

for an infinite sequence of n's

n 1 , n 2 , ••• then

= 0 so f'(x 0 ) (F

o

= 0.

f) '(x0 )

=

0.

Also F (f(x 0 + hnj)) - F (f(x0 )) Hence (F

o

f) '(x 0 )

=

= 0 so

F (f(x 0 ))f'(x 0 )

and the

proof is complete. THEOREM 3.19

(Rolle's Theorem). Let f be a continuous

real valued function whose domain is the closed interval a and which is such that f(a)

=

f(b)

=

0. Suppose that f is differen-

tiable at every point in the open interval a with a


0 such that O < \h\ < o implies f(t + hi - f(c) > 0. Hence f(( + h) > Call it M. Then f((;J

f((J for O

< h < o and this is a contradiction. If f'((;J < 0 we

apply the same argument with h negative. COROLLARY 1 (Mean value theorem.) If f is continuous and real valued with domain a

x with a

f(x, y 0 ) is differ-

= x 0 we say that the first partial derivative of f

exists at x 0 , y0 and that its value is the derivative of x f(x, y0 ) at x 0 .

We denote it by f /x0 , y0 ).

--->

The function f 1

thus defined is called the first partial derivative of f. The second partial derivative is defined similarly. THEOREM

domain is a subset Let z 0 = x 0 + iy0

4.4. Let f be a complex valued function whose

:D of C. Let € :D. If f'(z 0 )

f(x + y) = u(x, y) + iv(x, y). exists then u 1 (x0 , y 0 ),

v /x 0 , y 0), u 2 (x 0, y0 ), v2 (x 0, y0 ) all exist and C denotes the set of all complex numbers.

60

Theory of Functions of a Complex Variable

11i (xo, Yo) + ivl (xo, Yo) -iuixo, Yo) + v2 (xo, Yo) Hence in particular

11i (x0, y0 ) = v2 (x0, y0 ) and v1 (x0, y0 )

-uixo, Yo). Proof: f(x 0 + iy0 + h 1 + ih 2) - f(x 0 + iy0 ) hl + ih2 u(lo +I;_, y0 +h 2) + iv(x0 +h 1, y0 +h 2)- u(x 0, y0 )- iv(x 0, y0 ) hl + ih2 - u(xo+l\, Yo +h2) - u(xo, Yo) + i v(xo+h1, Yo +h2) - v(xo, Yo) -

hl + ih2

If h 1 = h,

h2

hl + ih2

= 0 this reduces to

u(x 0 + h, y0 )- u(lo, y0 )

(x 0 +_ h,__::; y0 ) - v (x 0, Yo ) + 1. v_..;:._ _ _~ _ ; : ~

h

Let h

= hn where h n

h ~ 0 then since f'(z 0 )

exists, our expres-

sion .converges to f'(z 0 ). Hence the real part converges to the real part of f'(z 0 ) and the imaginary part converges to the imagin• ary part of f'(z 0). Hence u/x 0, y0 ) ~d v1 (x 0, y0 ) both exist and f'(z 0) = u 1(x 0, y0 ) + iv1 (x 0, y0 ). Similarly setting h 1

= 0, h 2 = hn we show that

uix0 , y0 ) and v2 (x0, y0 ) exist and that f'(z 0) =

I (~ (x0, y0 ) + iv2 (x0, y0 ))

= v2 (x0, y0 ) - iuix0, y0 ) .

Setting these two different expressions for f'(z 0) equal to one another we conclude that

61

Differeniiation of Complex Valued Functions

= v2 (x0 , Yo) ( u i

z 0 then z 0

€

A 1 by the con-

tinuity of f. Hence no limit of members of A 1 can be in A 2 . Since A was connected and A1 is not empty, A2 must be empty and the proof is complete.

64

Theory of Functions of a Complex Variable Definition: Let ai, a2, ... be a sequence of complex num-

bers. Let s n = a1 + a2 . .. an. If sn converges to a limit s we say that s is the sum of the infinite series and that the infinite series converges.

a1 + a2 + a3 + ...

sn is called the n-th partial

sum of the series. Note: The fact that the sequence a 1 , a 2 , ... converges to a limit E does not imply that the corresponding series a 1 + a 2 + .'.. converges. Indeed, it is obvious that a 1 + a 2 + . .. cannot converge unless an converges to O and easy examples, such as 1 + ½ +

½ +... show that even this is not enough. THEOREM 4. 7. If a 1 + a2 + a3+ ... is a series of non nega-

tive real numbers and s n the sequence of partial sums is a bounded sequence, then a 1 + a 2 + · · · converges. Proof: Left to reader. THEOREM 4.8. If b1 + b2, ••• converges and Jani

< bn

for all n, then a 1 + a 2 + · · · converges. Proof: Let Sn = b 1 + ... + bn'

s

n

if n > m, Jam+l + am+2 ··· an I


S,

Sn is Cauchy. Since isn-sml .:S JSn-Sml' sn

is also Cauchy. Hence sn --" s

and the proof is complete.

COROLLARY. If I a 1 I + I a2 i ...

converges.

converges then a 1 + a 2 + ...

65

Differentiation of Complex Valued Functions Definition: If

Ia 1 I + Ia2 I +

·.. converges we say that

a 1+ a2 + ··· converges absolutely. LEMMA. If

\r\


1 + E where E > 0. Hence -~

= (1 +E)n =

(l+E)n-l + E

>

r

=

(1 +E)(l +E)n-l >

l+nE. Hence

1 1 = 1 ( -1- ) 0 such that

a 0 + a 1 z ... converges absolutely whenever Iz I

< r and

a0 + a 1 z ... has unbounded partial sums whenever

lzl

>

r.

Proof: lf (b) does not hold then it follows from Theorem 4.9 that 3 p

> 0 such that Iz I
n 0 ,

> 0 then there xn < e+ E and for all n 0 3 n > and

E

such that xn > E- E.

n0

e+ E.

Proof: Choose no so that n > no implies Un < Then

X

n -< un
no . On the other hand, given no

there exists n such that n > n 0 THEOREM

4.11.

and xn > un 0+l - E > E - E .

Let a0 , ai, a 2,

be any sequence of

...

complex numbers and let r be the radius of convergence of the seThen 1/r = lim y\an\

ries a 0 + a 1z+ a2 z 2 + ... terpret

1/oo

as O and 1/0 as

provided we in-

oo.

Proof: Clearly lim vlan I ?:: 0. Suppose that lim v1an1 = 0. Then, by the lemma,

n

yjan\

->

n

0. Hence, for any E > 0,

"

y\anl < E,

for n > n 0 (B). Hence \an\ 0 there exists a subsequence nk

via

nk

1/ka nk

I>

I>

M. If \zl

>

1.

Mnk

M was arbitrary, the radius of convergence is n

suppose that lim y an

l/ p

then

nk

_M._

Hence the series cannot converge absolutely for -

1

-M

= p where

O< p


1/M. Since

O = 1/ oo oo •

•

Finally,

Then for \ z \
no implies vlan hence \an\ 1/ p, p > 1/lz I and there exists n 1 , n 2 , ••• such "k nk that . Ia -+ p . Thus ya > 1/I z I for k sufticiently large. V nk nk Thus

Iz

n

k ank I

> 1. Thus a 0 + a 1 z+

···

does not converge so

r S 1/p . Since r S 1/p and r 2:. 1/p we conclude that r

=:

1/ p

as was to be proved. THEOREM 4.12. If a 0 , a 1, a 2, • • . are all non zero, then the radius of convergence of a 0 + a 1 z + a 2 z 2 + .. . is between

1

and

1 -

a

liml-"-1 an+l

Proof: Exercise II. Example:

The following series each of which is the formal deriva-

tive of the preceding

aJ.1

have radius of convergence 1.

1+ z+ z 2 /2+ z 3 /3+ ... z"/n 1 + z + z 2 + z 3 + .. · 1 + 2z + 3z2 + ... We shall see below that a power series and its formal derivative always have the same radius of convergence. (Weierstrass M test). Let f 1 , f 2 , .. . be a sequence of complex valued functions defined on a subset A of THEOREM 4.13

the complex plane. Let lf/z )I

S Mi for all z

€

A and suppose

that M1 + M2 + ... converges. Then f/z) + f2 (z) + ... converges uniformly on A. Proof: Let Fn(z)

IF,/z)- Fm(z)I

=:

=:

f/z) + f2(z) + ... + f.,(z). Then

lfn+/z) + ... ,- fm(z)I S Mn+l + ... Mm

69

Differentiation of Complex Valued Functions Hence !Fn(z)l isaCauchysequenceforall z. Let f(z) =

lim n4oo F n (z) = f 1 (z) + f2(z) + ··· . Then lf(z) - Fn (z)I -< Mn+l + ·· · = y n --> 0 as n --> oo. The conclusion of the theorem is now obvious. THEOREM 4.14. Let a be less than or equal to the radius

of convergence of a 0 + a 1 z + a2 z 2 + ... . verges uniformly in the circle

Iz I :S

a.

Proof: The series converges for z Ian Ian

for Iz I

Then the series con-

= a and

Ian~

I :S

:S a . Thus Theorem 4.14 follows from Theorem

4.13. LEMMA.

lim n-->oo

vn ½

Proof: Let bn

vn

1

> 1

for

where

Then bn = 1 + h n

n > 1. h n > 0.

Therefore ·V1n = (b n )n = (l+h n)n > 1 + nh n Therefore h < yn- l < 1 n n \fr1 Therefore 1

< 1+ 2 + -1 Since 1 + -:-r;:vn n

0

-->

(l+h n ) 2

< Vn = bn2

v~

< (1+ yn 1 )2

+ 1/n .

the lemma is proved.

THEOREM 4.15. The series a0 + a 1 z + a 2 z 2 ... and a 1 + 2a 2 z + 3a3 z 2 + ,, , have the same radius of convergence.

Proof: Let p vergence of

= lim

ao + alz +

...

vi an I . Then

1/p is the radius of conand lim nyn vlan I = 1/p. Hence

the two series have the same radius of convergence.

70

Theory of Functions of a Complex Variable Let r be the radius of convergence of

THEOREM 4.16.

a 0 + a 1 z + a 2 z 2 ··· and let f(z) = a 0 + a 1 z + ··· for every z with Iz I < r. Then f'(z) exists for every z such that Iz I < r and for all such z

Proof: Fix a z with lzl < r and r - Iz I - Then for O

< \h I
0

o
E(O) = 1 so E(l) = 1 + h where h > 0. Hence E (n)

= (1 + h)n > 1 + nh. Thus E (n) is unbounded.

73

Differentiation of Complex Valued Functions

Since E ( 1/n) = 1/E (n) we see that the range of E contains arbitrarily large positive numbers and arbitrarily small positive numbers. Hence, since it is connected, this range contains all positive real numbers. Definition: If a > 0 then such that

e10 g a

=

log a

is the unique real number

a.

Definition: If a > 0 and z is any complex number, then az

ez log a . THEOREM

over log (x 1x 2 ) for all

4.21.

d: log x

= 1/x for all x > 0. More-

= log x 1 + log x 2 and log (x/ x2 ) = log x 1 - log x2

x 1 > 0 and x 2 > 0. Proof: Exercise III. THEOREM

4.22.

If

> 0 and b and

a

c

are real then

ab> O, (ab)c = abc. Also, for all complex numbers z 1 and z2 and

zl - z2

a

zl

= a /a

z2

.

Proof: Exercise IV.

> 0 and b > 0 then log 8 b is the unique such that ax = b.

Definition: If a real number x

THEOREM

then

4.23. If a

log 8 b = log b/log a

> 0,

and

a -/. 1;· b

> 0 and c is real

log 8 be = c log 8 b.

Proof: Exercise V. THEOREM

4.24.

If

y is real then

/ eiy/

1.

74

Theory of Functions of a Complex Variable

= c(y) + is(y) where c(y) and s(y) are

Proof: Let eiy real. Then eiy eiy

c(y) 2 + s(y) 2

=

.

But e 2

=

ex+iy

=

ex(c(y) + is(y)) is analytic as a function of z. Hence, by the Cauchy-Riemann equations,

+ ex s'(y) -exc'(y) Thus

s'(y)

= c(y),

c'(y)

_!L (c(y) 2 + s(y) 2 ) dy

= -s(y) . Hence -2c(y)s(y) + 2s(y)c(y) -

0 .

Hence

THEOREM

that

eia = i

positive for O

4.25. There is a real positive number a such

and the real and imaginary parts of

eiy

are

both

< y < a.

Proof: Let c(y) and s(y) be as in the proof of Theorem 4.24. Suppose that c(y) is never zero for y > 0 . Then since s'(y) = c(y), O_::.;y1 0 Now c'(y)

Hence

for all y > 0.

= -sly) < 0 for ally> 0. Thus, bythelaw

.

of the mean, c(y)- c(l) < -s(l)(y-1) if y > 1.

Since c(y) > 0,

-c(l) < -s(l)(y-1). Hence for all y > 1, y-1
0. Let A be the

2: 0 for which c(y) = 0. Then A is closed and hence contains its greatest lower bound a. Since c (0) = 1, a > set of all y

0. By definition c(y)

t-

0 for O < y < a. Hence c(y) > 0 for

0 < y < a. Since s(O) = 0 and s'(y) = c(y), 0 < y < a plies s(y) > s(O)

= 0. Thus

a

im-

has all properties required.

75

Differentiation of Complex Valued Functions

Definition: rr is 2a where a is as in Theorem 4. 25. In other words, ~

is the least positive zero of c (x).

THEOREM

4.26.

w is any complex number with

If

0,

v

> 0. Since c is continuous, c (O)

c(p

= 0 we can find

Now

c(y)2 + s(y) 2

v 2: 0,

v

= c(y) + is(y)

s(y). Thus w

eiy.

v > 0. Then u + iv

Next suppose that u .:S O, i ( v- iu) = ie

u.

1 = u 2 + v 2 • Hence v = ± s(y). Since

s(y) 2: 0,

.

y .:S ~ with c(y)

.:S

y with O

1 and

,

by the argument just given. But

1Y

. ,

i!!.2

ie 1 Y

z'y

e

e

= -e-iy

1

= e

i(y+!Z) 2

If u < 0, v < 0 then u+ iv= -(-u-iv)

Finally, if u 2: 0 and v

·2

e iy'

< 0 then

u + iv = -i(-v+iu) = eiy'(-i) TtIEOREM

teger k such that

4.27. eiy = 1 if and only if there exists an in-

= 2rrk.

y

I!. i

Proof: Since e

2

=

1,

i4

=

1 = e 2 "i

.

Hence e 2 rrki

· lk = 1 for all k. Now let eiy = 1 and choose an integer n such that y -

;rr

nr .:S

y

,I, 0 then

< (n+ 1) ; . Then O

.:S y - ;rr < ;

iy

i (y-~)

\

e

this is a contradiction since

I \ lnl

2

1¼1 ,/, 1 and 1 Ieiy I = 1. Thus

,/, 1. Hence =

1

and

. If

76

Theory of Functions of a Comp/ ex Variable i77

y

; 77

•

But eiy = (e2 )n = in = 1. Hence n = 4k where

k is an integer. Hence y = COROLLARY 1.

t{ rr

= 2rrk.

ei (y+ Zrr) = eiy for all y.

COROLLARY 2. If \ w \

1 then there exists a unique

real number y such that O :S y < 2rr COROLLARY

and

eiy = w.

3. Every complex number z

f- 0 may be

written in one and only one way as z = re;e where 0 is real, 0

< 0 < 2rr and r > 0 . Definition: When z = re;e and r and 0 are as in Corol-

lary 3,. r is called the modulus of z and 0 is called the argument of z. Of course the modulus of z is equal to \ z \ .

Definition: If 0 is a real number then T 0 (z) = e;e z for all complex numbers z. It is clear that IT 0(z1 ) - T e(z2 ) \ lz1-z2I,

Te(O) = 0 and

Te (Te (z)) = Te +0 (z). 1

2

1

The

2

transformation T0 is called the rotation about zero through the "angle" 0. THEOREM

4. 28. If E1 and E2 are two lines through 0

then there exists a unique real number 0 such that O < y


0 and R > 0 such that [z [ > R implies [f(z)[

~ M[z[m.

Then f is a polynomial.

Proof: By Theorem 5.19 if f(z) = a0 +a1 z+a 2 z 2 ...

la n [ -
R implies

Mr /rm .

.·. [a [ < Ml" n - rn

= 0 for

.·. [an [

_M_ ,n-m

=

lim _M_ r->oo rn-m

If n>m

=

then

Mr -< Mrm 0

n >m

polynomial. THEOREM

5. 25. Let f be an entire function which is not a

polynomial. Then given any complex number c, any E > 0 R > 0 3 z

€

and any

C such that

[z [ > R ;

(1)

(2) [f(z).- c[ < E .

Proof: Let Ve be the set of all z such that f(z) = c. Case I: Ve

is infinite. Suppose Ve is bounded. Then it is totally

bounded. Hence 3 z 1, z 2 , •.. z0

•

Since f(zj)

=c

all different in Ve such that zj ->

it follows from Theorem 5.15 that f(z) ; c

and this contradicts our hypothesis that f is not a polynomial. Hence Ve is unbounded. Hence given R > 0 3 z

R that is [z [ > R such that [f(z) - c[ Case ll:

Ve

is empty. Let g(z)

=

€

Ve with [z [ >

= 0.

f(zl -c· Then g is entire.

If g were a constant f would also be a constant. Hence g is un1 > bounded. Hence given E > 0, R > 0 3 [z [ > R with [f(z)-c[

1/E i.e.,

[f(z)-c[ R

for all R.)

since f is

100

Theory of Functions of a Complex Variable

Case ll/:

Ve

is finite. By Theorem 5. 23 f(z) - c

where p is a polynomial and g is entire and

f

'=

p(z)g(z)

O for z

€

Ve .

Hence g is entire and never zero. Suppose that the theorem is false in this case. Then 3 R > 0 and E > 0 such that lp(z)g(z)I

2:

E

for Iz I > R. Hence 1

1\ R

Hence

1 I .:S _El gGJ

lzf m

j

for some m and all z with Iz I > R . Since lows from Theorem 5.24 that

kg,z,

1/g

*

is entire it fol-

is a polynomial. Since

is g,z1 never zero g must be constant. Thus f must be a polynomial and this is a contradiction.

Definition: Let ... a_2, a_ 1, a 0 , a 1, a 2 , ... be a double sequence of complex numbers. If for some n the series an+ an+l + an+ 2

...

converges and the series an-l + an_ 2 + .. . converges we say that

I".° a. convergesandisequalto 1= -oo J

(an +a+ n 1.... )+(an- 1 +an- 2+ ... ).

Cl early the convergence of the two half series and the sum is independent of our choice of n. We define absolute and uniform convergence of double series in the obvious way. THEOREM

5. 26. Let f be analytic in the open set A of all

z with r < Iz- z 0 I < R. Then 3 a unique double sequence ... a_ 2, a_l' a0 , a 1, a 2, .. . such that 00

f(z) =

I n=-oo

with

r

< lz-z0 1 < R.

for all z

101

Complex Integration

Proof: Choose r 1 and R 1 so that r < r1 < R 1 < R. Let i}0) = z 0 +r1 e ;0 , 1-/0) = z 0 +R 1 e ;0 .... Let A 1 betheset where r 1 < (z-z0 ) < R 1 . For each z analytic in ( for (

1:

1:

f(?)-_f;z)

A1

f(z)- t(z) z- z

A provided we define

is

to be

f'(z). Hence

dt; =

I

dt;

t(?),_ t(z) z

'f2

by the theorem preceding Cauchy's theorem. Hence f(z) [

I

'f2

But (~z r

A

J-5- ':, - z

,~'z -I ,~'z ] 'f1

= -

I ~

'f1

is analytic in ( for l(-z0 \

z

( -

+

I

'f2

< \z-z0 \.

fst:df

Hence

= 0 by Cauchy's theorem. On the other hand by the Cauchy

integral formula

J

'f2

_g_ =

27Ti

f;e~(

+

;; _:

Hence 277Nf(z) = -

f 'f1

Now on 1- 1 , 1(-z0 \

z

)n

n=-oo

where for n < 0 and for n

~

0

d(

103

Complex Integration Actually by the theorem preceding Cauchy's

where 'f(0)

Thus an r

= z + r' eie and

< r '< R. It remains to prove that the an are unique. Suppose that 00

n=-oo

00

an(z- z0)n

=

!, n=-oo

for all z with r < \z-z 0 \ < R.

Let en

a n -bn . Then

00

I

cn(z- zo)n

n=-oo

Hence

=

0

for all z

f

A.

-oo

- I n=-1

c/z- z 0 )n

For each z such that z - z 0 is in the circle of convergence of !, 00n=O an wn

let 00

For each z such that

I

n=l c-n ~

00

is in the circle of convergence of

let -oo

Then the intersection of the domains of f 1 and f 2 is an annulus including the annulus A. Since f 1 = f 2 in A it follows that

f1 = f 2

in the intersection of the domains of these two functions.

Hence _3 a unique analytic function f defined on the union of these domains which agrees with f 1 and f 2 on their respective domains. Clearly f is an entire function. Hence I ;=O en (z- z 0 )n

104

Theory of Functions of a Complex Variable

converges for all z and Z

t' z 0

-

l;:'_ 1

cn(z-z0 )n

converges for all

•

Thus f 1 is entire and 00

fiz) = -n:1 c_n ( tends to O as

Izl

--, oo.

1 ) z-zo

Since f/z) = f 2(z)

for z -/c z 0 it fol-

1ows that f 1 is entire and bounded and hence a constant. Since f2 (z)

->

0 the constant is zero. Thus f/z) = 0 .·.

n 2: 0 by the uniqueness theorem for power series. as en= 0 for n z 0 .

If

z0

is an essential singularity then for each complex number c and each r

> 0 and ea.ch E > 0 there exists z with \z-z 0 \


R we say that

> 0 such that f is defined oo

is an isolated singularity

of f. Definition: Let f have

oo

as an isolated singularity. By

the theorem on the existence of a Laurent series f(z) =

I

:=-oo

If an

anzn

where the series converges at least for

= 0 for all

n

> 0 we say that

f is analytic at

Iz I > R. and let

oo

= 0 for all but a finite number of positive has a pole at oo • If an t- 0 for infinitely many

f( 00 ) = a 0 • If an

we say that f

points n we say that f has an essential singularity at Note that f(z)

=I :=l

an zn + I

:=o

n

oo.

a

-:.Zn where the z first series converges for all z and represents an entire function

while the second series rep resents a function defined for

Iz I >

R

107

Complex Integration

which tends to zero as z THEOREM

rz:r- -,

O,

n

f(zn)

1

whenever

Tz:f

given any

c

5.30. ->

If f

then whenever

oo

If f has an essential singularity at

:3 {z l with n

THEOREM

is analytic at

f(oo). If f has a pole at oo then j f(;n)j

0.

--->

We conclude at once the truth of

-> oo.

JZII .:.. 0 and f(z n ) z n

0

then

c.

·.

5.31. Let f be analytic everywhere except for a

finite number of poles and let f be analytic at there.

->

oo

--->

or have a pole

00

Then f is a quotient of two polynomials. Proof: If f is analytic at oo then f(z)

al

a2

= a0 + -

+ ~ , + ···

z

z

where the series converges for all z with I z I > R. Let an cf O. Then zn+l f(z)

has a pole at

Thus we need only consider the

oo.

case in which f has a pole at oo. Let z 1 , z 2 , of f (other than oo ). As we have seen,

=

f(z)

••• , zr

be the poles

g/z)

(z- zl{l

where g 1 is analytic at z 1 and g 1 (z 1 ) cf O. Thus by induction f(z)

=

g(z) (z-z { 1 ··· (z-z

1

r

{r

where g is entire and g(z.) cf O for any pole z. of f. Thus J

g(z)

Let

=

f(z)P(z)

r;-r-

--->

0 . Then

lznl

is

J

for all z not a pole off,

£tzT

->

0

and

P

p (; l

n

a polynomial and f has a pole at oo. Hence

Hence g has a pole at entire. Hence

oo.

being a polynomial. --->

0 since P

n

I g(!) I

--->

0.

Hence g is a polynomial since g is

108

Theory of Functions of a Complex Variable

f(z) =

~ P(z)

is a quotient of two polynomials. We now turn our attention back from the application of Cauchy's theorem to the theorem itself and derive a converse and some refinements. Definition: Let R be the rectangle: x + iy

if a


r

f

= -21ri9{

00

then 00

f %

f(z )dz =

l n=-oo

an

f ~ dz %

= a_ 1

~z = 21ria_ 1

%

.·. 21ri(g{l + 9{2 +···+ g{n) = -277ig{oo

and the truth of the theorem follows at once.

00

122

Theory of Functions of a Complex Variable THEOREM 5.39. Let f be analytic eJ!Cept for a finite num-

ber of isolated singularities. Suppose that none of these singulari-

ties is on the real axis and that f is zero at . A hm f f(x )dx A-+

oo

oo •

Then

:J{

-A

exists and is equal to 21ri [:R 1 +· :R 2 ... :Rn + 200 ] where

:R.J

is the

residue at zi and z 1, z2, ... , zn are the singularities of f in the upper half plane. .

Proof: Let ~A(t) = -A+ 4At for O .:S t :S ½ Let ~ A (t) = Ae 277 i(t-½) for ½ .:S t ,S 1 . ~A

Then the inside of plane with \ z \


oo -A

f(x)dx if the limit

exists,

Remark: If

cp is analytic at z 0 then the residue of _L_ at z 0 (z-zo)n

may be computed as follows:

Hence the residue is

cpn-l(zo) (n-1)!

Hence if P and Q are polynomials ·with degree P < degree Q and if Q. has no roots on the real axis then

J

00

-00

pf J Q

x dx

X

exists and can be computed in a routine manner as soon as ·the roots of Q above the real axis have been found.

Example:

~ = J.,. (l-•-l. + 1+ z

z"'

z"'

4 ,.,) .·. z'+

g{00

= 0,

The only singularity in the upper half plane is at z = i and 1

1

~=z77

g{_ 1

. 1 = value of ....l., at i i.e., 2i x+z

z-i 00

J

-00

dx 1 + x2

277i

2i

17.

Exercise I. Using the theory of residues, compute

124

Theory of Functions of a Complex Variable 00

xdx

00

J

J

and

l+x+x2

-oo

S:

5.40. Let

THEOREM

S: (sin 0,

ables. Let g(0) =

f(z) = j' (

xdx

-oo

be a rational function of two vari-

cos 0). Let

½ (z + ½) fz(z-½)) iz

Then if g (0) is defined for all 0 we have

J20

g(0)d0 = 2rri(Sl 1 +Sl 2 +···+Sln)

11

where z 1 ..• zn are the singularities of f inside the unit circle and

Sl 1

is the residue of f at z 1 • Proof: Let i-(0) = ei8

0 .:5 0 .:5 211. Then

= Sl(cos 0_,

f(i-(0))

sin 0)

ie 18

since z + .!

z

=

cos 0 + i sin 0 + cos 0 - i sin 0 2

and

z _ .!

z

=

cos 0 + i sin 0 - cos 0 + i ·sin 0 2

Hence

f

i-

f(z )dz

= .f. 217

Cl) (

J\

o

211

J

cos 0. sin 0 iei(J

g(0)d0 =

0

Example: Let

Sl(s, t)

f

0 iei d0

)

= J211 g(0)d0 . .

.

f(z)dz = 2rri[Sl 1 +Sl 2 +···+Sln]

= __!_2- . Then g(0) = s

o

+

1

2 + cos 0

and

f(z)

1/z i(2+½(z+ 1/z))

1

2

. 2

i(z 2 + 4z + 1)

i(2z + ~ + ½) 2

Now z 2 + 4z+ 1 factors into (z+2+y'3)(z+2-y'3).

125

Complex Integration

Thus -2 + y3 is the only root inside the unit circle. Thus the required residue is the value of i (z + 22+ i.e., 2 1

i (2 \/3)

·J 3)

at z = -2 + y3

= "Ty'3

Hence

211

y3 Exercise ll. Using the theory of residues compute !0217

THEOREM

d0 l+ sin 2 0

5.41. Let f be analytic except for a finite num-

ber of isolated singularities and be zero at oo. Suppose that all singularities on the real axis are simple poles. Then f~00 f(x )dx exists and is equal to where the

211i [9l 1 + 9l 2 + ...

9ln + ½(S 1 + S 2

...

+ Sm)]

9li are the residues of the singularities in the upper half

plane and the S. are the residues at J

oo

and on the real axis.

Proof: The proof is an obvious modification of the proof of Theorem 5. 39 and will be left to the reader with the remark that the _parameterized path to use is one whose range is as outlined in the figure below. Each small semicircle has its center at one of the poles on the real axis.

126

Theory of Functions of a Complex Variable THEOREM

5.42. Let f be as in Theorem 5.41 and let m >

0. Let g(z) = eimxf(z). Then 2iri [:R 1 + :R 2 + ··· +

f_':, g(x )dx

:Rn -+ ½ (S 1 + S 2 + ···

exists and equals

Sm)] where the

:Rj

are the

residues of g in the upper half plane and the Sk are the residues of g on the real axis with

excluded.

oo

Proof: Same as proof of Theorem 5.41 except that it now

f

turns out that

~A

g(z )dz

0 as A

->

.t

where ~A (t) = Ae 1

-> oo

•

In fact for [z [ > M where M is sufficiently large we have

f(z) = a-} + a-2 z 22

+ ... =

1 [a z

-1

+ a_2 + ... ] z

so

(w))

2 -

(z-¢(w)) + ... )

We compute at once that the residue of 0 at ¢ (w) is ¢ (w) . Hence

130

Theory of Functions of a Complex Variable

¢(w)

= -1.. f 2Tri

-5'.-

zf'(z)dz

EG')-:-W-

That ¢ '( w) exists can now easily be shown by a direct computation.

CHAPTER VI ENTIRE AND MEROMORPHIC FUNCTIONS SUMMARY.

Every rational function may be written essentim

m

m

ally uniquely in the form c(z-z 1 ) 1 (z- z 2 ) 2 ••· (z-zn) n where the z. are distinct and the m. are non zero integers. Thus the J

J

zeros, the poles and their orders determine the function up to a multiplicative constant. Also every rational function may be written uniquely in the form m

j=l

n

k=l

a j (z-:.) k + P(z) J

where P is a polynomial and the z. are the distinct poles. A raJ

tional function is thus determined up to an additive polynomial by its poles and the coefficients of the Laurent expansion about these poles. Obviously the z. and m. may be assigned arbitrarily in the first J

case and the

zj

J

and the

aj

in the secon. This chapter is con-

cerned with the extent to which analogues of these results exist when "rational" is replaced by "analytic except for isolated poles." In the first half of the chapter no further restrictions are put upon the functions and uniqueness theorems are weak and complicated. In the second half we consider only functions which are doubly periodic with fixed periods. These are the celebrated elliptic functions which first arose as the inverse functions of indefinite elliptic integrals. For them one has a theory fully as complete as that for rational functions.

131

132

Theory of Functions of a Complex Variable Let f and g be entire functions_ Then f/ g is defined and

analytic everywhere except at the zeros of g . These zeros are isolated and hence define isolated singularities of

f/ g. These isolated

singularities are clearly poles since for each zero z 0 of g there exists n such that g(z) = (z- z0 )n h (z) where h (z0 ) -/,. 0.

Definition: A function F analytic in the entire complex 1 plane except for poles is said to be a meromorphic function. We have just seen that f/ g is meromorphic whenever f and

g are entire. As we shall see the converse of this result is also true. ~very meromorphic function is a quotient of two entire functions. The proof of this result depends upon showing that there exists an entire function having any prescribed discrete set of zeros. In general this chapter will be. devoted to the existence of entire and

meromorphic functions with prescdbed ·zeros and poles and to the study of certain important subclasses of such functions whose existence can thus be established. It is easy to produce a function with exactly one zero, namely z- z 0 where z 0 is the required zero. By taking products we can thus form a function having any finite number of zeros. We can do the same thing for infinite sets of zeros provided we have a suitable theory of infinite products and admit functions having only one zero other than linear polynomials. We turn our attention next to the appropriate theory of infinite products.

Definition: Let a 1, a 2, a3 , ent from zero and let P n

=

•••

a 1 a 2 .•• an.

P -1,. 0 we say that the infinite product

equal to P. if a 1, a 2 , ••.

be complex numbers differThen if limn->oo Pn

=

II;:'=l an converges and is

are complex numbers such that an 0 + 1 ,

an 0 +2 ' ••. are all :ifferent from zero for :ome n 0 then we say that a

n

133

Entire and Meromorphic Functions whenever the infinite product on the right converges in the sense given above. Remark: if IT"' n=l an converges then

lim a = 1 n-->oo n

for

an

= Ppn n-1

n

= 1.

1 + ·vn where now

For this reason we often set an V

p p

....

--> 0.

= log Iz I + i0 where

Log z

Let us set

-11

< 0 < 7T and

= z. THEOREM

and only if

6.1.

no an

If

1;=l Log an

Proof: If

rr;=l

= 0 then

rr;=l

an converges if

converges.

an

converges then the sequence

k ....

k 1 an converge s . Hence Iln= k

log

I II

an

n=l

converges where an = Ian Ie

converges. Also II!=l e n i(01+ ... +

0 ) n

l n=l

·fj

1

Hence e

k

I

i0n

converges. We cannot conclude immediately

that 0 1 + ... + 0n converges but only that there exists a sequence of integer kn such that 0 1 ... 0n + 27Tkn converges. But since 0n .... 0 it follows that kn = k for n 0 1 + 0 2 •••

~

some n 0 • Hence

converges. Thus 00

l

00

00

Log an

n=l

n=l

n=l

converges. Conversely if l Log an converges then 1;=l log Ianl converges an d ""'n=l "'"° 0n k

converges. Hence

lk

II lanl = e n=l n=l

log

I

la n

134

Theory olFunctions of a Complex Variable

converges to

i0

Moreover II!= 1 e n k

k

II la Ie n n=l

II a n n=l

Note that if II;:'= 1 an

n \an!

n=l i.e.,

=

lim

i0n

---> e

i00 !,°o 1 log la

e n=

n

I

-/- 0 .

converges then

t

n--->oo n=1

janI = n--->oo lim In=l t an I = I n=l IT an\

II;:'=l lanl also converges. On the other hand, if an = (-l)n

then II;;'=l an

does not converge while II;;'=l lanl

does converge.

In other words, the situation is exactly the opposite of what it is for series. Accordingly we make the definition

Definition: II;;'=l an

converges absolutely if l ;;'=l Log an

converges absolutely (after omitting any zero factors). LEMMJ)., If \z\

2

z3

< l then Log(l+z) = z- z2 + 3

Proof:

Jz ( z_z; ···)

1

1- z+ z 2

l+z

Moreover eLog ( l+z) = 1 + z so

_g_ eLog(l+z) = 1 = eLog(l+z)

d Log(l+ z) . dz

dz

Therefore

d~

Log(l+ z)

1

l-z+z2

l+z

Therefore Log(l + z) Since Log 1

z2

= c+z--+··· z

c + 0 - 0,... we have c = 0 .

•

135

Entire and Meromorphic Functions THEOREM

only if

I

;;'=l vn

6.2.

II ;;'= 1 (1 + vn) converges absolutely if and

converges absolutely.

Proof: If either the series or product converges, then vn ---> 0. Hence for n > n 0 , Jvnl < 1 so log~:+vn)

= 1 "'-

which approaches 1 as n tends to

~n

;,2

+ v1

00 •

Hence there exists n 1 , a and b such that O < ·a

n 1 a


z 0

140

Theory of Functions of a Complex Variable

implies \zn\ > 2r. Then (_r_)n+l < (½)n+l which is the nth term \zn\ of a convergent series. Thus the theorem is proved.

Remark 1. We can usually get by with a smaller subscript for E. Indeed, if m1 , m2 , . .. is any sequence of integers such that !, (r/\zn\)

m +1 n

converges for all r then II;;'=l Em (z/zn) may be n

Remark 2. We can, of course. add a zero of multiplicity n to F by simply multiplying F by zn. THEOREM

6. 6. 1f f is any meromorphic function then

3

en-

tire functions g and h which have no zero in common, such that f(z)

= (g (z ))/(h (z))

for all z where f(z) is defined.

Proof: Left to reader. Definition: The entire function f is said to be of finite order if there exists a positive real number. A such that

JfulL e\z\A

is bounded.. The greatest lower bound of all such A is called the

order of f. THEOREM

::; p . Then f + g

6. 7. Let f and g be entire functions of order and lg are also order ::; p .

Proof: If f and g are of order :S p then \f(z)\

:S K(A)e\z\A and g(z) :S K 1(a)e\z\A

for all A> p.

< (K(A) + K (A))e\z\A

for all A > p.

Hence \f(z) + g(z)\

141

Entire and Meromorphic Functions Hence f + g is of order :S p . Also lf(z)g(z)I


p and all A 1 .

For any ~ > p take p < A < A 1 , then 2 - lz !Al-A }s negative for all lzl sufficientlylarge. Hence lf(z)g(z)l/elzl 1 is bounded. Hence fg is of order :S p. THEOREM

6.8. Every polynomial is of order zero.

Proof: In view of Theorem 6. 7 it will suffice to prove that z is of order O i.e., that lzl/elzlf is bounded for all f > 0. But elzlf

= 1 + lz If+ lz 12 f/2! .... Choose elzlf > J..tl'.:".f

m

> 1/f then

for all lz I > 1

ml

Hence elzlf >

ltl ml

for all lzl > 1

and the result follows.

Exercise /. Show that ez der 1 and that cos LEMMA.

yz

!

is of order

sin z, and cos z are all of or-

½.

Let ¢ (z) be analytic in the circle lz I

never zero there. Then 3 a function tjJ analytic in that e1/I (z)

='

¢ (z).

Proof: Exercise II. LEMMA.

Let r > 0 and let lal f


p. Hence

Hence by the last lemma

145

Entire and Meromorphic Functions

J 2r

< log K + 2f3rf3 - loglf(O)I

n~x) dx

r 0

But

n ~x) dx

> n (,)

r

J'

n (r) log 2 .

~x

f

Therefore , n ( r)


r so that jz-z0 j , i0 Let ~ (e) = z O + r e . r

:S r' implies z

€

0.

148

Theory of Functions of a Complex Variable

Then f.

f

_l_ 2rri

(z)

J

fj

(C::)dC:: ;:- z

%

and

f. '(z) J

Hence n+q

n+q

n

But since

I;;'=l

\fj(c;;)\

converges uniformly on \C::-z 0 1

oo 277 (n + l)n+l/2 e-(n+ 1)

1

can be obtained by applying complex variable methods to r(z). See, for example, Ahlfors, Complex Analysis, p. 162. Let z -> r(z) be a rational function. Then r has only a finite number of singularities, z 1 , z 2 , ... , zn each of which is a pole. The Laurent development about zj has only a finite number of negative terms and hence is of the form a~2 2 + ... + ai_njn. [ a~l + z-zj (z-zj) (z-z) 1

J

Let us call the part in brackets hi.

a{(z:...zj) + ...

hi is called the principal

part of r at z. . Note that h j has z. as its only singularity and J

J

154

Theory of Functions of a Complex Variable

that r - h j has z. as a removable singularity. Hence J

r -

(h 1+ ... + hn) has only removable singularities and so coincides except at the z. with an entire function P . Since P is also rational, J

it must be a polynomial. Hence there exists a polynomial P such

that r(z)

+ P(z). (z-z.l J

This is called the decomposition of r into partial fractions. It is useful in finding indefinite integrals of rational functions. Our next goal is an analogous formula for meromorphic functions and a theorem stating that the poles and principal parts of a meromorphic function may be chosen more or less arbitrarily. THEOREM

6.14. Let z 1, z 2 ,

complex numbers such that 1/lzjl the rational function such that

-->

...

be a sequence of distinct

0 and for each j let hj be

ai

-1

h .(z) = z - z. J

+ ... + (z-z.)Pi

J

J

Then there exist polynomials pl' p 2 , p 3 , ••• such that l;:'=l (h/z)-p/z)) converges for all z in C - Iz 1 U z 2 ... l, absolutely and uniformly

on compact subsets to a meromorphic function f whose only singularities are at z 1, z 2 ,... and such that the principal part of f at zj IS

h .. J

Proof: We may assume without loss of generality that no zj = 0 since the pole at zero can always be added later .. Then h .(z) is analytic for J

h/z)

=

•

c6 +

I

Iz I < Iz.J I

. 2

cf z+ cJz

side the circle

Ii n.+1

.

+... where the convergence is uniform in-

Iz I : : ; Izj 1/2. c~znl


form> n 0 . Thenfor \z[ < 9{, \h.(z)-p.(z)[ n 0 . Hence

\z \

I

J

J

+l (h.J (z) - p.J (z )) converges uniformly in

00

n=n 0

< 9{. Since (h. (z) - p. (z) is analytic in J

J

\z [

9l




9{. Therefore identical, inThe conclusion

If f is any meromorphic function z 1 , z 2 , ... are its poles and hi is the principle point at zj' then there exists COROLLARY.

an entire function h such that 00

f(z)

= I n

=1

(h.(z)-p.(z)) + h(z) 1

J

Theorem 6.14 is knowrt as the Mittag-Leffler's theorem. Definition: Let f be a meromorphic function.

Then the complex number w is a period for f if f(z+w) = f(z) whenever z is not a pole of f. THEOREM 6.15.

If w is a period for f and z 0 is a pole

z 0 + w is also a pole of f and the coefficients in the

of f, then

Laurent expansion about z 0 and z 0 + w are identical. Proof: Left to the reader. Definition: We shall denote the set of all periods of f by

Of course O f Pf for all f. If Pf contains non-zero elements we shall say that f is periodic.

Pf"

THEOREM 6.16. For any f,

w1 + w2

f

Pf,

-w 1

f

Pf.

Pf

w1

= C

f Pf, w 2 f Pf implies if and only if f is a constant.

156

Theory of Functions of a Complex Variable

If f is not constant and ill 1, ill 2,... is any infinite sequence of distinct non-zero members of Pf then 1/ Iill n I . . 0 .

Proof: All statements but the last are obvious. Suppose that 1/lw 0. Then 3 a subsequence w n. which converges . nI

+

to a point w. Let z 0 be any regular point of J. Then f(z) f(z 0 ) will have zeros at z 0 + illn;. Hence f(z) = f(z 0 ) and we have a contradiction. COROLLARY 1. If f is not constant Pf has only a finite number of members in any compact set. COROLLARY 2. If

f is periodic and non-constant, then the periods may be put in one-to-one correspondence with the integers; i.e., the set of periods is countable.

3. If f is not constant then any family of non-zero periods has a member of least modulus. COROLLARY

Definition: Let f be periodic and non-constant. Let illo be a non-zero member of Pt such that ill ,/, 0 and ill f Pf implies Jcuj ?. w 0 . If the periods 0, ± ill 0 , ± 2ill 0 , ... are the only periods of f we say that f is simply periodic with primitive periods ± illo· Let f be periodic, non-constant and not simply periodic. Let illo be a non-zero period of smallest possible modulus and let w 1 be a period of smallest possible. modulus amongst the periods other than nw 0 . Then every period of f is uniquely the form n(L) 0 + mw 1 where n and m are ·integers. Moreover w/w 0 is not real. · THEOREM 6.17.

Prooi: If w 1 = ,\w 0 where ,\ is a real number > 1, then (,\ - n ) w O is a period for all integral n . Hence if n is the biggest integer less than ,\, J(A-n)(,) 0 1 < Jw 0 J. This contradiction shows that ill/w 0 is not real. Hence 3 real numbers ,\ and µ such that

Entire and Meromorphic Functions ,\w 1 -µw 0

157

= 1 and,\' and w' so that

,\'w 1 -w'w 0 1. Hence every complex number is uniquely of the form C0 w 0 + C 1w 1 where

C0 and C 1 are real. Let C 0 w 0 + C1w 1 be a period and suppose that C 0 and C 1 are both integers. Choose integers n and m so

n < 1, 0 ::; C1 - m < 1. Then (C 0 - n) w 0 + (C 1 - m)w 1 is a period. Setting C 0 - n = C~, C 1 - m = C{ we see that C{ w 0 + C{ w 1 is a period where O ::;: C~ < 1, 0 ::;: C{ < 1 and C~ and C{ are not both zero. Clearly, -(1-C~)w 0 + C{ wl' (l-C~)w 0 + (1-C{)w 1 and C~w 0 - (1-C{)w 1 are also that O

:S C0

-

periods. Hence 3 a period C~w 0 + C~w 1, where O :S JC~'I

S.

:S JC{'I :S 1/2 and C~' and C{' are not both zero. But IC~wo + C{w1l :S 1/2(lwol + lw1l). If lwol < lw1l This says that JC~'w 0 + C{'w 11 < lcu 11 and hence that C~'w 0 + C{'w1 is an integral multiple of w 0 . This contradiction shows that Jw 0 I = 1/2,

0

Jw 11 and hence that JC~'I = JC{'I = 1/2 and JC~' w 0 + C{' w 11 = Jw 0 l = Jw 1J. Therefore JC~'w 0 + IC{'w 1 = JC~'w 0 + C{'w 1 J. 1

1

Hence Wo is a real multiple of w l. This contradiction completes the proof of the theorem.

Definition: A periodic function which is not constant and not simply periodic is said to be doubly periodic. Definition: A function is said to be elliptic if it is a constant or is meromorphic and doubly periodic.

Definition: It follows from Theorem 6.17 that for any doubly periodic function f 3 periods w O and w 1 such that every period is uniquely expendable in the form nw 0 + mw 1 where n and m are integers. We call w 0 and w 1 primitive periods. THEOREM

6.18. If an elliptic function f is entire it is a

constant.

µw1

Proof: Let z be any complex number. Then z = ,\w 0 + where ,\ and µ are real and w O and w 1 are periods for f.

158

Theory of Functions of

a

Complex Variable

Choose integers m and n such that O $ 1. Then

IA-nl


az + bz

f(z) is

f(zei 8 ) = aeie z + be-ie z. Thus f is conformal

if and only if aeie z + be-ie z = eie az + eie bz; that is, if and only if b z (e-ie - ei 8 ) = 0. But this is so if and only if b = 0. Let f(z + iy) = u(x, y) + iv(x, y) be a complex valued

function defined in some open subset 0 of the complex plane such that u and v have continuous first partial derivatives there. Let x 0 +iy0 f 0. Then f(x 0 + Li x + i(y0 + 1'.iy)) - f(x 0 + iy0 ) = u(x 0 + Lix, y 0 + l'.iy) + iv(x 0+ Lix, y 0 +1'.iy) - u(x 0 , y 0 ) - iv(x 0 , y 0 ) and by the law of the mean (Theorem 4.5) this is equal in turn to l'.ixu 1 (x 0 , y 0 ) + l'.iyu 2 (x 0 , y 0 ) + il'.ixv 1 (x 0 , y 0 ) + il'.iyv 2 (x 0 , y 0 ) + h 11'.ix + h 2 Liy where h 1 and h 2 depend on x 0 , y 0 , Lix and l'.iy and tend to zero as l'.ix and l'.iy tend to zero. Let

177

Conformal Mapping

Then the linear function g

is called the differential of f at xo,Yo the point x0 + iy0 . It follows at once from Theorem 7.2 that gx 0 ,Y 0 is conformal if and only if u 1 (x 0 , y 0 ) = v 2 (x 0 , y 0 ) and v1 (x 0 , y 0) =-u2 (x 0 , y 0 ).

We define f to be conformal at x 0 , Yo

if its differential at x 0 , y0 is a conformal linear function. We note for its suggestive value that the differential is a first approximation to the function z - z 0

--->

f(z) - f(z 0 )

which becomes better the

smaller that z - z 0 is. As an immediate consequence of the foregoing discussion and the theorem relating analyticly and the CauchyRiemann equation, we conclude the truth of THEOREM 7.3. The function f considered above is conformal at every point of its domain if and only if it is analytic in this domain. THEOREM 7.4. The differential of the analytic function f at the point z 0 = x 0 + iy0 is the composite of a rotation through the angle 0 with multiplication by r where f'(z 0 ) = reie

Proof: Exercise I. For the proof of the next theorem we shall need a lemma strengthening Theorem 5.45. LEMMA.

Let f be analytic in [z - z 0 [
0) power series centered at z 0 , p 0 = z,::'=o an(z-z 0 )n. By a continuation of p 0 along 1- we shall mean an assignment to each t in [O, 1] of a non-trivial power series pt such that: (a) (b)

The center of Pt is 1-(t). For each t 0 € [O, 1] there exists E.

t-to < E implies that Pt and Pt

< 0 such that

agree in the intersection of their circles of converge~ce.

218

Theory of Functions of a Complex Variable

THEOREM 8.1. Let t -+ Pt and t -+ p/ be two continqations of p 0 along the same parameterized path i-. Then pt = p / for all t.

Proof: Choose E > 0 so that It I < E implies that pt and p/ both agree with p0 in the intersections of their circles of convergence with that of p 0 and also so that h(t) - z 0 1 < r(p 0 ). Then for \ t I < E , the center of pt and pt' are in the circle of convergence of p 0 and both agree with p 0 in /a neighborhood of i-(t). Hence Pt and p/ are identical. We have thus proved that pt = pt' for all t with It I < E . The same argument shows that . the set of all t for which pt = pt' is an open set. We show next that it is also closed. Let t1 where Pt

n-> oo

for all n. Choose n so that \i-(tn)-i-(t 1 ) I is

p/ n

= lim tn

n

less than the minimum of the radii of convergence of Pt

and p/ . 1

Then pt

P/ n

and pt n

1

and p/. Since Pt 1

1

have a common region of convergence and so do

= p/,

n

n

Pt

and p/ agree on an open 1

1

.

= p/ . Since 1 1 1 is connected and the set of all t for which pt = pt'

subset of their region of convergence. Hence Pt

0 :S t :S is open and closed and not empty it must be all of [O, 1].

THEOREM 8.2. Let p 0 be a non-trivial power series with center z 0 and let i- be a parameterized path with i-(0) = z 0 , i--(1) = z. Suppose there exist O < t 1 < t 2 < ... < tn < 1 and nontrivial power series Pt , Pt , ... , Pt p 1 with centers at i-(t 1) , ... , 1

- 2

i-(tn) i-(1) respectively such that the circle of convergence of Pt . . o f th eit . circles . J-i . th e mtersectzon m

a unique continuation of t

P/J

for all j.

->

n

for tj-l :S t :S ti i-(t) is in and that Pt agrees with Pt 1 J-;t o f convergence. Th en th ere exists

p/ of p 0 along i- such that Pt

J

Analytic Continuation and Riemann Surfaces

219

Proof: For each t f [O, 1] there exists j such that tj-l ~ t < ti. Let p/ be the power series expansion of the analytic function defined by pt about the point 'f (t) . Clearly this is a con1-1

tinuation and by the previous theorem it must be unique. THEOREM 8.3. Let t --> Pt be a continuation of the nontrivia-Z power series p0 along the parameterized path -w-. Then r(pt), the radius of convergence of pt is a continuous function of t.

Proof: If c (pt ) , the center of the circle of convergence of 1

pt , is in the circle of convergence of pt , then 1

2

r(pt 1)?:. r(pt 2 ) - lz 1 -z2 1 where zj

= c(pt1)

If r(pt )) ~ r(pt ) then 1

while if

2

r (pt ) > r (pt ) then c (p 2 ) is in the circle of convergence 1

of Pt

2

so that 1

so

in any case. Thus

and the desired result follows.

8.4.

Pt is a continuation of p 0 along the parameterized path -w- with domain [O, 1] then there exist O < t 1 < t 2 < ... < tn < 1 such that Pt , ... , Pt satisfies the hypoTHEOREM

thesis of Theorem 8.2.

If

1

n

220

Theory of Functions of a Complex Variable Proof: Since r(pt) is continuous and never zero, there ex-

ists r 0 > 0 such that r(pt) ~ r 0 for all t.

Now choose E > 0

so that l1r(t)-1r(t')I < r 0 for Jt-t'J 1/E and let t. = j/(n+ 1).

Then choose n >

< E.

J

Definition: If p1 and p2

are two non-trivial power series,

we shall say that p 2 is a continuation of p 1 if there is a parameterized path % such that 1r(O) = c(p 1), 1r(l) = c(p 2 ) and a conof p1 along % such that p/ = p 2 . Obviously, p 1 is a continuation p 2 whenever p 2 is a continuation of p 1 and p 1 is always a continuation of itself. Almost equally obvious is the fact that p 3 is a continuation of p 1 whentinuation p /

ever p 3 is a continuation of p 2 and p 2 is a continuation of p 1 . Thus the relation "p 2 is a continuation of p 1 ," is an equivalence relation and divides the.set of all non-trivial power series into mutually disjoint classes so that any two members of the same class are continuations of one another and no two members of different classes are continuations of one another.

Definition: A complete analytic function is the set of all power series in some equivalence class under continuation, i.e., the set of all continuations of some fixed non-trivial power series. If 1 is a complete analytic function, then :Dj= , the domain of 1, is the set of all c (p) for p € 1 . :Dj= is obviously open. If c (p) = c (q) implies p = q for all p and q in 1, we say that 1 is

single-valued otherwise multiple-valued. Suppose that c(p) for a unique p =

a 0 where p

1 €

= ~;0

is single-valued. For each z f :Dj=, z = 1. Let fj=(z) = p(c(p)), i.e., let fj=(z) a/z-c(p))n. Clearly fj= is an analytic func-

tion in :Dj= and is maximal in the sense that there is no analytic function with domain not contained in :Dj= which agrees with fj= in every open subset of its domain.

Definition: Let 1 be a complete analytic function and let f be an analytic function defined in some connected open subset 0

Analytic Continuation and Riemann Surfaces

221

of 'D:f. If, for each point z 0 E 0, the Taylor's series expansion of f about z 0 is in :f, we shall call f a single-valued branch of the complete analytic function

:f .

Convention: In order to distinguish between the family of power series t -> Pt (one for each t) obtained by continuing p 0 along the path 15- and the end result-the single power series p 1 -we shall refer to the former as the continuation of p0 along 15- and to the latter as the continuation of p 0 via 15-. THEOREM 8.5. (Monodromy theorem). Let 0 be an open, simply connected subset of the complex plane. Let p0 be a nontrivial power series with c(p)

E

0 and suppose that there exists a

continuation of p0 along every parameterized path lying wholly in

0 .. Then the continuations of p0 via these paths depend only upon the end point of the path and not upon the path itself. Thus they constitute the expansions about the points of 0 of a single-valued analytic function defined in 0. Proof: Let %o and 15- 1 be two parameterized paths joining z 0 = c (p0 ) to the same point z 1 c 0, both paths lying in 0 .. Let 15-2 be the closed path obtained by following 15-0 in reverse from z 1 back to z 0 and then 15- 1 from z 0 to z 1 . Let t -> Pt and t .... p/ be the continuations of p0 along 15-0 and % 1 respectively. Then, p 1 ' will be the continuation of p 1 via 15-2 . We wish to prove that p 1 = p 1 '. Thus it will suffice to prove that the continuation of a non-trivial power series p0 via a closed parameterized path in 0 is al ways equal to p0 • Let 15- be any closed parameterized path with domain [O, 1] ;y(O) = c(p 0 ) = ;y(l) andrange in 0. By the definition of simple connecticity there exists a continuous function g from the unit square O $ t $ 1, 0 $ s $ 1 to 0 such that g (0, t) = c(p 0 ), g(l,t) = 15-(t), g(s,O) g(s,1) c(p 0 ). Let us denote the closed parameterized path t -> g (s, t) by 15-s and the continuation of p 0 along 15- by t -> p/. It follows from the proof of Theorem 8.3 that r(p/) is a continuous function of s and t and hence

=

=

222

Theory of Functions of a Complex Variable

that there exists r O such that r (pt) ~ r O for all s and t with 0 :s; s :s; 1 and O :S: t :s; 1. Choose E > 0 so that Is C s2 I < E implies

That this can be done follows from the uniform continuity of g. We show now that whenever is 1-s 2 1 < S then pt1 and pt2 agree in the common part of their circles of convergence. To do so, let As1 , 82 be the set of all t for which agreement holds. It is trivial that O

€

A 81 , 82 • Let t 0 be any point in A 81 , 82 • Choose

o>

0

so that It- t 0 I < o implies that pti and pt~ agree in the common part B of their circle of convergence for j = 1, 2 and that B is not

Pt:

1 and Pt:2 agree in B and empty. Let It- t0 I < o. Since 1 and pt1 agree in and and pt2 agree in it follows that pt 1 and pt2 agree in B. Hence pt 1 and pt2 agree throughout the intersection of their circles of convergence. Hence,

Pt:

B

Pt;

B

t € A 81 ,s2 ., In other words, A 81 ,s2 is open. Now let tn --> t0 where each tn € A 81 , 82 • Choose o as before and define B as before. Choose n so that ltn - t 0 1
0.

sufficiently small, as follows:

(a) If p is a regular point of '.RJ defined by a power or Laurent series centered at a point c (p) of the finite plane and if E < radius of annulus or circle of convergence, then NE (p) is the

~

set of all q

E

1

with center c(q) such that \c(p) - c(q)\ < E

and such that p and q agree in their common region of conver-

gence. (b) If p is a regular point of '.RJ defined by a Laurent series centered at oo and E < 1/radius of convergence, then NE (p) is the set of all q E 1 with center c(q) such that \c(q)\ > 1/E and such that q and p agree in their common region of convergence. (c) If p is a branch point of '.RJ defined by a fractional Laurent series centered at a finite point c (p) and E < radius of annulus A in which family 1 '(p,c(p)) is defined, then NE (p) is the set of all members q of 1'(p,c(p))with \c(q) - c(p)\ < E. (d) If p is a branch point of '.RJ defined by a fractional Laurent series centered at oo and E < 1/(radius of circle in exterior of which the family 1 '(p, A) is defined) then NE (p) is the set of all q

E

1 '(p,

with \c ( q) \ > 1/E

oo)

•

Definition: Let G be a subset of '.RJ. We shall say that G is open if for each p E G there exists E > 0 such that NE (p) is defined and N (p) C G. E

THEOREM

-

8.6.

'.RJ

nected Hausdorff space. Proof: Exercise I.

with the topology just defined is a con-

Analytic Continuation and Riemann Surfaces

22.9

In the future, in speaking of the Riemann surface of a complete analytic function, we shall always assume it equipped with the above described topology.

8. 7 Let :Rj= be the Riemann surface of the complete analytic function 1. Then for each point p of :Rj= (branch THEOREM

points included) there exists a one-to-one bicontinuous map of an open set containing p onto an open set of the complex plane. Proof: Consider the function c, q --> c ( q) mapping each point in :Rj= onto the center of the power or Laurent series defining it. If p is not a branch point and c (p) /, oo. then it is clear that c itself restricted to NE (p) for a suitable E is the desired mapping. If p is not a branch point and c (p) = oo, then q --> 1/c ( q) will do the trick. If c (p) /, and p is an m fold branch point, then q --> c ( q) maps NE (p) in an m to one manner onto an open circle in the complex plane and will not do. However, this m to one mapping may be "unwound" to a one-to-one mapping by means of the m-th root function in the following manner. 00

Let A be the image in C under c of NE (p ). Then A will be the set of all complex numbers of such that Iz - c (p) I < E . Let A ' be the set of all members of A such that the real part of

z - c (p) is not positive or such that the imaginary part is not zero. Let B 0 ,Bp···, Bm-l be the m subsets of N/p) which c maps in a one-to-one continuous manner on A ' obtained by taking the subsets which are the power series expansions of the functions f o, f 1 , ... fm-l of our preliminary discussion of branch points. For each point z in A' let Arg (z - c (p )) be the unique real number A between O and 277 such that eiA _ p) - I z-c P )I

z-cf

Let d(q) = lc(q)Jei Arg(cfn7)-c(p)) ei

where j(g)=k for all q

f

2;:

j(q)

Bk and d(p) = c (p ). Then d is one-

230

Theory of Functions di a Complex Variable

to-one and maps B 0 U B 1 ... U Bm-l onto A with the "spokes"

e

2rrii m

removed. It has an obvious extension to a map of Ne (p) onto A which is one-to-one and bicontinuous. If c (p) = oo and p is an m fold branch point, an obvious combination of the methods used in the last two cases accomplishes our purpose.

Definition: Let 1 be a complete analytic function and let :R1 be its Riemann surface. For each p € :R1 let i1(P) be defined as follows: (a) If c(p) = 00 and p has terms with positive index, then t1(p) = 00. (b) If c(p) =

00

and p=ion=-oo aznlm then t1(p) n '

=

ao·

I=

oo

and p has terms with negative index,

then i1(p) = 00. (d) If c(p) /,

oo

and P

(c) If c(p)

t1(p)

= !.;;'=O

an(z-c1(P)rlm, then

= ao· THEOREM

8.8. t1 is a meromorphic function on :R1 .

Proof: Left to reader. THEOREM

:R5

8.9. Let

at which neither

9l

11 nor

be the set of all regular points of c1 =

oo,

For each p

€

9l

choose

E so that Ne (p) exists and consider the restriction d of c to N/p ). Then 11 (d- 1(z )) is analytic and its power series expan-

sion about c(p) is in 1. Moreover, every member of 1 may be so obtained. Proof: Obvious. It is clear from Theorem 8.9 that 1 may be recovered from :R1. The two single valued functions, c and 11, with domain :Jl1 may be regarded as giving a parametric representation of the multiple

valued function defined by

1.

231

Analytic Continuation and Riemann Surfaces

THEOREM 8.10. Let g{1 be the Riemann surface of the co complete analytic function 1. Then g{1 satisfies the "second axiom of countability", i.e., there exists a sequence (\, 0 2 , 0 3 , ••• of open subsets of g{1 such that every open subset of g{1 is a union of a subset of the 0 f

Proof: Exercise III. THEOREM 8.11. Let g{1 and c1 be as above. Then for each z in the extended plane, there is at most a denumerable number of points p in g{1 such that c5= (p) = z.

Proof: Exercise IV. We turn our attention now to an important class of special cases in which the structure of g{j= may be made quite transparent.

Definition: The complete analytic function 1 is almost freely continuable if there exists a finite subset S of the complex plane C such that every p f 1 may be contained along every parameterized path in C - S. Suppose now that

'

1

is almost freely continuable. Let zl'

z 2 , z 3, ... , zn be the points of S arranged in "lexicographic" order; i.e., for each j either xi < xi+ 1 or xi = xi+ 1 and Yi < Y·+i (z. = x. + y.). Then no two non-successive line segments 1 1 1 1 z .z1·+1 intersect. Let L be the union of the line segments 1 1 z.z.+ 1 and of the half line e : y = y, x > x . C-L is then 11 n n -n

e.

easily seen to be simply connected. In fact, it is obvious that there is a one-to-one continuous map of C onto itself which takes L into the positive real axis and we have shown that the complex plane minus the positive real axis is conformally equivalent to and hence homeomorphic to the interior of the unit circle. Hence any member p of 1, not centered on L defines a single-valued analytic function f P when continued along paths lying wholly on C - L.

232

Theory of Functions of a Complex Variable

Since f P and fq are either identical or else have no power series expansion in common, it follows from Theorem 8.11 that there exists a sequence (possibly terminating) of functions f 1, f2 , ... each of which is of the form f q and such that the power series expansions of f 1 , f 2 , •.. about a point z 0 of C-L constitute exactly all members p of :f centered at z 0 Let ms== 1, 2, ... , ~ 0 be the number of terms in the sequence f 1 , £2 , .... Let % be a parameterized o

path which cuts the set L just once and cuts it in a point other than one of the z., i.e., in the interior of the segment L Let 1-(t 0 ) be J J the intersection point. Let r be a positive number less than the distance from 1-U 0 ) to S and choose E so small that It- t 0 I < E implies \1-(t) -1- (t 0 )J < .r. The circle \z - z 0 I < r is cut into two connected halves by fr Clearly the 1-(t) for t0 < t < t0 + E and the 1-U) for t 0 - E < t < t 0 lie in different halves. We shall call the half in which the imaginary part of z achieves the greatest value the positive half. If e. is vertical so that J z has the same imaginary points in both halves, we take the half with the greater real point. We call the other half the negative half.

@®

If the half in which 1-(t) with t 0 - E < t < t 0 is the negative half, we say that % crosses L at % (t 0 ) in the positive direction, otherwise in the negative direction. Now let 1- 1 and 1-2 cross L through f. in the positive direction, each cutting L only once. J Let p 1 and p 2 be the power series expansions of fk about 1- 1(0) and 1-2 (0) respectively. Let p 1 ' and p 2 ' be the continuations of p 1 and p 2 via 1- 1 and 1-2 respectively. We show now that p 2 ' is the continuation of p 1 ,. via a path in C - L. Indeed, the two

233

Analytic Continuation and Riemann Surfaces

crossing points may be enclosed in a rectangular box as shown in the diagram.

Let %3 join a point of %1 lying in the box on the

, Pz

negative side of Ej to a point of %2 lying in the box on the negative side of Ej and let %4 be similarly constructed on the positive side of ej . Let %1 J J a(p.) there exists a meromorphic function f such that f = f. near J

J

pj. Now if we identify functions which differ by an analytic function

p. the functions f. with order J

J

> a(p.) form a -a(p.) dimensional J . J

vector space. Hence e(a) will be -a(p 1 ) - a(p 2 ) - ··· -a(pn) + 1

=

- n (a) + 1. If the genus g is not zero, then e (a) will be less than - n (a) + 1 since each analytic differential puts a linear condition on the f/s which can occur and cuts down the dimension by one. We might think then, that e(a) = -n (a) + 1-g. However, this is not

262

Theory of Functions of a Complex Variable

true either for if 0 d