zotero-db/storage/K9PSWHIV/.zotero-ft-cache

COMPLEX VARIABLES AND APPLICATIONS
Eighth Edition
James Ward Brown
Professor of Mathematics The University of Michigan–Dearborn
Ruel V. Churchill
Late Professor of Mathematics The University of Michigan

COMPLEX VARIABLES AND APPLICATIONS, EIGHTH EDITION
Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright  2009 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions  2004, 1996, 1990, 1984, 1974, 1960, 1948 No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
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Brown, James Ward. Complex variables and applications / James Ward Brown, Ruel V. Churchill.—8th ed.
p. cm. Includes bibliographical references and index. ISBN 978–0–07–305194–9—ISBN 0–07–305194–2 (hard copy : acid-free paper) 1. Functions of complex variables. I. Churchill, Ruel Vance, 1899- II. Title. QA331.7.C524 2009 515 .9—dc22
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ABOUT THE AUTHORS
JAMES WARD BROWN is Professor of Mathematics at The University of Michigan– Dearborn. He earned his A.B. in physics from Harvard University and his A.M. and Ph.D. in mathematics from The University of Michigan in Ann Arbor, where he was an Institute of Science and Technology Predoctoral Fellow. He is coauthor with Dr. Churchill of Fourier Series and Boundary Value Problems, now in its seventh edition. He has received a research grant from the National Science Foundation as well as a Distinguished Faculty Award from the Michigan Association of Governing Boards of Colleges and Universities. Dr. Brown is listed in Who’s Who in the World. RUEL V. CHURCHILL was, at the time of his death in 1987, Professor Emeritus of Mathematics at The University of Michigan, where he began teaching in 1922. He received his B.S. in physics from the University of Chicago and his M.S. in physics and Ph.D. in mathematics from The University of Michigan. He was coauthor with Dr. Brown of Fourier Series and Boundary Value Problems, a classic text that he ﬁrst wrote almost 70 years ago. He was also the author of Operational Mathematics. Dr. Churchill held various ofﬁces in the Mathematical Association of America and in other mathematical societies and councils.
iii

To the Memory of My Father George H. Brown
and of My Long-Time Friend and Coauthor Ruel V. Churchill
These Distinguished Men of Science for Years Inﬂuenced The Careers of Many People, Including Myself.
JWB

CONTENTS

Preface

x

1 Complex Numbers

1

Sums and Products 1 Basic Algebraic Properties 3 Further Properties 5 Vectors and Moduli 9 Complex Conjugates 13 Exponential Form 16 Products and Powers in Exponential Form 18 Arguments of Products and Quotients 20 Roots of Complex Numbers 24 Examples 27 Regions in the Complex Plane 31

2 Analytic Functions

35

Functions of a Complex Variable 35 Mappings 38 Mappings by the Exponential Function 42 Limits 45 Theorems on Limits 48

v

vi contents
Limits Involving the Point at Inﬁnity 50 Continuity 53 Derivatives 56 Differentiation Formulas 60 Cauchy–Riemann Equations 63 Sufﬁcient Conditions for Differentiability 66 Polar Coordinates 68 Analytic Functions 73 Examples 75 Harmonic Functions 78 Uniquely Determined Analytic Functions 83 Reﬂection Principle 85
3 Elementary Functions
The Exponential Function 89 The Logarithmic Function 93 Branches and Derivatives of Logarithms 95 Some Identities Involving Logarithms 98 Complex Exponents 101 Trigonometric Functions 104 Hyperbolic Functions 109 Inverse Trigonometric and Hyperbolic Functions 112
4 Integrals
Derivatives of Functions w(t) 117 Deﬁnite Integrals of Functions w(t) 119 Contours 122 Contour Integrals 127 Some Examples 129 Examples with Branch Cuts 133 Upper Bounds for Moduli of Contour Integrals 137 Antiderivatives 142 Proof of the Theorem 146 Cauchy–Goursat Theorem 150 Proof of the Theorem 152

89 117

contents vii

Simply Connected Domains 156 Multiply Connected Domains 158 Cauchy Integral Formula 164 An Extension of the Cauchy Integral Formula 165 Some Consequences of the Extension 168 Liouville’s Theorem and the Fundamental Theorem of Algebra 172 Maximum Modulus Principle 175

5 Series
Convergence of Sequences 181 Convergence of Series 184 Taylor Series 189 Proof of Taylor’s Theorem 190 Examples 192 Laurent Series 197 Proof of Laurent’s Theorem 199 Examples 202 Absolute and Uniform Convergence of Power Series 208 Continuity of Sums of Power Series 211 Integration and Differentiation of Power Series 213 Uniqueness of Series Representations 217 Multiplication and Division of Power Series 222

181

6 Residues and Poles
Isolated Singular Points 229 Residues 231 Cauchy’s Residue Theorem 234 Residue at Inﬁnity 237 The Three Types of Isolated Singular Points 240 Residues at Poles 244 Examples 245 Zeros of Analytic Functions 249 Zeros and Poles 252 Behavior of Functions Near Isolated Singular Points 257

229

viii contents
7 Applications of Residues
Evaluation of Improper Integrals 261 Example 264 Improper Integrals from Fourier Analysis 269 Jordan’s Lemma 272 Indented Paths 277 An Indentation Around a Branch Point 280 Integration Along a Branch Cut 283 Deﬁnite Integrals Involving Sines and Cosines 288 Argument Principle 291 Rouche´’s Theorem 294 Inverse Laplace Transforms 298 Examples 301
8 Mapping by Elementary Functions
Linear Transformations 311 The Transformation w = 1/z 313 Mappings by 1/z 315 Linear Fractional Transformations 319 An Implicit Form 322 Mappings of the Upper Half Plane 325 The Transformation w = sin z 330 Mappings by z2 and Branches of z1/2 336 Square Roots of Polynomials 341 Riemann Surfaces 347 Surfaces for Related Functions 351
9 Conformal Mapping
Preservation of Angles 355 Scale Factors 358 Local Inverses 360 Harmonic Conjugates 363 Transformations of Harmonic Functions 365 Transformations of Boundary Conditions 367

261 311 355

contents ix

10 Applications of Conformal Mapping
Steady Temperatures 373 Steady Temperatures in a Half Plane 375 A Related Problem 377 Temperatures in a Quadrant 379 Electrostatic Potential 385 Potential in a Cylindrical Space 386 Two-Dimensional Fluid Flow 391 The Stream Function 393 Flows Around a Corner and Around a Cylinder 395

373

11 The Schwarz–Christoffel Transformation
Mapping the Real Axis Onto a Polygon 403 Schwarz–Christoffel Transformation 405 Triangles and Rectangles 408 Degenerate Polygons 413 Fluid Flow in a Channel Through a Slit 417 Flow in a Channel With an Offset 420 Electrostatic Potential About an Edge of a Conducting Plate 422

403

12 Integral Formulas of the Poisson Type
Poisson Integral Formula 429 Dirichlet Problem for a Disk 432 Related Boundary Value Problems 437 Schwarz Integral Formula 440 Dirichlet Problem for a Half Plane 441 Neumann Problems 445

429

Appendixes
Bibliography 449 Table of Transformations of Regions 452

449

Index

461

PREFACE
This book is a revision of the seventh edition, which was published in 2004. That edition has served, just as the earlier ones did, as a textbook for a one-term introductory course in the theory and application of functions of a complex variable. This new edition preserves the basic content and style of the earlier editions, the ﬁrst two of which were written by the late Ruel V. Churchill alone.
The ﬁrst objective of the book is to develop those parts of the theory that are prominent in applications of the subject. The second objective is to furnish an introduction to applications of residues and conformal mapping. With regard to residues, special emphasis is given to their use in evaluating real improper integrals, ﬁnding inverse Laplace transforms, and locating zeros of functions. As for conformal mapping, considerable attention is paid to its use in solving boundary value problems that arise in studies of heat conduction and ﬂuid ﬂow. Hence the book may be considered as a companion volume to the authors’ text “Fourier Series and Boundary Value Problems,” where another classical method for solving boundary value problems in partial differential equations is developed.
The ﬁrst nine chapters of this book have for many years formed the basis of a three-hour course given each term at The University of Michigan. The classes have consisted mainly of seniors and graduate students concentrating in mathematics, engineering, or one of the physical sciences. Before taking the course, the students have completed at least a three-term calculus sequence and a ﬁrst course in ordinary differential equations. Much of the material in the book need not be covered in the lectures and can be left for self-study or used for reference. If mapping by elementary functions is desired earlier in the course, one can skip to Chap. 8 immediately after Chap. 3 on elementary functions.
In order to accommodate as wide a range of readers as possible, there are footnotes referring to other texts that give proofs and discussions of the more delicate results from calculus and advanced calculus that are occasionally needed. A bibliography of other books on complex variables, many of which are more advanced, is provided in Appendix 1. A table of conformal transformations that are useful in applications appears in Appendix 2.
x

preface xi
The main changes in this edition appear in the ﬁrst nine chapters. Many of those changes have been suggested by users of the last edition. Some readers have urged that sections which can be skipped or postponed without disruption be more clearly identiﬁed. The statements of Taylor’s theorem and Laurent’s theorem, for example, now appear in sections that are separate from the sections containing their proofs. Another signiﬁcant change involves the extended form of the Cauchy integral formula for derivatives. The treatment of that extension has been completely rewritten, and its immediate consequences are now more focused and appear together in a single section.
Other improvements that seemed necessary include more details in arguments involving mathematical induction, a greater emphasis on rules for using complex exponents, some discussion of residues at inﬁnity, and a clearer exposition of real improper integrals and their Cauchy principal values. In addition, some rearrangement of material was called for. For instance, the discussion of upper bounds of moduli of integrals is now entirely in one section, and there is a separate section devoted to the deﬁnition and illustration of isolated singular points. Exercise sets occur more frequently than in earlier editions and, as a result, concentrate more directly on the material at hand.
Finally, there is an Student’s Solutions Manual (ISBN: 978-0-07-333730-2; MHID: 0-07-333730-7) that is available upon request to instructors who adopt the book. It contains solutions of selected exercises in Chapters 1 through 7, covering the material through residues.
In the preparation of this edition, continual interest and support has been provided by a variety of people, especially the staff at McGraw-Hill and my wife Jacqueline Read Brown.
James Ward Brown

Brown-chap01-v3 10/29/07 3:32pm 1
CHAPTER
1
COMPLEX NUMBERS

In this chapter, we survey the algebraic and geometric structure of the complex number system. We assume various corresponding properties of real numbers to be known.

1. SUMS AND PRODUCTS

Complex numbers can be deﬁned as ordered pairs (x, y) of real numbers that are to be interpreted as points in the complex plane, with rectangular coordinates x and y, just as real numbers x are thought of as points on the real line. When real numbers x are displayed as points (x, 0) on the real axis, it is clear that the set of complex numbers includes the real numbers as a subset. Complex numbers of the form (0, y) correspond to points on the y axis and are called pure imaginary numbers when y = 0. The y axis is then referred to as the imaginary axis.
It is customary to denote a complex number (x, y) by z, so that (see Fig. 1)

(1)

z = (x, y).

The real numbers x and y are, moreover, known as the real and imaginary parts of z, respectively; and we write

(2)

x = Re z, y = Im z.

Two complex numbers z1 and z2 are equal whenever they have the same real parts and the same imaginary parts. Thus the statement z1 = z2 means that z1 and z2
correspond to the same point in the complex, or z, plane.

1

Brown-chap01-v3 10/29/07 3:32pm 2

2 Complex Numbers
y z = (x, y)

chap. 1

i = (0, 1)

O

x = (x, 0)

x FIGURE 1

The sum z1 + z2 and product z1z2 of two complex numbers z1 = (x1, y1) and z2 = (x2, y2)

are deﬁned as follows:

(3)

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2),

(4)

(x1, y1)(x2, y2) = (x1x2 − y1y2, y1x2 + x1y2).

Note that the operations deﬁned by equations (3) and (4) become the usual operations of addition and multiplication when restricted to the real numbers:

(x1, 0) + (x2, 0) = (x1 + x2, 0), (x1, 0)(x2, 0) = (x1x2, 0).

The complex number system is, therefore, a natural extension of the real number
system. Any complex number z = (x, y) can be written z = (x, 0) + (0, y), and it is
easy to see that (0, 1)(y, 0) = (0, y). Hence

z = (x, 0) + (0, 1)(y, 0);

and if we think of a real number as either x or (x, 0) and let i denote the pure imaginary number (0,1), as shown in Fig. 1, it is clear that∗

(5)

z = x + iy.

Also, with the convention that z2 = zz, z3 = z2z, etc., we have i2 = (0, 1)(0, 1) = (−1, 0),

or

(6)

i2 = −1.

∗In electrical engineering, the letter j is used instead of i.

Brown-chap01-v3 10/29/07 3:32pm 3

sec. 2

Basic Algebraic Properties 3

Because (x, y) = x + iy, deﬁnitions (3) and (4) become

(7)

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2),

(8)

(x1 + iy1)(x2 + iy2) = (x1x2 − y1y2) + i(y1x2 + x1y2).

Observe that the right-hand sides of these equations can be obtained by formally
manipulating the terms on the left as if they involved only real numbers and by replacing i2 by −1 when it occurs. Also, observe how equation (8) tells us that any complex number times zero is zero. More precisely,

z · 0 = (x + iy)(0 + i0) = 0 + i0 = 0

for any z = x + iy.

2. BASIC ALGEBRAIC PROPERTIES
Various properties of addition and multiplication of complex numbers are the same as for real numbers. We list here the more basic of these algebraic properties and verify some of them. Most of the others are veriﬁed in the exercises.
The commutative laws

(1)

z1 + z2 = z2 + z1, z1z2 = z2z1

and the associative laws

(2)

(z1 + z2) + z3 = z1 + (z2 + z3), (z1z2)z3 = z1(z2z3)

follow easily from the deﬁnitions in Sec. 1 of addition and multiplication of complex numbers and the fact that real numbers obey these laws. For example, if

z1 = (x1, y1) and z2 = (x2, y2),

then z1 + z2 = (x1 + x2, y1 + y2) = (x2 + x1, y2 + y1) = z2 + z1.

Veriﬁcation of the rest of the above laws, as well as the distributive law

(3)

z(z1 + z2) = zz1 + zz2,

is similar. According to the commutative law for multiplication, iy = yi. Hence one can
write z = x + yi instead of z = x + iy. Also, because of the associative laws, a sum z1 + z2 + z3 or a product z1z2z3 is well deﬁned without parentheses, as is the case with real numbers.

Brown-chap01-v3 10/29/07 3:32pm 4

4 Complex Numbers

chap. 1

The additive identity 0 = (0, 0) and the multiplicative identity 1 = (1, 0) for real numbers carry over to the entire complex number system. That is,

(4)

z + 0 = z and z · 1 = z

for every complex number z. Furthermore, 0 and 1 are the only complex numbers with such properties (see Exercise 8).
There is associated with each complex number z = (x, y) an additive inverse

(5)

−z = (−x, −y),

satisfying the equation z + (−z) = 0. Moreover, there is only one additive inverse for any given z, since the equation

(x, y) + (u, v) = (0, 0)

implies that

u = −x and v = −y.

For any nonzero complex number z = (x, y), there is a number z−1 such that zz−1 = 1. This multiplicative inverse is less obvious than the additive one. To ﬁnd
it, we seek real numbers u and v, expressed in terms of x and y, such that

(x, y)(u, v) = (1, 0).

According to equation (4), Sec. 1, which deﬁnes the product of two complex numbers, u and v must satisfy the pair

xu − yv = 1, yu + xv = 0

of linear simultaneous equations; and simple computation yields the unique solution

u

=

x2

x +

y2 ,

v

=

−y x2 + y2 .

So the multiplicative inverse of z = (x, y) is

(6)

z−1 =

x

−y

x2 + y2 , x2 + y2

(z = 0).

The inverse z−1 is not deﬁned when z = 0. In fact, z = 0 means that x2 + y2 = 0 ; and this is not permitted in expression (6).

Brown-chap01-v3 10/29/07 3:32pm 6

6 Complex Numbers

chap. 1

in Sec. 2. Inasmuch as such properties continue to be anticipated because they also apply to real numbers, the reader can easily pass to Sec. 4 without serious disruption.
We begin with the observation that the existence of multiplicative inverses enables us to show that if a product z1z2 is zero, then so is at least one of the factors z1 and z2. For suppose that z1z2 = 0 and z1 = 0. The inverse z1−1 exists; and any complex number times zero is zero (Sec. 1). Hence

z2 = z2 · 1 = z2(z1z1−1) = (z1−1z1)z2 = z1−1(z1z2) = z1−1 · 0 = 0.

That is, if z1z2 = 0, either z1 = 0 or z2 = 0; or possibly both of the numbers z1 and z2 are zero. Another way to state this result is that if two complex numbers z1 and z2 are nonzero, then so is their product z1z2.
Subtraction and division are deﬁned in terms of additive and multiplicative
inverses:

(1)

z1 − z2 = z1 + (−z2),

(2)

z1 z2

= z1z2−1

(z2 = 0).

Thus, in view of expressions (5) and (6) in Sec. 2,

(3)

z1 − z2 = (x1, y1) + (−x2, −y2) = (x1 − x2, y1 − y2)

and

(4)

z1 z2

=

(x1, y1)

x22

x2 +

y22

,

−y2 x22 + y22

=

x1x2 x22

+ +

y1y2 y22

,

y1x2 x22

− +

x1y2 y22

(z2 = 0)

when z1 = (x1, y1) and z2 = (x2, y2). Using z1 = x1 + iy1 and z2 = x2 + iy2, one can write expressions (3) and (4)
here as

(5)

z1 − z2 = (x1 − x2) + i(y1 − y2)

and

(6)

z1 z2

=

x1x2 + y1y2 x22 + y22

+

i

y1x2 x22

− +

x1y2 y22

(z2 = 0).

Although expression (6) is not easy to remember, it can be obtained by writing (see Exercise 7)

(7)

z1 z2

=

(x1 (x2

+ iy1)(x2 + iy2)(x2

− −

iy2) , iy2)

Brown-chap01-v3 10/29/07 3:32pm 5

sec. 3

Further Properties 5

EXERCISES

1. Verify that

√

√

(a) ( 2 − i) − i(1 − 2i) = −2i;

(b) (2, −3)(−2, 1) = (−1, 8);

(c) (3, 1)(3, −1)

11 ,

= (2, 1).

5 10

2. Show that

(a) Re(iz) = − Im z; (b) Im(iz) = Re z.

3. Show that (1 + z)2 = 1 + 2z + z2. 4. Verify that each of the two numbers z = 1 ± i satisﬁes the equation z2 − 2z + 2 = 0.
5. Prove that multiplication of complex numbers is commutative, as stated at the beginning of Sec. 2.
6. Verify
(a) the associative law for addition of complex numbers, stated at the beginning of Sec. 2;
(b) the distributive law (3), Sec. 2.
7. Use the associative law for addition and the distributive law to show that

z(z1 + z2 + z3) = zz1 + zz2 + zz3.
8. (a) Write (x, y) + (u, v) = (x, y) and point out how it follows that the complex number 0 = (0, 0) is unique as an additive identity.
(b) Likewise, write (x, y)(u, v) = (x, y) and show that the number 1 = (1, 0) is a unique multiplicative identity.
9. Use −1 = (−1, 0) and z = (x, y) to show that (−1)z = −z.
10. Use i = (0, 1) and y = (y, 0) to verify that −(iy) = (−i)y. Thus show that the additive inverse of a complex number z = x + iy can be written −z = −x − iy without ambiguity.
11. Solve the equation z2 + z + 1 = 0 for z = (x, y) by writing

(x, y)(x, y) + (x, y) + (1, 0) = (0, 0)

and then solving a pair of simultaneous equations in x and y.

Suggestion: Use the fact that no real number x satisﬁes the given equation to

show that y = 0.

√

Ans. z = − 1 , ± 3 .

22

3. FURTHER PROPERTIES
In this section, we mention a number of other algebraic properties of addition and multiplication of complex numbers that follow from the ones already described

Brown-chap01-v3 10/29/07 3:32pm 7

sec. 3

Further Properties 7

multiplying out the products in the numerator and denominator on the right, and then using the property

(8)

z1 + z2 z3

= (z1 + z2)z3−1

= z1z3−1 + z2z3−1

=

z1 z3

+

z2 z3

(z3 = 0).

The motivation for starting with equation (7) appears in Sec. 5.

EXAMPLE. The method is illustrated below:

4+i 2 − 3i

=

(4 + i)(2 + 3i) (2 − 3i)(2 + 3i)

=

5 + 14i 13

=

5 13

+

14 i.
13

There are some expected properties involving quotients that follow from the relation

(9)

1 z2

= z2−1

(z2 = 0),

which is equation (2) when z1 = 1. Relation (9) enables us, for instance, to write equation (2) in the form

(10)

z1 z2

= z1

1 z2

Also, by observing that (see Exercise 3)

(z2 = 0).

(z1z2)(z1−1z2−1) = (z1z1−1)(z2z2−1) = 1 (z1 = 0, z2 = 0),

and hence that z1−1z2−1 = (z1z2)−1, one can use relation (9) to show that

(11)

1 z1

1 z2

=

z1−1z2−1

=

(z1z2)−1

=

1 z1z2

(z1 = 0, z2 = 0).

Another useful property, to be derived in the exercises, is

(12)

z1 z2 = z1z2

z3 z4

z3z4

(z3 = 0, z4 = 0).

Finally, we note that the binomial formula involving real numbers remains valid with complex numbers. That is, if z1 and z2 are any two nonzero complex numbers, then

(13)

n
(z1 + z2)n =
k=0

n k

z1k z2n−k

(n = 1, 2, . . .)

where

n = n! k k!(n − k)!

(k = 0, 1, 2, . . . , n)

and where it is agreed that 0! = 1. The proof is left as an exercise.

Brown-chap01-v3 10/29/07 3:32pm 8

8 Complex Numbers

chap. 1

EXERCISES
1. Reduce each of these quantities to a real number:

(a)

1 + 2i 3 − 4i

+

2−i ;
5i

5i (b) (1 − i)(2 − i)(3 − i) ;

(c) (1 − i)4.

Ans. (a) −2/5; 2. Show that

(b) −1/2; (c) −4.

1 =z 1/z

(z = 0).

3. Use the associative and commutative laws for multiplication to show that

(z1z2)(z3z4) = (z1z3)(z2z4).
4. Prove that if z1z2z3 = 0, then at least one of the three factors is zero. Suggestion: Write (z1z2)z3 = 0 and use a similar result (Sec. 3) involving two
factors.
5. Derive expression (6), Sec. 3, for the quotient z1/z2 by the method described just after it.
6. With the aid of relations (10) and (11) in Sec. 3, derive the identity

z1 z2 = z1z2

z3 z4

z3z4

(z3 = 0, z4 = 0).

7. Use the identity obtained in Exercise 6 to derive the cancellation law

z1z = z1 z2z z2

(z2 = 0, z = 0).

8. Use mathematical induction to verify the binomial formula (13) in Sec. 3. More pre-

cisely, note that the formula is true when n = 1. Then, assuming that it is valid

when n = m where m denotes any positive integer, show that it must hold when

n = m + 1. Suggestion: When n = m + 1, write

m

(z1 + z2)m+1 = (z1 + z2)(z1 + z2)m = (z2 + z1)

m k

z1k z2m−k

k=0

m

m

=

m k

z1kz2m+1−k +

m k

z1k+1z2m−k

k=0

k=0

and replace k by k − 1 in the last sum here to obtain

m
(z1 + z2)m+1 = z2m+1 +
k=1

m k

+

m k−1

z1k z2m+1−k + z1m+1.

Brown-chap01-v3 10/29/07 3:32pm 9

sec. 4

Vectors and Moduli 9

Finally, show how the right-hand side here becomes

m
z2m+1 +
k=1

m+1 k

m+1
z1k z2m+1−k + z1m+1 =
k=0

m+1 k

z1k z2m+1−k .

4. VECTORS AND MODULI
It is natural to associate any nonzero complex number z = x + iy with the directed line segment, or vector, from the origin to the point (x, y) that represents z in the complex plane. In fact, we often refer to z as the point z or the vector z. In Fig. 2 the numbers z = x + iy and −2 + i are displayed graphically as both points and radius vectors.

(–2, 1) –2

–2 + i

y

1

z = (x, y)

z = x + iy

O

x FIGURE 2

When z1 = x1 + iy1 and z2 = x2 + iy2, the sum
z1 + z2 = (x1 + x2) + i(y1 + y2)
corresponds to the point (x1 + x2, y1 + y2). It also corresponds to a vector with those coordinates as its components. Hence z1 + z2 may be obtained vectorially as shown in Fig. 3.

y

z2 O

z1+ z2 z1

z2 x FIGURE 3

Although the product of two complex numbers z1 and z2 is itself a complex number represented by a vector, that vector lies in the same plane as the vectors for
z1 and z2. Evidently, then, this product is neither the scalar nor the vector product used in ordinary vector analysis.

Brown-chap01-v3 10/29/07 3:32pm 10

10 Complex Numbers

chap. 1

The vector interpretation of complex numbers is especially helpful in extending the concept of absolute values of real numbers to the complex plane. The modulus, or absolute value, of a complex number z = x + iy is deﬁned as the nonnegative real number x2 + y2 and is denoted by |z|; that is,

(1)

|z| = x2 + y2.

Geometrically, the number |z| is the distance between the point (x, y) and
the origin, or the length of the radius vector representing z. It reduces to the usual absolute value in the real number system when y = 0. Note that while the inequality z1 < z2 is meaningless unless both z1 and z2 are real, the statement |z1| < |z2| means that the point z1 is closer to the origin than the point z2 is.

√

√

EXAMPLE 1. Since |− 3 + 2i| = 13 and |1 + 4i| = 17, we know that

the point −3 + 2i is closer to the origin than 1 + 4i is.

The distance between two points (x1, y1) and (x2, y2) is |z1 − z2|. This is clear from Fig. 4, since |z1 − z2| is the length of the vector representing the number
z1 − z2 = z1 + (−z2);
and, by translating the radius vector z1 − z2, one can interpret z1 − z2 as the directed line segment from the point (x2, y2) to the point (x1, y1). Alternatively, it follows from the expression
z1 − z2 = (x1 − x2) + i(y1 − y2)
and deﬁnition (1) that
|z1 − z2| = (x1 − x2)2 + (y1 − y2)2.

y

(x2, y2) z2

|z1 – z2|

(x1, y1)

z1

O

z1 – z2

– z2

x

FIGURE 4

The complex numbers z corresponding to the points lying on the circle with
center z0 and radius R thus satisfy the equation |z − z0| = R, and conversely. We refer to this set of points simply as the circle |z − z0| = R.

EXAMPLE 2. The equation |z − 1 + 3i| = 2 represents the circle whose center is z0 = (1, −3) and whose radius is R = 2.

Brown-chap01-v3 10/29/07 3:32pm 11

sec. 4

Vectors and Moduli 11

It also follows from deﬁnition (1) that the real numbers |z|, Re z = x, and Im z = y are related by the equation

(2)

|z|2 = (Re z)2 + (Im z)2.

Thus

(3)

Re z ≤ |Re z| ≤ |z| and Im z ≤ |Im z| ≤ |z|.

We turn now to the triangle inequality, which provides an upper bound for the modulus of the sum of two complex numbers z1 and z2:

(4)

|z1 + z2| ≤ |z1| + |z2|.

This important inequality is geometrically evident in Fig. 3, since it is merely a statement that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. We can also see from Fig. 3 that inequality (4) is actually an equality when 0, z1, and z2 are collinear. Another, strictly algebraic, derivation is given in Exercise 15, Sec. 5.
An immediate consequence of the triangle inequality is the fact that

(5)

|z1 + z2| ≥ ||z1| − |z2||.

To derive inequality (5), we write

|z1| = |(z1 + z2) + (−z2)| ≤ |z1 + z2| + |− z2|,

which means that

(6)

|z1 + z2| ≥ |z1| − |z2|.

This is inequality (5) when |z1| ≥ |z2|. If |z1| < |z2|, we need only interchange z1 and z2 in inequality (6) to arrive at

|z1 + z2| ≥ −(|z1| − |z2|),

which is the desired result. Inequality (5) tells us, of course, that the length of one side of a triangle is greater than or equal to the difference of the lengths of the other two sides.
Because |− z2| = |z2|, one can replace z2 by −z2 in inequalities (4) and (5) to summarize these results in a particularly useful form:

(7)

|z1 ± z2| ≤ |z1| + |z2|,

(8)

|z1 ± z2| ≥ ||z1| − |z2||.

When combined, inequalities (7) and (8) become

(9)

||z1| − |z2|| ≤ |z1 ± z2| ≤ |z1| + |z2|.

Brown-chap01-v3 10/29/07 3:32pm 12

12 Complex Numbers

chap. 1

EXAMPLE 3. If a point z lies on the unit circle |z| = 1 about the origin, it follows from inequalities (7) and (8) that
|z − 2| ≤ |z| + 2 = 3
and |z − 2| ≥ ||z| − 2| = 1.
The triangle inequality (4) can be generalized by means of mathematical induction to sums involving any ﬁnite number of terms:
(10) |z1 + z2 + · · · + zn| ≤ |z1| + |z2| + · · · + |zn| (n = 2, 3, . . .).
To give details of the induction proof here, we note that when n = 2, inequality (10) is just inequality (4). Furthermore, if inequality (10) is assumed to be valid when n = m, it must also hold when n = m + 1 since, by inequality (4),
|(z1 + z2 + · · · + zm) + zm+1| ≤ |z1 + z2 + · · · + zm| + |zm+1| ≤ (|z1| + |z2| + · · · + |zm|) + |zm+1|.

EXERCISES

1. Locate the numbers z1 + z2 and z1 − z2 vectorially when

(a) z1 = 2i,

z2

=

2 3

−

i;

√

√

(b) z1 = (− 3, 1), z2 = ( 3, 0);

(c) z1 = (−3, 1), z2 = (1, 4); (d) z1 = x1 + iy1, z2 = x1 − iy1.

2. Verify inequalities (3), Sec. 4, involving Re z, Im z, and |z|. 3. Use established properties of moduli to show that when |z3| = |z4|,

Re(z1 + z2) |z3 + z4|

≤

|z1| ||z3|

+ −

|z2| |z4||

.

√ 4. Verify that 2 |z| ≥ |Re z| + |Im z|.

Suggestion: Reduce this inequality to (|x| − |y|)2 ≥ 0.

5. In each case, sketch the set of points determined by the given condition:

(a) |z − 1 + i| = 1; (b) |z + i| ≤ 3 ; (c) |z − 4i| ≥ 4.

6. Using the fact that |z1 − z2| is the distance between two points z1 and z2, give a geometric argument that
(a) |z − 4i| + |z + 4i| = 10 represents an ellipse whose foci are (0, ±4) ; (b) |z − 1| = |z + i| represents the line through the origin whose slope is −1.

Brown-chap01-v3 10/29/07 3:32pm 13

sec. 5

Complex Conjugates 13

5. COMPLEX CONJUGATES
The complex conjugate, or simply the conjugate, of a complex number z = x + iy is deﬁned as the complex number x − iy and is denoted by z ; that is,

(1)

z = x − iy.

The number z is represented by the point (x, −y), which is the reﬂection in the real axis of the point (x, y) representing z (Fig. 5). Note that

z = z and |z| = |z|

for all z.

y

z O
–z

(x, y) (x, –y)

x FIGURE 5

If z1 = x1 + iy1 and z2 = x2 + iy2, then

z1 + z2 = (x1 + x2) − i(y1 + y2) = (x1 − iy1) + (x2 − iy2).

So the conjugate of the sum is the sum of the conjugates:

(2)

z1 + z2 = z1 + z2.

In like manner, it is easy to show that

(3)

z1 − z2 = z1 − z2,

(4)

z1z2 = z1 z2,

and

(5)

z1 = z1

z2

z2

(z2 = 0).

The sum z + z of a complex number z = x + iy and its conjugate z = x − iy is the real number 2x, and the difference z − z is the pure imaginary number 2iy. Hence

(6)

Re z = z + z

and

Im z

=

z−z .

2

2i

Brown-chap01-v3 10/29/07 3:32pm 14

14 Complex Numbers

chap. 1

An important identity relating the conjugate of a complex number z = x + iy to its modulus is

(7)

z z = |z|2,

where each side is equal to x2 + y2. It suggests the method for determining a
quotient z1/z2 that begins with expression (7), Sec. 3. That method is, of course, based on multiplying both the numerator and the denominator of z1/z2 by z2, so that the denominator becomes the real number |z2|2.

EXAMPLE 1. As an illustration, −1 + 3i (−1 + 3i)(2 + i) −5 + 5i −5 + 5i 2 − i = (2 − i)(2 + i) = |2 − i|2 = 5 = −1 + i.
See also the example in Sec. 3.

Identity (7) is especially useful in obtaining properties of moduli from properties of conjugates noted above. We mention that

(8)

|z1z2| = |z1||z2|

and

(9)

z1 z2

=

|z1| |z2|

Property (8) can be established by writing

(z2 = 0).

|z1z2|2 = (z1z2)(z1z2) = (z1z2)(z1 z2) = (z1z1)(z2z2) = |z1|2|z2|2 = (|z1||z2|)2

and recalling that a modulus is never negative. Property (9) can be veriﬁed in a similar way.

EXAMPLE 2. Property (8) tells us that |z2| = |z|2 and |z3| = |z|3. Hence if z is a point inside the circle centered at the origin with radius 2, so that |z| < 2, it follows from the generalized triangle inequality (10) in Sec. 4 that
|z3 + 3z2 − 2z + 1| ≤ |z|3 + 3|z|2 + 2|z| + 1 < 25.

EXERCISES

1. Use properties of conjugates and moduli established in Sec. 5 to show that

(a) z + 3i = z − 3i; (c) (2 + i)2 = 3 − 4i;

(b) iz = −iz;

√

√

(d) |(2z + 5)( 2 − i)| = 3 |2z + 5|.

2. Sketch the set of points determined by the condition (a) Re(z − i) = 2; (b) |2z + i| = 4.

Brown-chap01-v3 10/29/07 3:32pm 15

sec. 5

Exercises 15

3. Verify properties (3) and (4) of conjugates in Sec. 5.

4. Use property (4) of conjugates in Sec. 5 to show that

(a) z1z2z3 = z1 z2 z3 ;

(b) z4 = z4.

5. Verify property (9) of moduli in Sec. 5.

6. Use results in Sec. 5 to show that when z2 and z3 are nonzero,

(a) z1 = z1 ;

z2z3

z2 z3

(b) z1 = |z1| . z2z3 |z2||z3|

7. Show that

|Re(2 + z + z3)| ≤ 4 when |z| ≤ 1.

8. It is shown in Sec. 3 that if z1z2 = 0, then at least one of the numbers z1 and z2 must be zero. Give an alternative proof based on the corresponding result for real numbers and using identity (8), Sec. 5.
9. By factoring z4 − 4z2 + 3 into two quadratic factors and using inequality (8), Sec. 4, show that if z lies on the circle |z| = 2, then

1 z4 − 4z2 + 3

≤

1 .

3

10. Prove that (a) z is real if and only if z = z; (b) z is either real or pure imaginary if and only if z2 = z2.
11. Use mathematical induction to show that when n = 2, 3, . . . , (a) z1 + z2 + · · · + zn = z1 + z2 + · · · + zn; (b) z1z2 · · · zn = z1 z2 · · · zn.

12. Let a0, a1, a2, . . . , an (n ≥ 1) denote real numbers, and let z be any complex number. With the aid of the results in Exercise 11, show that

a0 + a1z + a2z2 + · · · + anzn = a0 + a1z + a2z2 + · · · + anz n.

13. Show that the equation |z − z0| = R of a circle, centered at z0 with radius R, can be
written |z|2 − 2 Re(zz0) + |z0|2 = R2.

14. Using expressions (6), Sec. 5, for Re z and Im z, show that the hyperbola x2 − y2 = 1
can be written z2 + z2 = 2.

15. Follow the steps below to give an algebraic derivation of the triangle inequality (Sec. 4)

|z1 + z2| ≤ |z1| + |z2|.

(a) Show that |z1 + z2|2 = (z1 + z2)(z1 + z2) = z1z1 + (z1z2 + z1z2) + z2z2.

Brown-chap01-v3 10/29/07 3:32pm 16

16 Complex Numbers

chap. 1

(b) Point out why

z1z2 + z1z2 = 2 Re(z1z2) ≤ 2|z1||z2|.

(c) Use the results in parts (a) and (b) to obtain the inequality

|z1 + z2|2 ≤ (|z1| + |z2|)2,

and note how the triangle inequality follows.

6. EXPONENTIAL FORM

Let r and θ be polar coordinates of the point (x, y) that corresponds to a nonzero complex number z = x + iy. Since x = r cos θ and y = r sin θ , the number z can be written in polar form as

(1)

z = r(cos θ + i sin θ ).

If z = 0, the coordinate θ is undeﬁned; and so it is understood that z = 0 whenever polar coordinates are used.
In complex analysis, the real number r is not allowed to be negative and is the length of the radius vector for z ; that is, r = |z|. The real number θ represents the angle, measured in radians, that z makes with the positive real axis when z is interpreted as a radius vector (Fig. 6). As in calculus, θ has an inﬁnite number of possible values, including negative ones, that differ by integral multiples of 2π. Those values can be determined from the equation tan θ = y/x, where the quadrant containing the point corresponding to z must be speciﬁed. Each value of θ is called an argument of z, and the set of all such values is denoted by arg z. The principal value of arg z, denoted by Arg z, is that unique value such that −π < ≤ π. Evidently, then,

(2)

arg z = Arg z + 2nπ (n = 0, ±1, ±2, . . .).

Also, when z is a negative real number, Arg z has value π, not −π.

y
z = x + iy r

x FIGURE 6
EXAMPLE 1. The complex number −1 − i, which lies in the third quadrant, has principal argument −3π/4. That is,
Arg(−1 − i) = − 3π . 4

Brown-chap01-v3 10/29/07 3:32pm 17

sec. 6

Exponential Form 17

It must be emphasized that because of the restriction −π < argument , it is not true that Arg(−1 − i) = 5π/4.
According to equation (2),

≤ π of the principal

arg(−1 − i) = − 3π + 2nπ 4

(n = 0, ±1, ±2, . . .).

Note that the term Arg z on the right-hand side of equation (2) can be replaced by any particular value of arg z and that one can write, for instance,

arg(−1 − i) = 5π + 2nπ 4

(n = 0, ±1, ±2, . . .).

The symbol eiθ , or exp(iθ ), is deﬁned by means of Euler’s formula as

(3)

eiθ = cos θ + i sin θ,

where θ is to be measured in radians. It enables one to write the polar form (1) more compactly in exponential form as

(4)

z = reiθ .

The choice of the symbol eiθ will be fully motivated later on in Sec. 29. Its use in Sec. 7 will, however, suggest that it is a natural choice.

EXAMPLE 2. The number −1 − i in Example 1 has exponential form

(5)

√ −1 − i = 2 exp i

− 3π

.

4

With

the

agreement

that

e−iθ

=

ei (−θ ) ,

this

can

also

be

written

−1

−

i

=

√ 2

e−i

3π/4.

Expression (5) is, of course, only one of an inﬁnite number of possibilities for the

exponential form of −1 − i:

(6)

√ −1 − i = 2 exp i

− 3π + 2nπ

4

(n = 0, ±1, ±2, . . .).

Note how expression (4) with r = 1 tells us that the numbers eiθ lie on the circle centered at the origin with radius unity, as shown in Fig. 7. Values of eiθ are, then, immediate from that ﬁgure, without reference to Euler’s formula. It is, for instance, geometrically obvious that
eiπ = −1, e−iπ/2 = −i, and e−i4π = 1.

Brown-chap01-v3 10/29/07 3:32pm 18

18 Complex Numbers
y

1

O

x

chap. 1

FIGURE 7

Note, too, that the equation

(7)

z = Reiθ (0 ≤ θ ≤ 2π )

is a parametric representation of the circle |z| = R, centered at the origin with radius R. As the parameter θ increases from θ = 0 to θ = 2π, the point z starts from the positive real axis and traverses the circle once in the counterclockwise direction. More generally, the circle |z − z0| = R, whose center is z0 and whose radius is R, has the parametric representation

(8)

z = z0 + Reiθ (0 ≤ θ ≤ 2π ).

This can be seen vectorially (Fig. 8) by noting that a point z traversing the circle |z − z0| = R once in the counterclockwise direction corresponds to the sum of the ﬁxed vector z0 and a vector of length R whose angle of inclination θ varies from θ = 0 to θ = 2π.

y

z
z0 O

x FIGURE 8

7. PRODUCTS AND POWERS IN EXPONENTIAL FORM
Simple trigonometry tells us that eiθ has the familiar additive property of the exponential function in calculus:
eiθ1 eiθ2 = (cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = (cos θ1 cos θ2 − sin θ1 sin θ2) + i(sin θ1 cos θ2 + cos θ1 sin θ2) = cos(θ1 + θ2) + i sin(θ1 + θ2) = ei(θ1+θ2).

Brown-chap01-v3 10/29/07 3:32pm 19

sec. 7

Products and Powers in Exponential Form 19

Thus, if z1 = r1eiθ1 and z2 = r2eiθ2 , the product z1z2 has exponential form

(1)

z1z2 = r1eiθ1 r2eiθ2 = r1r2eiθ1 eiθ2 = (r1r2)ei(θ1+θ2).

Furthermore,

(2)

z1 z2

=

r1eiθ1 r2eiθ2

=

r1 r2

eiθ1 e−iθ2 · eiθ2 e−iθ2

=

r1 r2

ei (θ1 −θ2 ) ·
ei0

=

r1 ei(θ1−θ2). r2

Note how it follows from expression (2) that the inverse of any nonzero complex number z = reiθ is

(3)

z−1

=

1 z

=

1ei0 r eiθ

=

1 ei(0−θ) r

=

1 e−iθ . r

Expressions (1), (2), and (3) are, of course, easily remembered by applying the usual algebraic rules for real numbers and ex.
Another important result that can be obtained formally by applying rules for real numbers to z = reiθ is

(4)

zn = rneinθ (n = 0, ±1, ±2, . . .).

It is easily veriﬁed for positive values of n by mathematical induction. To be speciﬁc, we ﬁrst note that it becomes z = reiθ when n = 1. Next, we assume that it is valid when n = m, where m is any positive integer. In view of expression (1) for the product of two nonzero complex numbers in exponential form, it is then valid for n = m + 1:
zm+1 = zmz = rmeimθ reiθ = (rmr)ei(mθ+θ) = rm+1ei(m+1)θ .

Expression (4) is thus veriﬁed when n is a positive integer. It also holds when n = 0, with the convention that z0 = 1. If n = −1, −2, . . . , on the other hand, we deﬁne zn in terms of the multiplicative inverse of z by writing
zn = (z−1)m where m = −n = 1, 2, . . . .

Then, since equation (4) is valid for positive integers, it follows from the exponential form (3) of z−1 that

zn =

1 ei(−θ)

m
=

r

1

m
eim(−θ) =

r

1

−n
ei(−n)(−θ) = r neinθ

r

(n = −1, −2, . . .).

Expression (4) is now established for all integral powers. Expression (4) can be useful in ﬁnding powers of complex numbers even when
they are given in rectangular form and the result is desired in that form.

Brown-chap01-v3 10/29/07 3:32pm 20

20 Complex Numbers

chap. 1

√ EXAMPLE 1. In order to put ( 3 + 1)7 in rectangular form, one need only

write

√

√

( 3 + i)7 = (2eiπ/6)7 = 27ei7π/6 = (26eiπ )(2eiπ/6) = −64( 3 + i).

Finally, we observe that if r = 1, equation (4) becomes

(5)

(eiθ )n = einθ (n = 0, ±1, ±2, . . .).

When written in the form

(6)

(cos θ + i sin θ )n = cos nθ + i sin nθ (n = 0, ±1, ±2, . . .),

this is known as de Moivre’s formula. The following example uses a special case of it.

EXAMPLE 2. Formula (6) with n = 2 tells us that (cos θ + i sin θ )2 = cos 2θ + i sin 2θ,
or cos2 θ − sin2 θ + i2 sin θ cos θ = cos 2θ + i sin 2θ.
By equating real parts and then imaginary parts here, we have the familiar trigonometric identities
cos 2θ = cos2 θ − sin2 θ, sin 2θ = 2 sin θ cos θ. (See also Exercises 10 and 11, Sec. 8.)

8. ARGUMENTS OF PRODUCTS AND QUOTIENTS

If z1 = r1eiθ1 and z2 = r2eiθ2 , the expression

(1)

z1z2 = (r1r2)ei(θ1+θ2)

in Sec. 7 can be used to obtain an important identity involving arguments:

(2)

arg(z1z2) = arg z1 + arg z2.

This result is to be interpreted as saying that if values of two of the three (multiplevalued) arguments are speciﬁed, then there is a value of the third such that the equation holds.
We start the veriﬁcation of statement (2) by letting θ1 and θ2 denote any values of arg z1 and arg z2, respectively. Expression (1) then tells us that θ1 + θ2 is a value of arg(z1z2). (See Fig. 9.) If, on the other hand, values of arg(z1z2) and

Brown-chap01-v3 10/29/07 3:32pm 21

sec. 8

y

z1z2

Arguments of Products and Quotients 21

z2 z1
O

x FIGURE 9

arg z1 are speciﬁed, those values correspond to particular choices of n and n1 in the expressions

arg(z1z2) = (θ1 + θ2) + 2nπ (n = 0, ±1, ±2, . . .)

and arg z1 = θ1 + 2n1π (n1 = 0, ±1, ±2, . . .).

Since

(θ1 + θ2) + 2nπ = (θ1 + 2n1π ) + [θ2 + 2(n − n1)π ],

equation (2) is evidently satisﬁed when the value

arg z2 = θ2 + 2(n − n1)π

is chosen. Veriﬁcation when values of arg(z1z2) and arg z2 are speciﬁed follows by symmetry.
Statement (2) is sometimes valid when arg is replaced everywhere by Arg (see Exercise 6). But, as the following example illustrates, that is not always the case.

EXAMPLE 1. When z1 = −1 and z2 = i,

Arg(z1z2)

=

Arg(−i)

=

−

π 2

but

Arg z1

+ Arg z2

=

π

+

π 2

=

3π .
2

If, however, we take the values of arg z1 and arg z2 just used and select the value

Arg(z1z2) + 2π

=

−π 2

+ 2π

=

3π 2

of arg(z1z2), we ﬁnd that equation (2) is satisﬁed.

Statement (2) tells us that

arg z1 z2

= arg(z1z2−1) = arg z1 + arg(z2−1);

Brown-chap01-v3 10/29/07 3:32pm 22

22 Complex Numbers

chap. 1

and, since (Sec. 7)

one can see that

z2−1

=

1 e−iθ2 , r2

(3)

arg(z2−1) = −arg z2.

Hence

(4)

arg z1 z2

= arg z1 − arg z2.

Statement (3) is, of course, to be interpreted as saying that the set of all values on the left-hand side is the same as the set of all values on the right-hand side. Statement (4) is, then, to be interpreted in the same way that statement (2) is.

EXAMPLE 2. In order to ﬁnd the principal argument Arg z when

z = −√2 , 1 + 3i

observe that

√ arg z = arg(−2) − arg(1 + 3i).

Since

Arg(−2) = π

and

Arg(1

+

√ 3i)

=

π ,

3

one value of arg z is 2π/3; and, because 2π/3 is between −π and π, we ﬁnd that Arg z = 2π/3.

EXERCISES

1. Find the principal argument Arg z when

(a)

z

=

−2

i −

; 2i

√ (b) z = ( 3 − i)6.

Ans. (a) −3π/4; (b) π .

2. Show that (a) |eiθ | = 1; (b) eiθ = e−iθ .

3. Use mathematical induction to show that

eiθ1 eiθ2 · · · eiθn = ei(θ1+θ2+···+θn)

(n = 2, 3, . . .).

4. Using the fact that the modulus |eiθ − 1| is the distance between the points eiθ and 1 (see Sec. 4), give a geometric argument to ﬁnd a value of θ in the interval 0 ≤ θ < 2π that satisﬁes the equation |eiθ − 1| = 2.
Ans. π .

Brown-chap01-v3 10/29/07 3:32pm 23

sec. 8

Exercises 23

5. By writing the individual factors on the left in exponential form, performing the needed

operations, and ﬁnally changing back to rectangular coordinates, show that

√√

√

(a) i(1 − 3i)( 3 + i) = 2(1 + 3i); (b) 5i/(2 + i) = 1 + 2i;

(c) (−1 + i)7 = −8(1 + i);

(d)

(1

+

√ 3i

)−10

=

2−11(−1

+

√ 3i).

6. Show that if Re z1 > 0 and Re z2 > 0, then

Arg(z1z2) = Arg z1 + Arg z2,

where principal arguments are used.
7. Let z be a nonzero complex number and n a negative integer (n = −1, −2, . . .). Also, write z = reiθ and m = −n = 1, 2, . . . . Using the expressions

zm = rmeimθ and z−1 = 1 ei(−θ), r

verify that (zm)−1 = (z−1)m and hence that the deﬁnition zn = (z−1)m in Sec. 7 could have been written alternatively as zn = (zm)−1.
8. Prove that two nonzero complex numbers z1 and z2 have the same moduli if and only if there are complex numbers c1 and c2 such that z1 = c1c2 and z2 = c1c2. Suggestion: Note that

exp i θ1 + θ2 2

exp i θ1 − θ2 2

= exp(iθ1)

and [see Exercise 2(b)]

exp i θ1 + θ2 2

exp i θ1 − θ2 2

= exp(iθ2).

9. Establish the identity

1 + z + z2 + · · · + zn = 1 − zn+1 1−z

(z = 1)

and then use it to derive Lagrange’s trigonometric identity:

1 + cos θ + cos 2θ + · · · + cos nθ = 1 + sin[(2n + 1)θ/2]

2

2 sin(θ/2)

(0 < θ < 2π ).

Suggestion: As for the ﬁrst identity, write S = 1 + z + z2 + · · · + zn and consider the difference S − zS. To derive the second identity, write z = eiθ in the ﬁrst one.
10. Use de Moivre’s formula (Sec. 7) to derive the following trigonometric identities:
(a) cos 3θ = cos3 θ − 3 cos θ sin2 θ ; (b) sin 3θ = 3 cos2 θ sin θ − sin3 θ .

Brown-chap01-v3 10/29/07 3:32pm 24

24 Complex Numbers

chap. 1

11. (a) Use the binomial formula (Sec. 3) and de Moivre’s formula (Sec. 7) to write

n
cos nθ + i sin nθ =

n cosn−k θ (i sin θ )k

k

k=0

(n = 0, 1, 2, . . .).

Then deﬁne the integer m by means of the equations

m=

n/2 (n − 1)/2

if n is even, if nis odd

and use the above summation to show that [compare with Exercise 10(a)]

m
cos nθ =

n (−1)k cosn−2k θ sin2k θ

2k

k=0

(n = 0, 1, 2, . . .).

(b) Write x = cos θ in the ﬁnal summation in part (a) to show that it becomes a polynomial

m

Tn(x) =

n 2k

(−1)kxn−2k (1 − x2)k

k=0

of degree n (n = 0, 1, 2, . . .) in the variable x.∗

9. ROOTS OF COMPLEX NUMBERS
Consider now a point z = reiθ , lying on a circle centered at the origin with radius r (Fig. 10). As θ is increased, z moves around the circle in the counterclockwise direction. In particular, when θ is increased by 2π, we arrive at the original point; and the same is true when θ is decreased by 2π. It is, therefore, evident from Fig. 10 that two nonzero complex numbers
z1 = r1eiθ1 and z2 = r2eiθ2
y

r

O

x

FIGURE 10 ∗These are called Chebyshev polynomials and are prominent in approximation theory.

Brown-chap01-v3 10/29/07 3:32pm 25

sec. 9

Roots of Complex Numbers 25

are equal if and only if

r1 = r2 and θ1 = θ2 + 2kπ,

where k is some integer (k = 0, ±1, ±2, . . .).

This observation, together with the expression zn = rneinθ in Sec. 7 for integral

powers of complex numbers z = reiθ , is useful in ﬁnding the nth roots of any

nonzero complex number z0 = r0eiθ0 , where n has one of the values n = 2, 3, . . . .

The method starts with the fact that an nth root of z0 is a nonzero number z = reiθ

such that zn = z0, or

rneinθ = r0eiθ0 .

According to the statement in italics just above, then,

rn = r0 and nθ = θ0 + 2kπ,

where

k

is

any

integer

(k

=

0, ±1, ±2,

. . .).

So

r

=

√ n r0,

where

this

radical

denotes

the unique positive nth root of the positive real number r0, and

θ = θ0 + 2kπ = θ0 + 2kπ

n

nn

(k = 0, ±1, ±2, . . .).

Consequently, the complex numbers

z

=

√ n r0

exp

i

θ0 + 2kπ nn

(k = 0, ±1, ±2, . . .)

are the nth of the roots

roots of z0. We that they all lie

are on

tahbelecirtoclese|ez|im=m√nedr0iaatebloyuftrtohme

this exponential form origin and are equally

spaced every 2π/n radians, starting with argument θ0/n. Evidently, then, all of the

distinct roots are obtained when k = 0, 1, 2, . . . , n − 1, and no further roots arise

with other values of k. We let ck (k = 0, 1, 2, . . . , n − 1) denote these distinct roots

and write

(1)

√ ck = n r0 exp i

θ0 + 2kπ nn

(k = 0, 1, 2, . . . , n − 1).

(See Fig. 11.)

ck–1 y

ck

n

O

√n r0 x

FIGURE 11

Brown-chap01-v3 10/29/07 3:32pm 26

26 Complex Numbers

chap. 1

√ The number n r0 is the length of each of the radius vectors representing the

n roots. The the opposite

ﬁrst ends

root of a

c0 has argument θ0/n; and diameter of the circle |z| =

√the two roots when n = 2 n r0, the second root being

lie at −c0.

When n ≥ 3, the roots lie at the vertices of a regular polygon of n sides inscribed

in that circle. We shall let z01/n denote the set of nth roots of z0. If, in particular, z0 is a
positive √real number r0, the symbol r01/n denotes the entire set of roots; and the symbol n r0 in expression (1) is reserved for the one positive root. When the value of θ0 that is used in expression (1) is the principal value of arg z0 (−π < θ0 ≤ π ),
the number c0 is referred to as √the principal root. Thus when z0 is a positive real number r0, its principal root is n r0.

Observe that if we write expression (1) for the roots of z0 as

√ ck = n r0 exp

i θ0 n

2kπ exp i
n

(k = 0, 1, 2, . . . , n − 1),

and also write (2)

ωn = exp

2π i
n

,

it follows from property (5), Sec. 7. of eiθ that

(3)

ωnk = exp

2kπ i
n

(k = 0, 1, 2, . . . , n − 1)

and hence that

(4)

ck = c0ωnk (k = 0, 1, 2, . . . , n − 1).

The number c0 here can, of course, be replaced by any particular nth root of z0, since ωn represents a counterclockwise rotation through 2π/n radians.
Finally, a convenient way to remember expression (1) is to write z0 in its most general exponential form (compare with Example 2 in Sec. 6)

(5)

z0 = r0 ei(θ0+2kπ) (k = 0, ±1, ±2, . . .)

and to formally apply laws of fractional exponents involving real numbers, keeping in mind that there are precisely n roots:

z01/n =

r0 ei(θ0+2kπ)

1/n

=

√ n r0 exp

i(θ0 + 2kπ ) n

√ = n r0 exp i

θ0 + 2kπ nn

(k = 0, 1, 2, . . . , n − 1).

Brown-chap01-v3 10/29/07 3:32pm 27

sec. 10

Examples 27

The examples in the next section serve to illustrate this method for ﬁnding roots of complex numbers.

10. EXAMPLES
In each of the examples here, we start with expression (5), Sec. 9, and proceed in the manner described just after it.

EXAMPLE 1. Let us ﬁnd all values of (−8i)1/3, or the three cube roots of the number −8i. One need only write

−8i = 8 exp i − π + 2kπ 2

(k = 0, ±1, ±2, . . .)

to see that the desired roots are

(1)

ck = 2 exp

i

− π + 2kπ 63

(k = 0, 1, 2).

They lie at the vertices of an equilateral triangle, inscribed in the circle |z| = 2, and

are equally spaced around that circle every 2π/3 radians, starting with the principal

root (Fig. 12)

π

π

π√

c0 = 2 exp

i

− 6

= 2 cos − i sin = 3 − i.

6

6

Without any further calculations, it is then evident that c1 = 2i; and, since c2 is s√ymmetric to c0 with respect to the imaginary axis, we know that c2 = − 3 − i.
Note how it follows from expressions (2) and (4) in Sec. 9 that these roots can
be written

c0, c0ω3, c0ω32

where

ω3

= exp

2π i
3

.

y c1

2x

c2

c0

FIGURE 12

Brown-chap01-v3 10/29/07 3:32pm 28

28 Complex Numbers

chap. 1

EXAMPLE 2. In order to determine the nth roots of unity, we start with

1 = 1 exp[i(0 + 2kπ)] (k = 0, ±1, ±2 . . .)

and ﬁnd that

(2)

√ 11/n = n 1 exp i

0 + 2kπ

nn

= exp

2kπ i

n

(k = 0, 1, 2, . . . , n − 1).

When n = 2, these roots are, of course, ±1. When n ≥ 3, the regular polygon at whose vertices the roots lie is inscribed in the unit circle |z| = 1, with one vertex corresponding to the principal root z = 1 (k = 0). In view of expression (3), Sec. 9,
these roots are simply

1, ωn, ωn2, . . . , ωnn−1

where

2π ωn = exp i n

.

See Fig. 13, where the cases n = 3, 4, and 6 are illustrated. Note that ωnn = 1.

y

y

y

1x

1x

1x

FIGURE 13

√ EXAMPLE√3. The two values ck (k = 0, 1) of ( 3 + i)1/2, which are the square roots of 3 + i, are found by writing

√ 3 + i = 2 exp i

π + 2kπ

6

(k = 0, ±1, ±2, . . .)

and (see Fig. 14) (3)

√ ck = 2 exp i

π + kπ 12

(k = 0, 1).

Euler’s formula tells us that

√

π√

π

π

c0 =

2 exp i 12

=

2 cos + i sin ,

12

12

and the trigonometric identities

Brown-chap01-v3 10/29/07 3:32pm 29

sec. 10 y

Exercises 29

c1 = – c0

c0 √2 x

FIGURE 14

(4)

cos2

α

=

1

+ cos α ,

sin2 α = 1 − cos α

2

2

2

2

enable us to write

√

√

cos2 π = 1 1 + cos π

= 1 1+

3

= 2+

3 ,

12 2

62

2

4

√

√

sin2 π = 1 1 − cos π

= 1 1−

3

= 2−

3 .

12 2

62

2

4

Consequently,



c0

=

√ 2



√ 2+ 3 +i
4

√

2 − 3  = √1

4

2

√

√

2+ 3+i 2− 3 .

√ Since c1 = −c0, the two square roots of 3 + i are, then,

(5)

± √1

√

√

2+ 3+i 2− 3 .

2

EXERCISES
√ 1. Find the square roots of (a) 2i; (b) 1 − 3i and express them in rectangular coordi-

nates.

√

Ans. (a) ± (1 + i); (b) ± √3 − i .

2

2. In each case, ﬁnd all the roots in rectangular coordinates, exhibit them as vertices of

certain squares, and point out which is the principal root:

√

(a) (−16)1/4;

(b) (−8 − 8 3i)1/4.

√

√

√

√

Ans. (a) ± 2(1 + i), ± 2(1 − i); (b) ±( 3 − i), ±(1 + 3i).

Brown-chap01-v3 10/29/07 3:32pm 30

30 Complex Numbers

chap. 1

3. In each case, ﬁnd all the roots in rectangular coordinates, exhibit them as vertices of certain regular polygons, and identify the principal root:

(a) (−1)1/3;

(b) 81/6.

Ans.

(b)

√ ± 2,

√

±

1

+ √

3i ,

√

±

1

− √

3i .

2

2

4. According to Sec. 9, the three cube roots of a nonzero complex number z0 can be

written

c0, c0ω3, c0ω32

where

c0

is

the

principal

cube

root √

of

z0

and

ω3 = exp

2π i
3

= −1 + 2

3i .

√√

√

Show that if z0 = −4 2 + 4 2i, then c0 = 2(1 + i) and the other two cube roots

are, in rectangular form, the numbers

√

√

−( 3 + 1) + ( 3 − 1)i

c0ω3 =

√

,

2

√

√

c0ω32 = (

3 − 1) √− ( 2

3 + 1)i .

5. (a) Let a denote any ﬁxed real number and show that the two square roots of a + i

are

√ ± A exp

α i

√

2

where A = a2 + 1 and α = Arg(a + i).

(b) With the aid of the trigonometric identities (4) in Example 3 of Sec. 10, show that the square roots obtained in part (a) can be written

± √1

√

√

A+a+i A−a .

2

√

(Note that this becomes the ﬁnal result in Example 3, Sec. 10, when a = 3.)

6. Find the four zeros of the polynomial z4 + 4, one of them being √
z0 = 2 eiπ/4 = 1 + i.

Then use those zeros to factor z2 + 4 into quadratic factors with real coefﬁcients.

Ans. (z2 + 2z + 2)(z2 − 2z + 2).

7. Show that if c is any nth root of unity other than unity itself, then 1 + c + c2 + · · · + cn−1 = 0.

Suggestion: Use the ﬁrst identity in Exercise 9, Sec. 8. 8. (a) Prove that the usual formula solves the quadratic equation
az2 + bz + c = 0 (a = 0)

when the coefﬁcients a, b, and c are complex numbers. Speciﬁcally, by completing the square on the left-hand side, derive the quadratic formula

z

=

−b

+

(b2

−

4ac)1/2 ,

2a

where both square roots are to be considered when b2 − 4ac = 0,

Brown-chap01-v3 10/29/07 3:32pm 31

sec. 11

Regions in the Complex Plane 31

(b) Use the result in part (a) to ﬁnd the roots of the equation z2 + 2z + (1 − i) = 0.

Ans. (b) −1 + √1 + √i ,

2

2

−1 − √1 − √i .

2

2

9. Let z = reiθ be a nonzero complex number and n a negative integer (n = −1, −2, . . .). Then deﬁne z1/n by means of the equation z1/n = (z−1)1/m where m = −n. By showing that the m values of (z1/m)−1 and (z−1)1/m are the same, verify that z1/n = (z1/m)−1. (Compare with Exercise 7, Sec. 8.)

11. REGIONS IN THE COMPLEX PLANE

In this section, we are concerned with sets of complex numbers, or points in the z plane, and their closeness to one another. Our basic tool is the concept of an ε neighborhood

(1)

|z − z0| < ε

of a given point z0. It consists of all points z lying inside but not on a circle centered at z0 and with a speciﬁed positive radius ε (Fig. 15). When the value of ε is understood or is immaterial in the discussion, the set (1) is often referred to as just a neighborhood. Occasionally, it is convenient to speak of a deleted neighborhood, or punctured disk,

(2)

0 < |z − z0| < ε

consisting of all points z in an ε neighborhood of z0 except for the point z0 itself.

y
|z – z0| ε
z z0

O

x FIGURE 15

A point z0 is said to be an interior point of a set S whenever there is some neighborhood of z0 that contains only points of S; it is called an exterior point of S when there exists a neighborhood of it containing no points of S. If z0 is neither of these, it is a boundary point of S. A boundary point is, therefore, a point all of
whose neighborhoods contain at least one point in S and at least one point not in S. The totality of all boundary points is called the boundary of S. The circle |z| = 1, for instance, is the boundary of each of the sets

(3)

|z| < 1 and |z| ≤ 1.

Brown-chap01-v3 10/29/07 3:32pm 32

32 Complex Numbers

chap. 1

A set is open if it contains none of its boundary points. It is left as an exercise to show that a set is open if and only if each of its points is an interior point. A set is closed if it contains all of its boundary points, and the closure of a set S is the closed set consisting of all points in S together with the boundary of S. Note that the ﬁrst of the sets (3) is open and that the second is its closure.
Some sets are, of course, neither open nor closed. For a set to be not open, there must be a boundary point that is contained in the set; and if a set is not closed, there exists a boundary point not contained in the set. Observe that the punctured disk 0 < |z| ≤ 1 is neither open nor closed. The set of all complex numbers is, on the other hand, both open and closed since it has no boundary points.
An open set S is connected if each pair of points z1 and z2 in it can be joined by a polygonal line, consisting of a ﬁnite number of line segments joined end to end, that lies entirely in S. The open set |z| < 1 is connected. The annulus 1 < |z| < 2 is, of course, open and it is also connected (see Fig. 16). A nonempty open set that is connected is called a domain. Note that any neighborhood is a domain. A domain together with some, none, or all of its boundary points is referred to as a region.

y

O z1

z2 12 x

FIGURE 16
A set S is bounded if every point of S lies inside some circle |z| = R; otherwise, it is unbounded. Both of the sets (3) are bounded regions, and the half plane Re z ≥ 0 is unbounded.
A point z0 is said to be an accumulation point of a set S if each deleted neighborhood of z0 contains at least one point of S. It follows that if a set S is closed, then it contains each of its accumulation points. For if an accumulation point z0 were not in S, it would be a boundary point of S; but this contradicts the fact that a closed set contains all of its boundary points. It is left as an exercise to show that the converse is, in fact, true. Thus a set is closed if and only if it contains all of its accumulation points.
Evidently, a point z0 is not an accumulation point of a set S whenever there exists some deleted neighborhood of z0 that does not contain at least one point of S. Note that the origin is the only accumulation point of the set zn = i/n (n = 1, 2, . . .).

Brown-chap01-v3 10/29/07 3:32pm 33

sec. 11

Exercises 33

EXERCISES

1. Sketch the following sets and determine which are domains:

(a) |z − 2 + i| ≤ 1; (c) Im z > 1; (e) 0 ≤ arg z ≤ π/4 (z = 0);

(b) |2z + 3| > 4; (d) Im z = 1; (f) |z − 4| ≥ |z|.

Ans. (b), (c) are domains.

2. Which sets in Exercise 1 are neither open nor closed? Ans. (e).

3. Which sets in Exercise 1 are bounded? Ans. (a).

4. In each case, sketch the closure of the set: (a) −π < arg z < π (z = 0); (b) |Re z| < |z|;

1 (c) Re

≤

1 ;

z2

(d) Re(z2) > 0.

5. Let S be the open set consisting of all points z such that |z| < 1 or |z − 2| < 1. State why S is not connected.

6. Show that a set S is open if and only if each point in S is an interior point.

7. Determine the accumulation points of each of the following sets:

(a) zn = in (n = 1, 2, . . .); (c) 0 ≤ arg z < π/2 (z = 0);

(b) zn = in/n (n = 1, 2, . . .);

(d)

zn

=

(−1)n(1 + i)

n−1 n

(n =

1, 2, . . .).

Ans. (a) None; (b) 0; (d) ±(1 + i).

8. Prove that if a set contains each of its accumulation points, then it must be a closed set.

9. Show that any point z0 of a domain is an accumulation point of that domain.

10. Prove that a ﬁnite set of points z1, z2, . . . , zn cannot have any accumulation points.

Brown-chap01-v3 10/29/07 3:32pm 34

CHAPTER
2
ANALYTIC FUNCTIONS
We now consider functions of a complex variable and develop a theory of differentiation for them. The main goal of the chapter is to introduce analytic functions, which play a central role in complex analysis.
12. FUNCTIONS OF A COMPLEX VARIABLE
Let S be a set of complex numbers. A function f deﬁned on S is a rule that assigns to each z in S a complex number w. The number w is called the value of f at z and is denoted by f (z); that is, w = f (z). The set S is called the domain of deﬁnition of f .∗
It must be emphasized that both a domain of deﬁnition and a rule are needed in order for a function to be well deﬁned. When the domain of deﬁnition is not mentioned, we agree that the largest possible set is to be taken. Also, it is not always convenient to use notation that distinguishes between a given function and its values.
EXAMPLE 1. If f is deﬁned on the set z = 0 by means of the equation w = 1/z, it may be referred to only as the function w = 1/z, or simply the function 1/z.
Suppose that w = u + iv is the value of a function f at z = x + iy, so that u + iv = f (x + iy).
∗Although the domain of deﬁnition is often a domain as deﬁned in Sec. 11, it need not be.
35

36 Analytic Functions

chap. 2

Each of the real numbers u and v depends on the real variables x and y, and it follows that f (z) can be expressed in terms of a pair of real-valued functions of the real variables x and y:

(1)

f (z) = u(x, y) + iv(x, y).

If the polar coordinates r and θ , instead of x and y, are used, then u + iv = f (reiθ )

where w = u + iv and z = reiθ . In that case, we may write

(2)

f (z) = u(r, θ ) + iv(r, θ ).

EXAMPLE 2. If f (z) = z2, then f (x + iy) = (x + iy)2 = x2 − y2 + i2xy.

Hence

u(x, y) = x2 − y2 and v(x, y) = 2xy.

When polar coordinates are used, f (reiθ ) = (reiθ )2 = r2ei2θ = r2 cos 2θ + ir2 sin 2θ.

Consequently,

u(r, θ ) = r2 cos 2θ and v(r, θ ) = r2 sin 2θ.

If, in either of equations (1) and (2), the function v always has value zero, then the value of f is always real. That is, f is a real-valued function of a complex variable.
EXAMPLE 3. A real-valued function that is used to illustrate some important concepts later in this chapter is
f (z) = |z|2 = x2 + y2 + i0.

If n is zero or a positive integer and if a0, a1, a2, . . . , an are complex constants, where an = 0, the function
P (z) = a0 + a1z + a2z2 + · · · + anzn
is a polynomial of degree n. Note that the sum here has a ﬁnite number of terms and that the domain of deﬁnition is the entire z plane. Quotients P (z)/Q(z) of

sec. 12

Exercises 37

polynomials are called rational functions and are deﬁned at each point z where Q(z) = 0. Polynomials and rational functions constitute elementary, but important, classes of functions of a complex variable.
A generalization of the concept of function is a rule that assigns more than one value to a point z in the domain of deﬁnition. These multiple-valued functions occur in the theory of functions of a complex variable, just as they do in the case of a real variable. When multiple-valued functions are studied, usually just one of the possible values assigned to each point is taken, in a systematic manner, and a (single-valued) function is constructed from the multiple-valued function.

EXAMPLE 4. Let z denote any nonzero complex number. We know from Sec. 9 that z1/2 has the two values

z1/2

=

√ ±r

exp

i

,

2

where r = |z| and (−π < ≤ π√) is the principal value of arg z. But, if we choose only the positive value of ± r and write

√

(3)

f (z) = r exp i

(r > 0, −π < ≤ π),

2

the (single-valued) function (3) is well deﬁned on the set of nonzero numbers in the z plane. Since zero is the only square root of zero, we also write f (0) = 0. The function f is then well deﬁned on the entire plane.

EXERCISES

1. For each of the functions below, describe the domain of deﬁnition that is understood:

(a)

f (z)

=

z2

1 +

; 1

(b) f (z) = Arg 1 ; z

(c)

f (z)

=

z

z +

z

;

(d)

f (z)

=

1

1 − |z|2 .

Ans. (a) z = ±i; (c) Re z = 0.

2. Write the function f (z) = z3 + z + 1 in the form f (z) = u(x, y) + iv(x, y). Ans. f (z) = (x3 − 3xy2 + x + 1) + i(3x2y − y3 + y).

3. Suppose that f (z) = x2 − y2 − 2y + i(2x − 2xy), where z = x + iy. Use the expres-

sions (see Sec. 5)

x = z + z and y = z − z

2

2i

to write f (z) in terms of z, and simplify the result. Ans. f (z) = z2 + 2iz.

38 Analytic Functions

chap. 2

4. Write the function

f (z) = z + 1 z

(z = 0)

in the form f (z) = u(r, θ) + iv(r, θ).

Ans. f (z) = r + 1 cos θ + i r − 1 sin θ.

r

r

13. MAPPINGS
Properties of a real-valued function of a real variable are often exhibited by the graph of the function. But when w = f (z), where z and w are complex, no such convenient graphical representation of the function f is available because each of the numbers z and w is located in a plane rather than on a line. One can, however, display some information about the function by indicating pairs of corresponding points z = (x, y) and w = (u, v). To do this, it is generally simpler to draw the z and w planes separately.
When a function f is thought of in this way, it is often referred to as a mapping, or transformation. The image of a point z in the domain of deﬁnition S is the point w = f (z), and the set of images of all points in a set T that is contained in S is called the image of T . The image of the entire domain of deﬁnition S is called the range of f . The inverse image of a point w is the set of all points z in the domain of deﬁnition of f that have w as their image. The inverse image of a point may contain just one point, many points, or none at all. The last case occurs, of course, when w is not in the range of f .
Terms such as translation, rotation, and reﬂection are used to convey dominant geometric characteristics of certain mappings. In such cases, it is sometimes convenient to consider the z and w planes to be the same. For example, the mapping
w = z + 1 = (x + 1) + iy,
where z = x + iy, can be thought of as a translation of each point z one unit to the right. Since i = eiπ/2, the mapping
w = iz = r exp i θ + π , 2
where z = reiθ , rotates the radius vector for each nonzero point z through a right angle about the origin in the counterclockwise direction; and the mapping
w = z = x − iy
transforms each point z = x + iy into its reﬂection in the real axis. More information is usually exhibited by sketching images of curves and
regions than by simply indicating images of individual points. In the following three examples, we illustrate this with the transformation w = z2. We begin by ﬁnding the images of some curves in the z plane.

sec. 13

Mappings 39

EXAMPLE 1. According to Example 2 in Sec. 12, the mapping w = z2 can be thought of as the transformation

(1)

u = x2 − y2, v = 2xy

from the xy plane into the uv plane. This form of the mapping is especially useful in ﬁnding the images of certain hyperbolas.
It is easy to show, for instance, that each branch of a hyperbola

(2)

x2 − y2 = c1 (c1 > 0)

is mapped in a one to one manner onto the vertical line u = c1. We start by noting from the ﬁrst of equations (1) that u = c1 when (x, y) is a point lying on either branch. When, in particular, it lies on the right-hand branch, the second of equations (1) tells us that v = 2y y2 + c1. Thus the image of the right-hand branch can be expressed parametrically as

u = c1, v = 2y y2 + c1 (−∞ < y < ∞);
and it is evident that the image of a point (x, y) on that branch moves upward along the entire line as (x, y) traces out the branch in the upward direction (Fig. 17). Likewise, since the pair of equations

u = c1, v = −2y y2 + c1 (−∞ < y < ∞)
furnishes a parametric representation for the image of the left-hand branch of the hyperbola, the image of a point going downward along the entire left-hand branch is seen to move up the entire line u = c1.

y

v

u = c1 > 0

v = c2 > 0

O

x

O

u

FIGURE 17 w = z2.

On the other hand, each branch of a hyperbola

(3)

2xy = c2 (c2 > 0)

is transformed into the line v = c2, as indicated in Fig. 17. To verify this, we note from the second of equations (1) that v = c2 when (x, y) is a point on either

40 Analytic Functions

chap. 2

branch. Suppose that (x, y) is on the branch lying in the ﬁrst quadrant. Then, since

y = c2/(2x), the ﬁrst of equations (1) reveals that the branch’s image has parametric

representation

u

=

x2

−

c22 4x2

,

v = c2

(0 < x < ∞).

Observe that

lim u = −∞ and lim u = ∞.

x→0

x→∞

x>0

Since u depends continuously on x, then, it is clear that as (x, y) travels down the
entire upper branch of hyperbola (3), its image moves to the right along the entire horizontal line v = c2. Inasmuch as the image of the lower branch has parametric representation

u

=

c22 4y2

−

y2,

v = c2

(−∞ < y < 0)

and since

lim u = −∞ and lim u = ∞,

y→−∞

y→0

y<0

it follows that the image of a point moving upward along the entire lower branch also travels to the right along the entire line v = c2 (see Fig. 17).

We shall now use Example 1 to ﬁnd the image of a certain region.

EXAMPLE 2. The domain x > 0, y > 0, xy < 1 consists of all points lying on the upper branches of hyperbolas from the family 2xy = c, where 0 < c < 2 (Fig. 18). We know from Example 1 that as a point travels downward along the entirety of such a branch, its image under the transformation w = z2 moves to the right along the entire line v = c. Since, for all values of c between 0 and 2, these upper branches ﬁll out the domain x > 0, y > 0, xy < 1, that domain is mapped
onto the horizontal strip 0 < v < 2.

y

A

D

v

D′

2i

E′

E

FIGURE 18

B

C x A′

B′

C ′ u w = z2.

sec. 13

Mappings 41

In view of equations (1), the image of a point (0, y) in the z plane is (−y2, 0). Hence as (0, y) travels downward to the origin along the y axis, its image moves
to the right along the negative u axis and reaches the origin in the w plane. Then, since the image of a point (x, 0) is (x2, 0), that image moves to the right from the
origin along the u axis as (x, 0) moves to the right from the origin along the x axis. The image of the upper branch of the hyperbola xy = 1 is, of course, the horizontal line v = 2. Evidently, then, the closed region x ≥ 0, y ≥ 0, xy ≤ 1 is mapped onto the closed strip 0 ≤ v ≤ 2, as indicated in Fig. 18.

Our last example here illustrates how polar coordinates can be useful in analyzing certain mappings.

EXAMPLE 3. The mapping w = z2 becomes

(4)

w = r2ei2θ

when z = reiθ . Evidently, then, the image w = ρeiφ of any nonzero point z is found by squaring the modulus r = |z| and doubling the value θ of arg z that is used:

(5)

ρ = r2 and φ = 2θ.

Observe that points z = r0eiθ on a circle r = r0 are transformed into points w = r02ei2θ on the circle ρ = r02. As a point on the ﬁrst circle moves counterclockwise from the positive real axis to the positive imaginary axis, its image on the
second circle moves counterclockwise from the positive real axis to the negative
real axis (see Fig. 19). So, as all possible positive values of r0 are chosen, the corresponding arcs in the z and w planes ﬁll out the ﬁrst quadrant and the upper half plane, respectively. The transformation w = z2 is, then, a one to one mapping of the ﬁrst quadrant r ≥ 0, 0 ≤ θ ≤ π/2 in the z plane onto the upper half ρ ≥ 0, 0 ≤ φ ≤ π of the w plane, as indicated in Fig. 19. The point z = 0 is, of course, mapped onto the point w = 0.

y

v

O

r0 x

O

r

2 0

u

FIGURE 19 w = z2.

The transformation w = z2 also maps the upper half plane r ≥ 0, 0 ≤ θ ≤ π onto the entire w plane. However, in this case, the transformation is not one to one since both the positive and negative real axes in the z plane are mapped onto the positive real axis in the w plane.

42 Analytic Functions

chap. 2

When n is a positive integer greater than 2, various mapping properties of the transformation w = zn, or w = rneinθ , are similar to those of w = z2. Such a transformation maps the entire z plane onto the entire w plane, where each nonzero point in the w plane is the image of n distinct points in the z plane. The circle r = r0 is mapped onto the circle ρ = r0n; and the sector r ≤ r0, 0 ≤ θ ≤ 2π/n is mapped onto the disk ρ ≤ r0n, but not in a one to one manner.
Other, but somewhat more involved, mappings by w = z2 appear in Example 1, Sec. 97, and Exercises 1 through 4 of that section.

14. MAPPINGS BY THE EXPONENTIAL FUNCTION

In Chap. 3 we shall introduce and develop properties of a number of elementary functions which do not involve polynomials. That chapter will start with the exponential function

(1)

ez = exeiy (z = x + iy),

the two factors ex and eiy being well deﬁned at this time (see Sec. 6). Note that deﬁnition (1), which can also be written
ex+iy = ex eiy ,

is suggested by the familiar additive property ex1+x2 = ex1 ex2

of the exponential function in calculus. The object of this section is to use the function ez to provide the reader with
additional examples of mappings that continue to be reasonably simple. We begin by examining the images of vertical and horizontal lines.

EXAMPLE 1. The transformation

(2)

w = ez

can be written w = exeiy, where z = x + iy, according to equation (1). Thus, if w = ρeiφ, transformation (2) can be expressed in the form

(3)

ρ = ex, φ = y.

The image of a typical point z = (c1, y) on a vertical line x = c1 has polar coordinates ρ = exp c1 and φ = y in the w plane. That image moves counterclockwise around the circle shown in Fig. 20 as z moves up the line. The image of the
line is evidently the entire circle; and each point on the circle is the image of an
inﬁnite number of points, spaced 2π units apart, along the line.

sec. 14

Mappings by the Exponential Function 43

y x = c1
y = c2

O

x

v

c2

O

exp c1 u

FIGURE 20 w = exp z.

A horizontal line y = c2 is mapped in a one to one manner onto the ray φ = c2. To see that this is so, we note that the image of a point z = (x, c2) has polar coordinates ρ = ex and φ = c2. Consequently, as that point z moves along the entire line from left to right, its image moves outward along the entire ray φ = c2,
as indicated in Fig. 20.

Vertical and horizontal line segments are mapped onto portions of circles and rays, respectively, and images of various regions are readily obtained from observations made in Example 1. This is illustrated in the following example.
EXAMPLE 2. Let us show that the transformation w = ez maps the rectangular region a ≤ x ≤ b, c ≤ y ≤ d onto the region ea ≤ ρ ≤ eb, c ≤ φ ≤ d. The two regions and corresponding parts of their boundaries are indicated in Fig. 21. The vertical line segment AD is mapped onto the arc ρ = ea, c ≤ φ ≤ d, which is labeled A D . The images of vertical line segments to the right of AD and joining the horizontal parts of the boundary are larger arcs; eventually, the image of the line segment BC is the arc ρ = eb, c ≤ φ ≤ d, labeled B C . The mapping is one to one if d − c < 2π. In particular, if c = 0 and d = π, then 0 ≤ φ ≤ π; and the rectangular region is mapped onto half of a circular ring, as shown in Fig. 8, Appendix 2.

y

d

D

c A

O

a

FIGURE 21

w = exp z.

C

B

b

x

v C′

D′ O

B′ A′
u

44 Analytic Functions

chap. 2

Our ﬁnal example here uses the images of horizontal lines to ﬁnd the image of a horizontal strip.
EXAMPLE 3. When w = ez, the image of the inﬁnite strip 0 ≤ y ≤ π is the upper half v ≥ 0 of the w plane (Fig. 22). This is seen by recalling from Example 1 how a horizontal line y = c is transformed into a ray φ = c from the origin. As the real number c increases from c = 0 to c = π, the y intercepts of the lines increase from 0 to π and the angles of inclination of the rays increase from φ = 0 to φ = π. This mapping is also shown in Fig. 6 of Appendix 2, where corresponding points on the boundaries of the two regions are indicated.

y

v

i

ci

O

x

FIGURE 22

w = exp z.

O

u

EXERCISES
1. By referring to Example 1 in Sec. 13, ﬁnd a domain in the z plane whose image under the transformation w = z2 is the square domain in the w plane bounded by the lines u = 1, u = 2, v = 1, and v = 2. (See Fig. 2, Appendix 2.)
2. Find and sketch, showing corresponding orientations, the images of the hyperbolas
x2 − y2 = c1 (c1 < 0) and 2xy = c2 (c2 < 0)
under the transformation w = z2.
3. Sketch the region onto which the sector r ≤ 1, 0 ≤ θ ≤ π/4 is mapped by the transformation (a) w = z2; (b) w = z3; (c) w = z4.
4. Show that the lines ay = x (a = 0) are mapped onto the spirals ρ = exp(aφ) under the transformation w = exp z, where w = ρ exp(iφ).
5. By considering the images of horizontal line segments, verify that the image of the rectangular region a ≤ x ≤ b, c ≤ y ≤ d under the transformation w = exp z is the region ea ≤ ρ ≤ eb, c ≤ φ ≤ d, as shown in Fig. 21 (Sec. 14).
6. Verify the mapping of the region and boundary shown in Fig. 7 of Appendix 2, where the transformation is w = exp z.
7. Find the image of the semi-inﬁnite strip x ≥ 0, 0 ≤ y ≤ π under the transformation w = exp z, and label corresponding portions of the boundaries.

sec. 15

Limits 45

8. One interpretation of a function w = f (z) = u(x, y) + iv(x, y) is that of a vector ﬁeld in the domain of deﬁnition of f . The function assigns a vector w, with components u(x, y) and v(x, y), to each point z at which it is deﬁned. Indicate graphically the vector ﬁelds represented by (a) w = iz; (b) w = z/|z|.

15. LIMITS
Let a function f be deﬁned at all points z in some deleted neighborhood (Sec. l1) of z0. The statement that the limit of f (z) as z approaches z0 is a number w0, or that

(1)

lim
z→z0

f

(z)

=

w0,

means that the point w = f (z) can be made arbitrarily close to w0 if we choose the point z close enough to z0 but distinct from it. We now express the deﬁnition of limit in a precise and usable form.
Statement (1) means that for each positive number ε, there is a positive number δ such that

(2)

|f (z) − w0| < ε whenever 0 < |z − z0| < δ.

Geometrically, this deﬁnition says that for each ε neighborhood |w − w0| < ε of w0, there is a deleted δ neighborhood 0 < |z − z0| < δ of z0 such that every point z in it has an image w lying in the ε neighborhood (Fig. 23). Note that even though all points in the deleted neighborhood 0 < |z − z0| < δ are to be considered, their images need not ﬁll up the entire neighborhood |w − w0| < ε. If f has the constant value w0, for instance, the image of z is always the center of that neighborhood. Note, too, that once a δ has been found, it can be replaced by any smaller positive
number, such as δ/2.

y

z0 z

O

x

v
wε w0
Ou

FIGURE 23

It is easy to show that when a limit of a function f (z) exists at a point z0, it is unique. To do this, we suppose that

lim
z→z0

f

(z)

=

w0

and

lim
z→z0

f

(z)

=

w1.

Then, for each positive number ε, there are positive numbers δ0 and δ1 such that

|f (z) − w0| < ε whenever 0 < |z − z0| < δ0

46 Analytic Functions

chap. 2

and |f (z) − w1| < ε whenever 0 < |z − z0| < δ1.
So if 0 < |z − z0| < δ, where δ is any positive number that is smaller than δ0 and δ1, we ﬁnd that
|w1 − w0| = |[f (z) − w0] − [f (z) − w1]| ≤ |f (z) − w0| + |f (z) − w1| < ε + ε = 2ε.
But |w1 − w0| is a nonnegative constant, and ε can be chosen arbitrarily small. Hence
w1 − w0 = 0, or w1 = w0.
Deﬁnition (2) requires that f be deﬁned at all points in some deleted neighborhood of z0. Such a deleted neighborhood, of course, always exists when z0 is an interior point of a region on which f is deﬁned. We can extend the deﬁnition of limit to the case in which z0 is a boundary point of the region by agreeing that the ﬁrst of inequalities (2) need be satisﬁed by only those points z that lie in both the region and the deleted neighborhood.

EXAMPLE 1. Let us show that if f (z) = iz/2 in the open disk |z| < 1, then

(3)

lim f (z) =

i ,

z→1

2

the point 1 being on the boundary of the domain of deﬁnition of f . Observe that when z is in the disk |z| < 1,

f (z) − i

=

iz − i

=

|z

−

1| .

2 22

2

Hence, for any such z and each positive number ε (see Fig. 24),

f (z) − i < ε whenever 0 < |z − 1| < 2ε. 2

Thus condition (2) is satisﬁed by points in the region |z| < 1 when δ is equal to 2ε or any smaller positive number.

y

z δ = 2ε

O

1

x

v

–2i ε

f(z)

O

u

FIGURE 24

sec. 15

Limits 47

If limit (1) exists, the symbol z → z0 implies that z is allowed to approach z0 in an arbitrary manner, not just from some particular direction. The next example emphasizes this.

EXAMPLE 2. If (4) the limit

f (z) =

z ,

z

(5)

lim f (z)

z→0

does not exist. For, if it did exist, it could be found by letting the point z = (x, y)

approach the origin in any manner. But when z = (x, 0) is a nonzero point on the

real axis (Fig. 25),

f (z)

=

x x

+ −

i0 i0

=

1;

and when z = (0, y) is a nonzero point on the imaginary axis,

f (z)

=

0 + iy 0 − iy

= −1.

Thus, by letting z approach the origin along the real axis, we would ﬁnd that the desired limit is 1. An approach along the imaginary axis would, on the other hand, yield the limit −1. Since a limit is unique, we must conclude that limit (5) does not exist.

y z = (0, y)

(0, 0)

z = (x, 0)

x FIGURE 25

While deﬁnition (2) provides a means of testing whether a given point w0 is a limit, it does not directly provide a method for determining that limit. Theorems on limits, presented in the next section, will enable us to actually ﬁnd many limits.

48 Analytic Functions

chap. 2

16. THEOREMS ON LIMITS
We can expedite our treatment of limits by establishing a connection between limits of functions of a complex variable and limits of real-valued functions of two real variables. Since limits of the latter type are studied in calculus, we use their deﬁnition and properties freely.

Theorem 1. Suppose that

f (z) = u(x, y) + iv(x, y) (z = x + iy)

and z0 = x0 + iy0, w0 = u0 + iv0.

Then

(1) if and only if

lim
z→z0

f

(z)

=

w0

(2)

lim u(x, y) = u0 and

lim v(x, y) = v0.

(x,y)→(x0,y0)

(x,y)→(x0,y0)

To prove the theorem, we ﬁrst assume that limits (2) hold and obtain limit (1).

Limits (2) tell us that for each positive number ε, there exist positive numbers δ1 and δ2 such that

(3)

ε |u − u0| < 2

whenever

0<

(x − x0)2 + (y − y0)2 < δ1

and

(4)

|v

−

v0|

<

ε 2

whenever

0<

(x − x0)2 + (y − y0)2 < δ2.

Let δ be any positive number smaller than δ1 and δ2. Since

|(u + iv) − (u0 + iv0)| = |(u − u0) + i(v − v0)| ≤ |u − u0| + |v − v0|

and

(x − x0)2 + (y − y0)2 = |(x − x0) + i(y − y0)| = |(x + iy) − (x0 + iy0)|,

it follows from statements (3) and (4) that

|(u

+

iv)

−

(u0

+

iv0)|

<

ε 2

+

ε 2

=

ε

whenever

0 < |(x + iy) − (x0 + iy0)| < δ.

That is, limit (1) holds.

sec. 16

Theorems on Limits 49

Let us now start with the assumption that limit (1) holds. With that assumption, we know that for each positive number ε, there is a positive number δ such that

(5)

|(u + iv) − (u0 + iv0)| < ε

whenever

(6)

0 < |(x + iy) − (x0 + iy0)| < δ.

But

|u − u0| ≤ |(u − u0) + i(v − v0)| = |(u + iv) − (u0 + iv0)|, |v − v0| ≤ |(u − u0) + i(v − v0)| = |(u + iv) − (u0 + iv0)|,

and

|(x + iy) − (x0 + iy0)| = |(x − x0) + i(y − y0)| = (x − x0)2 + (y − y0)2.

Hence it follows from inequalities (5) and (6) that

|u − u0| < ε and |v − v0| < ε

whenever

0 < (x − x0)2 + (y − y0)2 < δ.

This establishes limits (2), and the proof of the theorem is complete.

Theorem 2. Suppose that

(7)
Then (8)

lim
z→z0

f

(z)

=

w0

and

lim
z→z0

F (z)

=

W0.

lim [f
z→z0

(z)

+

F

(z)]

=

w0

+

W0,

(9) and, if W0 = 0 , (10)

lim [f
z→z0

(z)F

(z)]

=

w0W0

;

lim f (z) = w0 . z→z0 F (z) W0

This important theorem can be proved directly by using the deﬁnition of the limit of a function of a complex variable. But, with the aid of Theorem 1, it follows almost immediately from theorems on limits of real-valued functions of two real variables.

50 Analytic Functions

chap. 2

To verify property (9), for example, we write

f (z) = u(x, y) + iv(x, y), F (z) = U (x, y) + iV (x, y),

z0 = x0 + iy0, w0 = u0 + iv0, W0 = U0 + iV0.

Then, according to hypotheses (7) and Theorem 1, the limits as (x, y) approaches (x0, y0) of the functions u, v, U , and V exist and have the values u0, v0, U0, and V0, respectively. So the real and imaginary components of the product
f (z)F (z) = (uU − vV ) + i(vU + uV )

have the limits u0U0 − v0V0 and v0U0 + u0V0, respectively, as (x, y) approaches (x0, y0). Hence, by Theorem 1 again, f (z)F (z) has the limit
(u0U0 − v0V0) + i(v0U0 + u0V0)

as z approaches z0 ; and this is equal to w0W0. Property (9) is thus established. Corresponding veriﬁcations of properties (8) and (10) can be given.
It is easy to see from deﬁnition (2), Sec. 15, of limit that

lim c = c
z→z0

and

lim
z→z0

z

=

z0,

where z0 and c are any complex numbers; and, by property (9) and mathematical induction, it follows that

lim
z→z0

zn

=

z0n

(n = 1, 2, . . .).

So, in view of properties (8) and (9), the limit of a polynomial

P (z) = a0 + a1z + a2z2 + · · · + anzn

as z approaches a point z0 is the value of the polynomial at that point:

(11)

lim
z→z0

P

(z)

=

P

(z0).

17. LIMITS INVOLVING THE POINT AT INFINITY
It is sometimes convenient to include with the complex plane the point at inﬁnity, denoted by ∞, and to use limits involving it. The complex plane together with this point is called the extended complex plane. To visualize the point at inﬁnity, one can think of the complex plane as passing through the equator of a unit sphere centered at the origin (Fig. 26). To each point z in the plane there corresponds exactly one point P on the surface of the sphere. The point P is the point where the line through z and the north pole N intersects the sphere. In like manner, to each point P on the surface of the sphere, other than the north pole N , there corresponds exactly one

sec. 17

Limits Involving the Point at Inﬁnity 51

N P z
O

FIGURE 26

point z in the plane. By letting the point N of the sphere correspond to the point at inﬁnity, we obtain a one to one correspondence between the points of the sphere and the points of the extended complex plane. The sphere is known as the Riemann sphere, and the correspondence is called a stereographic projection.
Observe that the exterior of the unit circle centered at the origin in the complex plane corresponds to the upper hemisphere with the equator and the point N deleted. Moreover, for each small positive number ε, those points in the complex plane exterior to the circle |z| = 1/ε correspond to points on the sphere close to N . We thus call the set |z| > 1/ε an ε neighborhood, or neighborhood, of ∞.
Let us agree that in referring to a point z, we mean a point in the ﬁnite plane. Hereafter, when the point at inﬁnity is to be considered, it will be speciﬁcally mentioned.
A meaning is now readily given to the statement

lim
z→z0

f

(z)

=

w0

when either z0 or w0, or possibly each of these numbers, is replaced by the point at inﬁnity. In the deﬁnition of limit in Sec. 15, we simply replace the appropriate neighborhoods of z0 and w0 by neighborhoods of ∞. The proof of the following theorem illustrates how this is done.

Theorem. If z0 and w0 are points in the z and w planes, respectively, then

(1)

lim f (z) = ∞ if and only if lim 1 = 0

z→z0

z→z0 f (z)

and

(2)

lim
z→∞

f

(z)

=

w0

if and only if

lim f
z→0

1 z

= w0.

Moreover,

(3)

lim f (z) = ∞

1 if and only if lim

= 0.

z→∞

z→0 f (1/z)

52 Analytic Functions

chap. 2

We start the proof by noting that the ﬁrst of limits (1) means that for each positive number ε, there is a positive number δ such that

(4)

|f (z)| > 1 ε

whenever

0 < |z − z0| < δ.

That is, the point w = f (z) lies in the ε neighborhood |w| > 1/ε of ∞ whenever

z lies in the deleted neighborhood 0 < |z − z0| < δ of z0. Since statement (4) can

be written

1 −0 <ε f (z)

whenever

0 < |z − z0| < δ,

the second of limits (1) follows. The ﬁrst of limits (2) means that for each positive number ε, a positive number
δ exists such that

(5)

|f (z) − w0| < ε

whenever

|z|

>

1 .

δ

Replacing z by 1/z in statement (5) and then writing the result as

f

1 z

− w0 < ε

whenever

0 < |z − 0| < δ,

we arrive at the second of limits (2). Finally, the ﬁrst of limits (3) is to be interpreted as saying that for each positive
number ε, there is a positive number δ such that

(6)

|f (z)| > 1

whenever

|z|

>

1 .

ε

δ

When z is replaced by 1/z, this statement can be put in the form

1 − 0 < ε whenever 0 < |z − 0| < δ; f (1/z)

and this gives us the second of limits (3).

EXAMPLES. Observe that lim iz + 3 = ∞ z→−1 z + 1

since

lim
z→−1

z+1 iz + 3

=

0

and

lim 2z + i = 2 since lim (2/z) + i = lim 2 + iz = 2.

z→∞ z + 1

z→0 (1/z) + 1 z→0 1 + z

Furthermore,

lim
z→∞

2z3 − 1 z2 + 1

=

∞

since

lim
z→0

(1/z2) + 1 (2/z3) − 1

=

lim
z→0

z + z3 2 − z3

=

0.

sec. 18

Continuity 53

18. CONTINUITY
A function f is continuous at a point z0 if all three of the following conditions are satisﬁed:

(1)

lim f (z) exists,

z→z0

(2)

f (z0) exists,

(3)

lim
z→z0

f

(z)

=

f

(z0).

Observe that statement (3) actually contains statements (1) and (2), since the existence of the quantity on each side of the equation there is needed. Statement (3) says, of course, that for each positive number ε, there is a positive number δ such that

(4)

|f (z) − f (z0)| < ε whenever |z − z0| < δ.

A function of a complex variable is said to be continuous in a region R if it is continuous at each point in R.
If two functions are continuous at a point, their sum and product are also continuous at that point; their quotient is continuous at any such point if the denominator is not zero there. These observations are direct consequences of Theorem 2, Sec. 16. Note, too, that a polynomial is continuous in the entire plane because of limit (11) in Sec. 16.
We turn now to two expected properties of continuous functions whose veriﬁcations are not so immediate. Our proofs depend on deﬁnition (4) of continuity, and we present the results as theorems.

Theorem 1. A composition of continuous functions is itself continuous.

A precise statement of this theorem is contained in the proof to follow. We let w = f (z) be a function that is deﬁned for all z in a neighborhood |z − z0| < δ of a point z0 , and we let W = g(w) be a function whose domain of deﬁnition contains the image (Sec. 13) of that neighborhood under f . The composition W = g[f (z)] is, then, deﬁned for all z in the neighborhood |z − z0| < δ. Suppose now that f is continuous at z0 and that g is continuous at the point f (z0) in the w plane. In view of the continuity of g at f (z0), there is, for each positive number ε, a positive number γ such that
|g[f (z)] − g[f (z0)]| < ε whenever |f (z) − f (z0)| < γ .
(See Fig. 27.) But the continuity of f at z0 ensures that the neighborhood |z − z0| < δ can be made small enough that the second of these inequalities holds. The continuity of the composition g[f (z)] is, therefore, established.

54 Analytic Functions

y

v

chap. 2 V

z z0

g[ f(z)] ε

f (z0)

g[ f(z0)]

O xO

u

O

U

f (z)

FIGURE 27

Theorem 2. If a function f (z) is continuous and nonzero at a point z0 , then f (z) = 0 throughout some neighborhood of that point.

Assuming that f (z) is, in fact, continuous and nonzero at z0, we can prove Theorem 2 by assigning the positive value |f (z0)|/2 to the number ε in statement
(4). This tells us that there is a positive number δ such that

|f (z)

−

f (z0)|

<

|f (z0)| 2

whenever

|z − z0| < δ.

So if there is a point z in the neighborhood |z − z0| < δ at which f (z) = 0, we have the contradiction

|f (z0)|

<

|f (z0)| ; 2

and the theorem is proved.

The continuity of a function

(5)

f (z) = u(x, y) + iv(x, y)

is closely related to the continuity of its component functions u(x, y) and v(x, y). We note, for instance, how it follows from Theorem 1 in Sec. 16 that the function (5) is continuous at a point z0 = (x0, y0) if and only if its component functions are continuous there. Our proof of the next theorem illustrates the use of this statement. The theorem is extremely important and will be used often in later chapters, especially in applications. Before stating the theorem, we recall from Sec. 11 that a region R is closed if it contains all of its boundary points and that it is bounded if it lies inside some circle centered at the origin.

Theorem 3. If a function f is continuous throughout a region R that is both closed and bounded, there exists a nonnegative real number M such that

(6)

|f (z)| ≤ M for all points z inR,

where equality holds for at least one such z.

sec. 18

Exercises 55

To prove this, we assume that the function f in equation (5) is continuous and note how it follows that the function
[u(x, y)]2 + [v(x, y)]2
is continuous throughout R and thus reaches a maximum value M somewhere in R.∗ Inequality (6) thus holds, and we say that f is bounded on R.

EXERCISES

1. Use deﬁnition (2), Sec. 15, of limit to prove that

(a)

lim
z→z0

Re

z

=

Re

z0

;

(b)

lim z
z→z0

=

z0

;

(c) lim z2 = 0. z→0 z

2. Let a, b, and c denote complex constants. Then use deﬁnition (2), Sec. 15, of limit to

show that

(a)

lim (az
z→z0

+

b)

=

az0

+

b;

(b)

lim (z2
z→z0

+

c)

=

z02

+

c;

(c) lim [x + i(2x + y)] = 1 + i (z = x + iy).
z→1−i

3. Let n be a positive integer and let P (z) and Q(z) be polynomials, where Q(z0) = 0. Use Theorem 2 in Sec. 16, as well as limits appearing in that section, to ﬁnd

(a)

lim
z→z0

1 zn

(z0

= 0);

iz3 − 1

(b) lim
z→i

z+i

;

P (z)

(c) lim

.

z→z0 Q(z)

Ans. (a) 1/z0n; (b) 0; (c) P (z0)/Q(z0).

4. Use mathematical induction and property (9), Sec. 16, of limits to show that

lim
z→z0

zn

=

z0n

when n is a positive integer (n = 1, 2, . . .).

5. Show that the limit of the function

z2 f (z) =
z

as z tends to 0 does not exist. Do this by letting nonzero points z = (x, 0) and z = (x, x) approach the origin. [Note that it is not sufﬁcient to simply consider points z = (x, 0) and z = (0, y), as it was in Example 2, Sec. 15.]
6. Prove statement (8) in Theorem 2 of Sec. 16 using
(a) Theorem 1 in Sec. 16 and properties of limits of real-valued functions of two real variables;
(b) deﬁnition (2), Sec. 15, of limit.

∗See, for instance, A. E. Taylor and W. R. Mann, “Advanced Calculus,” 3d ed., pp. 125–126 and p. 529, 1983.

56 Analytic Functions

chap. 2

7. Use deﬁnition (2), Sec. 15, of limit to prove that

if

lim
z→z0

f

(z)

=

w0,

then

lim
z→z0

|f

(z)|

=

|w0|.

Suggestion: Observe how the ﬁrst of inequalities (9), Sec. 4, enables one to write

||f (z)| − |w0|| ≤ |f (z) − w0|.

8. Write z = z − z0 and show that

lim
z→z0

f

(z)

=

w0

if and only if

lim
z→0

f

(z0

+

z) = w0.

9. Show that

lim f (z)g(z) = 0 if lim f (z) = 0

z→z0

z→z0

and if there exists a positive number M such that |g(z)| ≤ M for all z in some neighborhood of z0.

10. Use the theorem in Sec. 17 to show that

4z2

(a)

lim
z→∞

(z

−

1)2

=

4;

1

(b)

lim
z→1

(z

−

1)3

= ∞;

z2 + 1

(c)

lim
z→∞

z

−

1

= ∞.

11. With the aid of the theorem in Sec. 17, show that when

T

(z)

=

az + b cz + d

(ad − bc = 0),

(a) lim T (z) = ∞ if c = 0;

z→∞

(b) lim T (z) = a and lim T (z) = ∞

z→∞

c

z→−d /c

if c = 0.

12. State why limits involving the point at inﬁnity are unique.

13. Show that a set S is unbounded (Sec. 11) if and only if every neighborhood of the point at inﬁnity contains at least one point in S.

19. DERIVATIVES

Let f be a function whose domain of deﬁnition contains a neighborhood |z − z0| < ε of a point z0. The derivative of f at z0 is the limit

(1)

f

(z0)

=

lim
z→z0

f (z) − z−

f (z0) , z0

and the function f is said to be differentiable at z0 when f (z0) exists. By expressing the variable z in deﬁnition (1) in terms of the new complex
variable
z = z − z0 (z = z0),

sec. 19

Derivatives 57

one can write that deﬁnition as

(2)

f (z0) = lim f (z0 +
z→0

z) − f (z0) . z

Because f is deﬁned throughout a neighborhood of z0, the number f (z0 + z) is always deﬁned for | z| sufﬁciently small (Fig. 28).

y ε

z0 O

x FIGURE 28

When taking form (2) of the deﬁnition of derivative, we often drop the subscript on z0 and introduce the number

w = f (z + z) − f (z),

which denotes the change in the value w = f (z) of f corresponding to a change z in the point at which f is evaluated. Then, if we write dw/dz for f (z), equation (2) becomes

dw

w

(3)

= lim .

dz z→0 z

EXAMPLE 1. Suppose that f (z) = z2. At any point z,

lim w = lim (z + z)2 − z2 = lim (2z + z) = 2z

z→0 z

z→0

z

z→0

since 2z + z is a polynomial in z. Hence dw/dz = 2z, or f (z) = 2z.

EXAMPLE 2. If f (z) = z, then

(4)

w = z+

z−z = z+

z−z =

z .

z

z

z

z

58 Analytic Functions

chap. 2

If the limit of w/ z exists, it can be found by letting the point z = ( x, y) approach the origin (0, 0) in the z plane in any manner. In par-
ticular, as z approaches (0, 0) horizontally through the points ( x, 0) on the real axis (Fig. 29),

z = x + i0 = x − i0 = x + i0 = z.

In that case, expression (4) tells us that
w = z = 1. zz Hence if the limit of w/ z exists, its value must be unity. However, when z approaches (0, 0) vertically through the points (0, y) on the imaginary axis, so that

z = 0 + i y = 0 − i y = −(0 + i y) = − z,

we ﬁnd from expression (4) that

w = − z = −1.

z

z

Hence the limit must be −1 if it exists. Since limits are unique (Sec. 15), it follows that dw/dz does not exist anywhere.

(0, 0)

FIGURE 29

EXAMPLE 3. Consider the real-valued function f (z) = |z|2. Here

(5)

w = |z +

z|2 − |z|2 = (z +

z)(z +

z) − zz = z +

z+z

z .

z

z

z

z

Proceeding as in Example 2, where horizontal and vertical approaches of z toward the origin gave us

z = z and z = − z,

sec. 19

Derivatives 59

respectively, we have the expressions
w = z + z + z when z = ( x, 0) z and w = z − z − z when z = (0, y). z Hence if the limit of w/ z exists as z tends to zero, the uniqueness of limits, used in Example 2, tells us that
z + z = z − z,
or z = 0. Evidently, then dw/dz cannot exist when z = 0. To show that dw/dz does, in fact, exist at z = 0, we need only observe that
expression (5) reduces to
w= z z when z = 0. We conclude, therefore, that dw/dz exists only at z = 0, its value there being 0.

Example 3 shows that a function f (z) = u(x, y) + iv(x, y) can be differentiable at a point z = (x, y) but nowhere else in any neighborhood of that point. Since

(6)

u(x, y) = x2 + y2 and v(x, y) = 0

when f (z) = |z|2, it also shows that the real and imaginary components of a function of a complex variable can have continuous partial derivatives of all orders at a point z = (x, y) and yet the function may not be differentiable there.
The function f (z) = |z|2 is continuous at each point in the plane since its components (6) are continuous at each point. So the continuity of a function at a point does not imply the existence of a derivative there. It is, however, true that the existence of the derivative of a function at a point implies the continuity of the function at that point. To see this, we assume that f (z0) exists and write

lim [f (z) −
z→z0

f (z0)]

=

lim
z→z0

f (z) z

− f (z0) − z0

lim (z −
z→z0

z0)

=

f

(z0) ·

0

=

0,

from which it follows that

lim
z→z0

f

(z)

=

f

(z0).

This is the statement of continuity of f at z0 (Sec. 18). Geometric interpretations of derivatives of functions of a complex variable are
not as immediate as they are for derivatives of functions of a real variable. We defer the development of such interpretations until Chap. 9.

60 Analytic Functions

chap. 2

20. DIFFERENTIATION FORMULAS

The deﬁnition of derivative in Sec. 19 is identical in form to that of the derivative of a real-valued function of a real variable. In fact, the basic differentiation formulas given below can be derived from the deﬁnition in Sec. 19 by essentially the same steps as the ones used in calculus. In these formulas, the derivative of a function f at a point z is denoted by either

d f (z) or f (z),
dz

depending on which notation is more convenient. Let c be a complex constant, and let f be a function whose derivative exists
at a point z. It is easy to show that

(1)

d c = 0, d z = 1, d [cf (z)] = cf (z).

dz

dz

dz

Also, if n is a positive integer,

(2)

d zn = nzn−1.

dz

This formula remains valid when n is a negative integer, provided that z = 0. If the derivatives of two functions f and g exist at a point z, then

(3)

d [f (z) + g(z)] = f (z) + g (z),

dz

(4)

d [f (z)g(z)] = f (z)g (z) + f (z)g(z) ;

dz

and, when g(z) = 0,

(5)

d dz

f (z) g(z)

=

g(z)f

(z) − f (z)g [g(z)]2

(z) .

Let us derive formula (4). To do this, we write the following expression for the change in the product w = f (z)g(z):

w = f (z + z)g(z + z) − f (z)g(z) = f (z)[g(z + z) − g(z)] + [f (z + z) − f (z)]g(z + z).

Thus

w = f (z) g(z + z) − g(z) + f (z + z) − f (z) g(z + z) ;

z

z

z

and, letting z tend to zero, we arrive at the desired formula for the derivative of f (z)g(z). Here we have used the fact that g is continuous at the point z, since

sec. 20

Differentiation Formulas 61

g (z) exists; thus g(z + z) tends to g(z) as z tends to zero (see Exercise 8, Sec. 18).
There is also a chain rule for differentiating composite functions. Suppose that f has a derivative at z0 and that g has a derivative at the point f (z0). Then the function F (z) = g[f (z)] has a derivative at z0, and

(6)

F (z0) = g [f (z0)]f (z0).

If we write w = f (z) and W = g(w), so that W = F (z), the chain rule becomes

dW

=

dW

dw .

dz dw dz

EXAMPLE. W = w5. Then

To ﬁnd the derivative of (2z2 + i)5, write w = 2z2 + i and d (2z2 + i)5 = 5w44z = 20z(2z2 + i)4. dz

To start the derivation of formula (6), choose a speciﬁc point z0 at which f (z0) exists. Write w0 = f (z0) and also assume that g (w0) exists. There is, then, some ε neighborhood |w − w0| < ε of w0 such that for all points w in that neighborhood, we can deﬁne a function having the values (w0) = 0 and

(7)

(w)

=

g(w) w

− −

g(w0) w0

−

g

(w0)

when

w = w0.

Note that in view of the deﬁnition of derivative,

(8)

lim (w) = 0.

w→w0

Hence is continuous at w0. Now expression (7) can be put in the form

(9)

g(w) − g(w0) = [g (w0) + (w)](w − w0) (|w − w0| < ε),

which is valid even when w = w0; and since f (z0) exists and f is therefore
continuous at z0, we can choose a positive number δ such that the point f (z) lies in the ε neighborhood |w − w0| < ε of w0 if z lies in the δ neighborhood |z − z0| < δ of z0. Thus it is legitimate to replace the variable w in equation (9) by f (z) when z is any point in the neighborhood |z − z0| < δ. With that substitution, and with w0 = f (z0), equation (9) becomes

(10)

g[f (z)] − g[f (z0)] z − z0

=

{g

[f (z0)] +

[f

(z)]}

f

(z) z

− −

f (z0) z0

(0 < |z − z0| < δ),

62 Analytic Functions

chap. 2

where we must stipulate that z = z0 so that we are not dividing by zero. As already noted, f is continuous at z0 and is continuous at the point w0 = f (z0). Hence the composition [f (z)] is continuous at z0; and since (w0) = 0,
lim [f (z)] = 0.
z→z0
So equation (10) becomes equation (6) in the limit as z approaches z0.

EXERCISES

1. Use results in Sec. 20 to ﬁnd f (z) when

(a) f (z) = 3z2 − 2z + 4;

(b) f (z) = (1 − 4z2)3 ;

(c)

f (z)

=

z−1 2z + 1

(z = −1/2);

(d)

f (z) =

(1 + z2)4 z2

(z = 0).

2. Using results in Sec. 20, show that (a) a polynomial

P (z) = a0 + a1z + a2z2 + · · · + anzn (an = 0)

of degree n (n ≥ 1) is differentiable everywhere, with derivative

P (z) = a1 + 2a2z + · · · + nanzn−1 ;

(b) the coefﬁcients in the polynomial P (z) in part (a) can be written

a0 = P (0),

a1

=

P (0) ,
1!

a2

=

P

(0) ,
2!

...,

an

=

P (n)(0) .
n!

3. Apply deﬁnition (3), Sec. 19, of derivative to give a direct proof that

dw dz

=

−

1 z2

when

w= 1 z

(z = 0).

4. Suppose that f (z0) = g(z0) = 0 and that f (z0) and g (z0) exist, where g (z0) = 0. Use deﬁnition (1), Sec. 19, of derivative to show that

lim f (z) = f (z0) . z→z0 g(z) g (z0)
5. Derive formula (3), Sec. 20, for the derivative of the sum of two functions. 6. Derive expression (2), Sec. 20, for the derivative of zn when n is a positive integer
by using
(a) mathematical induction and formula (4), Sec. 20, for the derivative of the product of two functions;
(b) deﬁnition (3), Sec. 19, of derivative and the binomial formula (Sec. 3).

sec. 21

Cauchy–Riemann Equations 63

7. Prove that expression (2), Sec. 20, for the derivative of zn remains valid when n is a negative integer (n = −1, −2, . . .), provided that z = 0. Suggestion: Write m = −n and use the formula for the derivative of a quotient of two functions.
8. Use the method in Example 2, Sec. 19, to show that f (z) does not exist at any point z when (a) f (z) = Re z; (b) f (z) = Im z.
9. Let f denote the function whose values are
z2/z when z = 0, f (z) =
0 when z = 0.
Show that if z = 0, then w/ z = 1 at each nonzero point on the real and imaginary axes in the z, or x y, plane. Then show that w/ z = −1 at each nonzero point ( x, x) on the line y = x in that plane. Conclude from these observations that f (0) does not exist. Note that to obtain this result, it is not sufﬁcient to consider only horizontal and vertical approaches to the origin in the z plane. (Compare with Example 2, Sec. 19.)

21. CAUCHY–RIEMANN EQUATIONS
In this section, we obtain a pair of equations that the ﬁrst-order partial derivatives of the component functions u and v of a function

(1)

f (z) = u(x, y) + iv(x, y)

must satisfy at a point z0 = (x0, y0) when the derivative of f exists there. We also show how to express f (z0) in terms of those partial derivatives.
We start by writing

z0 = x0 + iy0, z = x + i y,

and

w = f (z0 + z) − f (z0) = [u(x0 + x, y0 + y) − u(x0, y0)] + i[v(x0 + x, y0 + y) − v(x0, y0)].

Assuming that the derivative

(2)

f (z0) =

lim
z→0

w z

exists, we know from Theorem 1 in Sec. 16 that

(3)

f

(z0) =
(

lim
x, y)→(0,0)

Re

w z

+ i lim

Im

( x, y)→(0,0)

w z

.

64 Analytic Functions

chap. 2

Now it is important to keep in mind that expression (3) is valid as ( x, y) tends to (0, 0) in any manner that we may choose. In particular, we let ( x, y) tend to (0, 0) horizontally through the points ( x, 0), as indicated in Fig. 29 (Sec. 19). Inasmuch as y = 0, the quotient w/ z becomes

w = u(x0 + x, y0) − u(x0, y0) + i v(x0 + x, y0) − v(x0, y0) .

z

x

x

Thus

w

lim

Re

( x, y)→(0,0)

z

= lim u(x0 +
x→0

x, y0) x

−

u(x0, y0)

=

ux (x0, y0)

and

w

lim

Im

( x, y)→(0,0)

z

= lim v(x0 +
x→0

x, y0) x

−

v(x0, y0)

=

vx (x0,

y0),

where ux(x0, y0) and vx(x0, y0) denote the ﬁrst-order partial derivatives with respect to x of the functions u and v, respectively, at (x0, y0). Substitution of these limits into expression (3) tells us that

(4)

f (z0) = ux(x0, y0) + ivx (x0, y0).

We might have let z tend to zero vertically through the points (0, y). In that case, x = 0 and

w = u(x0, y0 + y) − u(x0, y0) + i v(x0, y0 + y) − v(x0, y0)

z

iy

iy

= v(x0, y0 + y) − v(x0, y0) − i u(x0, y0 + y) − u(x0, y0) .

y

y

Evidently, then,

lim

Re

( x, y)→(0,0)

w z

= lim v(x0, y0 +
y→0

y) y

−

v(x0, y0)

=

vy(x0, y0)

and

lim

Im

( x, y)→(0,0)

w z

= − lim u(x0, y0 +
y→0

y) y

−

u(x0, y0)

=

−uy (x0,

y0).

Hence it follows from expression (3) that

(5)

f (z0) = vy(x0, y0) − iuy (x0, y0),

where the partial derivatives of u and v are, this time, with respect to y. Note that equation (5) can also be written in the form

f (z0) = −i[uy (x0, y0) + ivy (x0, y0)].

sec. 21

Cauchy–Riemann Equations 65

Equations (4) and (5) not only give f (z0) in terms of partial derivatives of the component functions u and v, but they also provide necessary conditions for the existence of f (z0). To obtain those conditions, we need only equate the real parts and then the imaginary parts on the right-hand sides of equations (4) and (5) to see that the existence of f (z0) requires that

(6)

ux (x0, y0) = vy(x0, y0) and uy (x0, y0) = −vx(x0, y0).

Equations (6) are the Cauchy–Riemann equations, so named in honor of the French mathematician A. L. Cauchy (1789–1857), who discovered and used them, and in honor of the German mathematician G. F. B. Riemann (1826–1866), who made them fundamental in his development of the theory of functions of a complex variable.
We summarize the above results as follows.

Theorem. Suppose that

f (z) = u(x, y) + iv(x, y)

and that f (z) exists at a point z0 = x0 + iy0. Then the ﬁrst-order partial derivatives of u and v must exist at (x0, y0), and they must satisfy the Cauchy–Riemann equations

(7)

ux = vy, uy = −vx

there. Also, f (z0) can be written

(8)

f (z0) = ux + ivx ,

where these partial derivatives are to be evaluated at (x0, y0).

EXAMPLE 1. In Example 1, Sec. 19, we showed that the function f (z) = z2 = x2 − y2 + i2xy

is differentiable everywhere and that f (z) = 2z. To verify that the Cauchy–Riemann equations are satisﬁed everywhere, write
u(x, y) = x2 − y2 and v(x, y) = 2xy.

Thus

ux = 2x = vy, uy = −2y = −vx.

Moreover, according to equation (8),

f (z) = 2x + i2y = 2(x + iy) = 2z.

66 Analytic Functions

chap. 2

Since the Cauchy–Riemann equations are necessary conditions for the existence of the derivative of a function f at a point z0, they can often be used to locate points at which f does not have a derivative.
EXAMPLE 2. When f (z) = |z|2, we have
u(x, y) = x2 + y2 and v(x, y) = 0.
If the Cauchy–Riemann equations are to hold at a point (x, y), it follows that 2x = 0 and 2y = 0, or that x = y = 0. Consequently, f (z) does not exist at any nonzero point, as we already know from Example 3 in Sec. 19. Note that the theorem just proved does not ensure the existence of f (0). The theorem in the next section will, however, do this.

22. SUFFICIENT CONDITIONS FOR DIFFERENTIABILITY
Satisfaction of the Cauchy–Riemann equations at a point z0 = (x0, y0) is not sufﬁcient to ensure the existence of the derivative of a function f (z) at that point. (See Exercise 6, Sec. 23.) But, with certain continuity conditions, we have the following useful theorem.
Theorem. Let the function
f (z) = u(x, y) + iv(x, y)
be deﬁned throughout some ε neighborhood of a point z0 = x0 + iy0, and suppose that (a) the ﬁrst-order partial derivatives of the functions u and v with respect to x and
y exist everywhere in the neighborhood; (b) those partial derivatives are continuous at (x0, y0) and satisfy the Cauchy–
Riemann equations
ux = vy, uy = −vx
at (x0, y0). Then f (z0) exists, its value being
f (z0) = ux + ivx
where the right-hand side is to be evaluated at (x0, y0).
To prove the theorem, we assume that conditions (a) and (b) in its hypothesis are satisﬁed and write z = x + i y, where 0 < | z| < ε, as well as
w = f (z0 + z) − f (z0).

sec. 22

Sufﬁcient Conditions for Differentiability 67

Thus

(1)

w = u + i v,

where

u = u(x0 + x, y0 + y) − u(x0, y0)

and v = v(x0 + x, y0 + y) − v(x0, y0).

The assumption that the ﬁrst-order partial derivatives of u and v are continuous at the point (x0, y0) enables us to write∗

(2)

u = ux(x0, y0) x + uy(x0, y0) y + ε1 x + ε2 y

and

(3)

v = vx (x0, y0) x + vy(x0, y0) y + ε3 x + ε4 y,

where ε1, ε2, ε3, and ε4 tend to zero as ( x, y) approaches (0, 0) in the z plane. Substitution of expressions (2) and (3) into equation (1) now tells us that

(4)

w = ux (x0, y0) x + uy (x0, y0) y + ε1 x + ε2 y

+ i[vx (x0, y0) x + vy (x0, y0) y + ε3 x + ε4 y].

Because the Cauchy–Riemann equations are assumed to be satisﬁed at (x0, y0), one can replace uy (x0, y0) by −vx(x0, y0) and vy(x0, y0) by ux(x0, y0) in equation (4) and then divide through by the quantity z = x + i y to get

(5)

w z

=

ux (x0, y0)

+ ivx (x0, y0) + (ε1

+ iε3)

x z

+

(ε2

+

iε4)

y .
z

But | x| ≤ | z| and | y| ≤ | z|, according to inequalities (3) in Sec. 4, and so

x ≤ 1 and z

y ≤ 1. z

Consequently,

(ε1 + iε3)

x z

≤ |ε1 + iε3| ≤ |ε1| + |ε3|

and

(ε2 + iε4)

y z

≤ |ε2 + iε4| ≤ |ε2| + |ε4|;

∗See, for instance, W. Kaplan, “Advanced Calculus,” 5th ed., pp. 86ff, 2003.

68 Analytic Functions

chap. 2

and this means that the last two terms on the right in equation (5) tend to zero as the variable z = x + i y approaches zero. The expression for f (z0) in the statement of the theorem is now established.

EXAMPLE 1. Consider the exponential function
f (z) = ez = ex eiy (z = x + iy),
some of whose mapping properties were discussed in Sec. 14. In view of Euler’s formula (Sec. 6), this function can, of course, be written
f (z) = ex cos y + iex sin y,
where y is to be taken in radians when cos y and sin y are evaluated. Then
u(x, y) = ex cos y and v(x, y) = ex sin y.
Since ux = vy and uy = −vx everywhere and since these derivatives are everywhere continuous, the conditions in the above theorem are satisﬁed at all points in the complex plane. Thus f (z) exists everywhere, and
f (z) = ux + ivx = ex cos y + iex sin y.
Note that f (z) = f (z) for all z.
EXAMPLE 2. It also follows from our theorem that the function f (z) = |z|2, whose components are
u(x, y) = x2 + y2 and v(x, y) = 0,
has a derivative at z = 0. In fact, f (0) = 0 + i0 = 0. We saw in Example 2, Sec. 21, that this function cannot have a derivative at any nonzero point since the Cauchy–Riemann equations are not satisﬁed at such points. (See also Example 3, Sec. 19.)

23. POLAR COORDINATES

Assuming that z0 = 0, we shall in this section use the coordinate transformation

(1)

x = r cos θ, y = r sin θ

to restate the theorem in Sec. 22 in polar coordinates. Depending on whether we write
z = x + iy or z = reiθ (z = 0)

sec. 23

Polar Coordinates 69

when w = f (z), the real and imaginary components of w = u + iv are expressed

in terms of either the variables x and y or r and θ . Suppose that the ﬁrst-order

partial derivatives of u and v with respect to x and y exist everywhere in some

neighborhood of a given nonzero point z0 and are continuous at z0. The ﬁrst-order

partial derivatives of u and v with respect to r and θ also have those properties,

and the chain rule for differentiating real-valued functions of two real variables can

be used to write them in terms of the ones with respect to x and y. More precisely,

since

∂u

=

∂u

∂x

+

∂u

∂y ,

∂u

=

∂u

∂x

+

∂u

∂y ,

∂r ∂x ∂r ∂y ∂r ∂θ ∂x ∂θ ∂y ∂θ

one can write

(2)

ur = ux cos θ + uy sin θ, uθ = −ux r sin θ + uy r cos θ.

Likewise,

(3)

vr = vx cos θ + vy sin θ, vθ = −vx r sin θ + vy r cos θ.

If the partial derivatives of u and v with respect to x and y also satisfy the Cauchy–Riemann equations

(4)

ux = vy, uy = −vx

at z0, equations (3) become

(5)

vr = −uy cos θ + ux sin θ, vθ = uy r sin θ + ux r cos θ

at that point. It is then clear from equations (2) and (5) that

(6)

rur = vθ , uθ = −rvr

at z0. If, on the other hand, equations (6) are known to hold at z0, it is straightforward
to show (Exercise 7) that equations (4) must hold there. Equations (6) are, therefore, an alternative form of the Cauchy–Riemann equations (4).
In view of equations (6) and the expression for f (z0) that is found in Exercise 8, we are now able to restate the theorem in Sec. 22 using r and θ.

Theorem. Let the function
f (z) = u(r, θ ) + iv(r, θ )
be deﬁned throughout some ε neighborhood of a nonzero point z0 = r0 exp(iθ0), and suppose that (a) the ﬁrst-order partial derivatives of the functions u and v with respect to r and
θ exist everywhere in the neighborhood;

70 Analytic Functions

chap. 2

(b) those partial derivatives are continuous at (r0, θ0) and satisfy the polar form rur = vθ , uθ = −rvr
of the Cauchy–Riemann equations at (r0, θ0). Then f (z0) exists, its value being
f (z0) = e−iθ (ur + ivr ), where the right-hand side is to be evaluated at (r0, θ0).

EXAMPLE 1. Consider the function

f (z) =

1 z

=

1 r eiθ

=

1 e−iθ r

=

1 (cos θ
r

− i sin θ )

(z = 0).

Since

u(r, θ ) = cos θ and v(r, θ ) = − sin θ ,

r

r

the conditions in this theorem are satisﬁed at every nonzero point z = reiθ in the plane. In particular, the Cauchy–Riemann equations

r ur

= − cos θ r

= vθ

and

uθ

= − sin θ r

= −rvr

are satisﬁed. Hence the derivative of f exists when z = 0; and, according to the theorem,

f (z) = e−iθ

−

cos r2

θ

+

i

sin θ r2

= −e−iθ

e−iθ r2

=

−

(r

1 eiθ

)2

=

−

1 z2

.

EXAMPLE 2. The theorem can be used to show that when α is a ﬁxed real

number, the function

f

(z)

=

√ 3r

eiθ

/3

(r > 0, α < θ < α + 2π)

has a derivative everywhere in its domain of deﬁnition. Here

u(r,

θ)

=

√ 3r

cos

θ

and

v(r,

θ)

=

√ 3r

sin

θ

.

3

3

Inasmush as

√

√

rur =

3r 3

cos

θ 3

=

vθ

and

uθ

=

−

3r 3

sin

θ 3

=

−r vr

sec. 23

Exercises 71

and since the other conditions in the theorem are satisﬁed, the derivative f (z) exists at each point where f (z) is deﬁned. The theorem tells us, moreover, that

f (z) = e−iθ

√1

θ cos

+i

√1

θ sin ,

3( 3 r)2 3 3( 3 r)2 3

or

f (z) = e√−iθ eiθ/3 = √ 1

=

1 .

3( 3 r)2

3( 3 reiθ/3)2 3[f (z)]2

Note that when a speciﬁc point z is taken in the domain of deﬁnition of f, the value f (z) is one value of z1/3 (see Sec. 9). Hence this last expression for f (z) can
be put in the form

d z1/3 dz

=

1 3(z1/3)2

when that value is taken. Derivatives of such power functions will be elaborated on in Chap. 3 (Sec. 33).

EXERCISES

1. Use the theorem in Sec. 21 to show that f (z) does not exist at any point if

(a) f (z) = z ; (c) f (z) = 2x + ixy2 ;

(b) f (z) = z − z ; (d) f (z) = ex e−iy .

2. Use the theorem in Sec. 22 to show that f (z) and its derivative f (z) exist every-

where, and ﬁnd f (z) when

(a) f (z) = iz + 2; (c) f (z) = z3;

(b) f (z) = e−x e−iy ; (d) f (z) = cos x cosh y − i sin x sinh y.

Ans. (b) f (z) = f (z); (d) f (z) = −f (z).

3. From results obtained in Secs. 21 and 22, determine where f (z) exists and ﬁnd its value when
(a) f (z) = 1/z; (b) f (z) = x2 + iy2; (c) f (z) = z Im z.

Ans. (a) f (z) = −1/z2 (z = 0); (b) f (x + ix) = 2x; (c) f (0) = 0.

4. Use the theorem in Sec. 23 to show that each of these functions is differentiable in the indicated domain of deﬁnition, and also to ﬁnd f (z):

(a) f (z) = 1/z4 (z = 0);

(b)

f

(z)

=

√ r

eiθ

/2

(r > 0, α < θ < α + 2π );

(c) f (z) = e−θ cos(ln r) + ie−θ sin(ln r) (r > 0, 0 < θ < 2π ).

Ans. (b) f (z) =

1

;

(c)

f

(z)

=

i

f (z) .

2f (z)

z

72 Analytic Functions

chap. 2

5. Show that when f (z) = x3 + i(1 − y)3, it is legitimate to write

f (z) = ux + ivx = 3x2

only when z = i.
6. Let u and v denote the real and imaginary components of the function f deﬁned by means of the equations

f (z) = z2/z when z = 0, 0 when z = 0.
Verify that the Cauchy–Riemann equations ux = vy and uy = −vx are satisﬁed at the origin z = (0, 0). [Compare with Exercise 9, Sec. 20, where it is shown that f (0) nevertheless fails to exist.]
7. Solve equations (2), Sec. 23 for ux and uy to show that

ux

=

ur

cos θ

− uθ

sin θ ,
r

uy

=

ur

sin θ

+ uθ

cos θ .
r

Then use these equations and similar ones for vx and vy to show that in Sec. 23 equations (4) are satisﬁed at a point z0 if equations (6) are satisﬁed there. Thus complete the veriﬁcation that equations (6), Sec. 23, are the Cauchy–Riemann equations in polar form.
8. Let a function f (z) = u + iv be differentiable at a nonzero point z0 = r0 exp(iθ0). Use the expressions for ux and vx found in Exercise 7, together with the polar form (6), Sec. 23, of the Cauchy–Riemann equations, to rewrite the expression

f (z0) = ux + ivx

in Sec. 22 as

f (z0) = e−iθ (ur + ivr ),

where ur and vr are to be evaluated at (r0, θ0). 9. (a) With the aid of the polar form (6), Sec. 23, of the Cauchy–Riemann equations,
derive the alternative form
−i f (z0) = z0 (uθ + ivθ )

of the expression for f (z0) found in Exercise 8. (b) Use the expression for f (z0) in part (a) to show that the derivative of the function
f (z) = 1/z (z = 0) in Example 1, Sec. 23, is f (z) = −1/z2.
10. (a) Recall (Sec. 5) that if z = x + iy, then

x = z+z

and

y

=

z

−

z .

2

2i

sec. 24

Analytic Functions 73

By formally applying the chain rule in calculus to a function F (x, y) of two real variables, derive the expression

∂F = ∂F ∂x + ∂F ∂y = 1 ∂F + i ∂F . ∂z ∂x ∂z ∂y ∂z 2 ∂x ∂y

(b) Deﬁne the operator

∂ = 1 ∂ +i ∂ , ∂z 2 ∂x ∂y

suggested by part (a), to show that if the ﬁrst-order partial derivatives of the real and imaginary components of a function f (z) = u(x, y) + iv(x, y) satisfy the Cauchy–Riemann equations, then

∂f 1 ∂z = 2 [(ux − vy) + i(vx + uy )] = 0. Thus derive the complex form ∂f/∂z = 0 of the Cauchy–Riemann equations.

24. ANALYTIC FUNCTIONS
We are now ready to introduce the concept of an analytic function. A function f of the complex variable z is analytic at a point z0 if it has a derivative at each point in some neighborhood of z0.∗ It follows that if f is analytic at a point z0, it must be analytic at each point in some neighborhood of z0. A function f is analytic in an open set if it has a derivative everywhere in that set. If we should speak of a function f that is analytic in a set S which is not open, it is to be understood that f is analytic in an open set containing S.
Note that the function f (z) = 1/z is analytic at each nonzero point in the ﬁnite plane. But the function f (z) = |z|2 is not analytic at any point since its derivative exists only at z = 0 and not throughout any neighborhood. (See Example 3, Sec. 19.)
An entire function is a function that is analytic at each point in the entire ﬁnite plane. Since the derivative of a polynomial exists everywhere, it follows that every polynomial is an entire function.
If a function f fails to be analytic at a point z0 but is analytic at some point in every neighborhood of z0, then z0 is called a singular point, or singularity, of f . The point z = 0 is evidently a singular point of the function f (z) = 1/z. The function f (z) = |z|2, on the other hand, has no singular points since it is nowhere analytic.
A necessary, but by no means sufﬁcient, condition for a function f to be analytic in a domain D is clearly the continuity of f throughout D. Satisfaction of the Cauchy–Riemann equations is also necessary, but not sufﬁcient. Sufﬁcient conditions for analyticity in D are provided by the theorems in Secs. 22 and 23.
Other useful sufﬁcient conditions are obtained from the differentiation formulas in Sec. 20. The derivatives of the sum and product of two functions exist wherever
∗The terms regular and holomorphic are also used in the literature to denote analyticity.

74 Analytic Functions

chap. 2

the functions themselves have derivatives. Thus, if two functions are analytic in a domain D, their sum and their product are both analytic in D. Similarly, their quotient is analytic in D provided the function in the denominator does not vanish at any point in D. In particular, the quotient P (z)/Q(z) of two polynomials is analytic in any domain throughout which Q(z) = 0.
From the chain rule for the derivative of a composite function, we ﬁnd that a composition of two analytic functions is analytic. More precisely, suppose that a function f (z) is analytic in a domain D and that the image (Sec. 13) of D under the transformation w = f (z) is contained in the domain of deﬁnition of a function g(w). Then the composition g[f (z)] is analytic in D, with derivative
d g[f (z)] = g [f (z)]f (z). dz
The following property of analytic functions is especially useful, in addition to being expected.

Theorem. If f (z) = 0 everywhere in a domain D, then f (z) must be constant throughout D.

We start the proof by writing f (z) = u(x, y) + iv(x, y). Assuming that f (z) = 0 in D, we note that ux + ivx = 0 ; and, in view of the Cauchy–Riemann equations, vy − iuy = 0. Consequently,

ux = uy = 0 and vx = vy = 0

at each point in D. Next, we show that u(x, y) is constant along any line segment L extending
from a point P to a point P and lying entirely in D. We let s denote the distance along L from the point P and let U denote the unit vector along L in the direction of increasing s (see Fig. 30). We know from calculus that the directional derivative du/ds can be written as the dot product

(1)

du = (grad u) · U,

ds

y

U P O

L

s P′

D Q

x FIGURE 30

sec. 25

Examples 75

where grad u is the gradient vector

(2)

grad u = uxi + uy j.

Because ux and uy are zero everywhere in D, grad u is evidently the zero vector at all points on L. Hence it follows from equation (1) that the derivative du/ds is zero along L; and this means that u is constant on L.
Finally, since there is always a ﬁnite number of such line segments, joined end to end, connecting any two points P and Q in D (Sec. 11), the values of u at P and Q must be the same. We may conclude, then, that there is a real constant a such that u(x, y) = a throughout D. Similarly, v(x, y) = b ; and we ﬁnd that f (z) = a + bi at each point in D.

25. EXAMPLES
As pointed out in Sec. 24, it is often possible to determine where a given function is analytic by simply recalling various differentiation formulas in Sec. 20.

EXAMPLE 1. The quotient

f (z)

=

(z2

z3 + 4 − 3)(z2 +

1)

is evid√ently analytic throughout the z plane except for the singular points z = ± 3 and z = ± i. The analyticity is due to the existence of familiar differentiation formulas, which need to be applied only if the expression for f (z) is wanted.

When a function is given in terms of its component functions u(x, y) and v(x, y), its analyticity can be demonstrated by direct application of the Cauchy– Riemann equations.

EXAMPLE 2. If f (z) = cosh x cos y + i sinh x sin y,
the component functions are u(x, y) = cosh x cos y and v(x, y) = sinh x sin y.
Because ux = sinh x cos y = vy and uy = − cosh x sin y = −vx
everywhere, it is clear from the theorem in Sec. 22 that f is entire.

76 Analytic Functions

chap. 2

Finally, we illustrate how the theorem in Sec. 24 can be used to obtain other properties of analytic functions.

EXAMPLE 3. Suppose that a function

f (z) = u(x, y) + iv(x, y)

and its conjugate

f (z) = u(x, y) − iv(x, y)

are both analytic in a given domain D. It is now easy to show that f (z) must be constant throughout D.
To do this, we write f (z) as

f (z) = U (x, y) + iV (x, y)

where

(1)

U (x, y) = u(x, y) and V (x, y) = −v(x, y).

Because of the analyticity of f (z), the Cauchy–Riemann equations

(2)

ux = vy, uy = −vx

hold in D; and the analyticity of f (z) in D tells us that

(3)

Ux = Vy, Uy = −Vx .

In view of relations (1), equations (3) can also be written

(4)

ux = −vy, uy = vx .

By adding corresponding sides of the ﬁrst of equations (2) and (4), we ﬁnd that ux = 0 in D. Similarly, subtraction involving corresponding sides of the second of equations (2) and (4) reveals that vx = 0. According to expression (8) in Sec. 21, then,
f (z) = ux + ivx = 0 + i0 = 0 ;

and it follows from the theorem in Sec. 24 that f (z) is constant throughout D.

EXAMPLE 4. As in Example 3, we consider a function f that is analytic throughout a given domain D. Assuming further that the modulus |f (z)| is constant throughout D, one can prove that f (z) must be constant there too. This result is needed to obtain an important result later on in Chap. 4 (Sec. 54).
The proof is accomplished by writing

(5)

|f (z)| = c for all z in D,

sec. 25

Exercises 77

where c is a real constant. If c = 0, it follows that f (z) = 0 everywhere in D. If c = 0, the fact that (see Sec. 5)
f (z)f (z) = c2
tells us that f (z) is never zero in D. Hence f (z) = c2 for all z in D, f (z)
and it follows from this that f (z) is analytic everywhere in D. The main result in Example 3 just above thus ensures that f (z) is constant throughout D.

EXERCISES

1. Apply the theorem in Sec. 22 to verify that each of these functions is entire:

(a) f (z) = 3x + y + i(3y − x);

(b) f (z) = sin x cosh y + i cos x sinh y;

(c) f (z) = e−y sin x − ie−y cos x;

(d) f (z) = (z2 − 2)e−x e−iy .

2. With the aid of the theorem in Sec. 21, show that each of these functions is nowhere analytic:
(a) f (z) = xy + iy; (b) f (z) = 2xy + i(x2 − y2); (c) f (z) = eyeix .

3. State why a composition of two entire functions is entire. Also, state why any linear combination c1f1(z) + c2f2(z) of two entire functions, where c1 and c2 are complex constants, is entire.

4. In each case, determine the singular points of the function and state why the function

is analytic everywhere except at those points:

(a)

f

(z)

=

2z + 1 z(z2 + 1) ;

(b)

f

(z)

=

z2

z3 −

+ 3z

i +

2

;

(c)

f

(z)

=

(z

+

z2 + 1 2)(z2 + 2z

+

2)

.

Ans. (a) z = 0, ± i; (b) z = 1, 2 ; (c) z = −2, −1 ± i.

5. According to Exercise 4(b), Sec. 23, the function

g(z)

=

√ r

ei

θ

/2

(r > 0, −π < θ < π )

is analytic in its domain of deﬁnition, with derivative

g (z) =

1 .

2 g(z)

Show that the composite function G(z) = g(2z − 2 + i) is analytic in the half plane

x > 1, with derivative

G

(z)

=

g(2z

1 −2

+

. i)

Suggestion: Observe that Re(2z − 2 + i) > 0 when x > 1.

78 Analytic Functions

chap. 2

6. Use results in Sec. 23 to verify that the function

g(z) = ln r + iθ (r > 0, 0 < θ < 2π )

is analytic in the indicated domain of deﬁnition, with derivative g (z) = 1/z. Then

show that the composite function G(z) = g(z2 + 1) is analytic in the quadrant

x > 0, y > 0, with derivative

G

(z)

=

2z z2 +

. 1

Suggestion: Observe that Im(z2 + 1) > 0 when x > 0, y > 0.
7. Let a function f be analytic everywhere in a domain D. Prove that if f (z) is realvalued for all z in D, then f (z) must be constant throughtout D.

26. HARMONIC FUNCTIONS
A real-valued function H of two real variables x and y is said to be harmonic in a given domain of the xy plane if, throughout that domain, it has continuous partial derivatives of the ﬁrst and second order and satisﬁes the partial differential equation

(1)

Hxx (x, y) + Hyy (x, y) = 0,

known as Laplace’s equation. Harmonic functions play an important role in applied mathematics. For
example, the temperatures T (x, y) in thin plates lying in the xy plane are often harmonic. A function V (x, y) is harmonic when it denotes an electrostatic potential that varies only with x and y in the interior of a region of three-dimensional space that is free of charges.

EXAMPLE 1. It is easy to verify that the function T (x, y) = e−y sin x is harmonic in any domain of the xy plane and, in particular, in the semi-inﬁnite vertical strip 0 < x < π, y > 0. It also assumes the values on the edges of the strip that are indicated in Fig. 31. More precisely, it satisﬁes all of the conditions

y

T = 0 Txx + Tyy = 0 T = 0

O T = sin x

x FIGURE 31

sec. 26

Harmonic Functions 79

Txx(x, y) + Tyy (x, y) = 0,
T (0, y) = 0, T (π, y) = 0,
T (x, 0) = sin x, lim T (x, y) = 0,
y→∞
which describe steady temperatures T (x, y) in a thin homogeneous plate in the xy plane that has no heat sources or sinks and is insulated except for the stated conditions along the edges.

The use of the theory of functions of a complex variable in discovering solutions, such as the one in Example 1, of temperature and other problems is described in considerable detail later on in Chap. 10 and in parts of chapters following it.∗ That theory is based on the theorem below, which provides a source of harmonic functions.
Theorem 1. If a function f (z) = u(x, y) + iv(x, y) is analytic in a domain D, then its component functions u and v are harmonic in D.

To show this, we need a result that is to be proved in Chap. 4 (Sec. 52). Namely, if a function of a complex variable is analytic at a point, then its real and imaginary components have continuous partial derivatives of all orders at that point.
Assuming that f is analytic in D, we start with the observation that the ﬁrstorder partial derivatives of its component functions must satisfy the Cauchy–Riemann equations throughout D:

(2)

ux = vy , uy = −vx.

Differentiating both sides of these equations with respect to x, we have

(3)

uxx = vyx , uyx = −vxx .

Likewise, differentiation with respect to y yields

(4)

uxy = vyy , uyy = −vxy .

Now, by a theorem in advanced calculus,† the continuity of the partial derivatives of u and v ensures that uyx = uxy and vyx = vxy. It then follows from equations (3) and (4) that
uxx + uyy = 0 and vxx + vyy = 0.

That is, u and v are harmonic in D.

∗Another important method is developed in the authors’ “Fourier Series and Boundary Value Prob-
lems,” 7th ed., 2008. †See, for instance, A. E. Taylor and W. R. Mann, “Advanced Calculus,” 3d ed., pp. 199–201, 1983.

80 Analytic Functions

chap. 2

EXAMPLE 2. The function f (z) = e−y sin x − ie−y cos x is entire, as is
shown in Exercise 1 (c), Sec. 25. Hence its real component, which is the temperature function T (x, y) = e−y sin x in Example 1, must be harmonic in every domain of
the xy plane.

EXAMPLE 3. Since the function f (z) = i/z2 is analytic whenever z = 0

and since

i z2

=

i z2

·

z2 z2

=

iz2 (zz)2

=

iz2 |z|4

=

2xy + i(x2 − y2) ,
(x2 + y2)2

the two functions

u(x, y)

=

2xy (x2 + y2)2

and

v(x, y)

=

x2 − y2 (x2 + y2)2

are harmonic throughout any domain in the xy plane that does not contain the origin.

If two given functions u and v are harmonic in a domain D and their ﬁrst-order partial derivatives satisfy the Cauchy–Riemann equations (2) throughout D, then v is said to be a harmonic conjugate of u. The meaning of the word conjugate here is, of course, different from that in Sec. 5, where z is deﬁned.

Theorem 2. A function f (z) = u(x, y) + iv(x, y) is analytic in a domain D if and only if v is a harmonic conjugate of u.

The proof is easy. If v is a harmonic conjugate of u in D, the theorem in Sec. 22 tells us that f is analytic in D. Conversely, if f is analytic in D, we know from Theorem 1 that u and v are harmonic in D ; furthermore, in view of the theorem in Sec. 21, the Cauchy–Riemann equations are satisﬁed.
The following example shows that if v is a harmonic conjugate of u in some domain, it is not, in general, true that u is a harmonic conjugate of v there. (See also Exercises 3 and 4.)

EXAMPLE 4. Suppose that
u(x, y) = x2 − y2 and v(x, y) = 2xy.
Since these are the real and imaginary components, respectively, of the entire function f (z) = z2, we know that v is a harmonic conjugate of u throughout the plane. But u cannot be a harmonic conjugate of v since, as veriﬁed in Exercise 2(b), Sec. 25, the function 2xy + i(x2 − y2) is not analytic anywhere.

In Chap. 9 (Sec. 104) we shall show that a function u which is harmonic in a domain of a certain type always has a harmonic conjugate. Thus, in such domains, every harmonic function is the real part of an analytic function. It is also true (Exercise 2) that a harmonic conjugate, when it exists, is unique except for an additive constant.

sec. 26

Exercises 81

EXAMPLE 5. We now illustrate one method of obtaining a harmonic conjugate of a given harmonic function. The function

(5)

u(x, y) = y3 − 3x2y

is readily seen to be harmonic throughout the entire xy plane. Since a harmonic conjugate v(x, y) is related to u(x, y) by means of the Cauchy–Riemann equations

(6)

ux = vy , uy = −vx,

the ﬁrst of these equations tells us that

vy(x, y) = −6xy.

Holding x ﬁxed and integrating each side here with respect to y, we ﬁnd that

(7)

v(x, y) = −3xy2 + φ(x)

where φ is, at present, an arbitrary function of x. Using the second of equations (6),
we have 3y2 − 3x2 = 3y2 − φ (x),

or φ (x) = 3x2. Thus φ(x) = x3 + C, where C is an arbitrary real number. According to equation (7), then, the function

(8)

v(x, y) = −3xy2 + x3 + C

is a harmonic conjugate of u(x, y). The corresponding analytic function is

(9)

f (z) = (y3 − 3x2y) + i(−3xy2 + x3 + C).

The form f (z) = i(z3 + C) of this function is easily veriﬁed and is suggested by noting that when y = 0, expression (9) becomes f (x) = i(x3 + C).

EXERCISES

1. Show that u(x, y) is harmonic in some domain and ﬁnd a harmonic conjugate v(x, y)

when

(a) u(x, y) = 2x(1 − y);

(b) u(x, y) = 2x − x3 + 3xy2;

(c) u(x, y) = sinh x sin y; (d) u(x, y) = y/(x2 + y2).

Ans. (a) v(x, y) = x2 − y2 + 2y; (c) v(x, y) = − cosh x cos y;

(b) v(x, y) = 2y − 3x2y + y3; (d) v(x, y) = x/(x2 + y2).

2. Show that if v and V are harmonic conjugates of u(x, y) in a domain D, then v(x, y) and V (x, y) can differ at most by an additive constant.

82 Analytic Functions

chap. 2

3. Suppose that v is a harmonic conjugate of u in a domain D and also that u is a harmonic conjugate of v in D. Show how it follows that both u(x, y) and v(x, y) must be constant throughout D.

4. Use Theorem 2 in Sec. 26 to show that v is a harmonic conjugate of u in a domain D if and only if −u is a harmonic conjugate of v in D. (Compare with the result obtained in Exercise 3.) Suggestion: Observe that the function f (z) = u(x, y) + iv(x, y) is analytic in D if and only if −if (z) is analytic there.
5. Let the function f (z) = u(r, θ) + iv(r, θ) be analytic in a domain D that does not include the origin. Using the Cauchy–Riemann equations in polar coordinates (Sec. 23) and assuming continuity of partial derivatives, show that throughout D the function u(r, θ) satisﬁes the partial differential equation

r2urr (r, θ ) + rur (r, θ ) + uθθ (r, θ ) = 0,

which is the polar form of Laplace’s equation. Show that the same is true of the function v(r, θ).

6. Verify that the function u(r, θ) = ln r is harmonic in the domain r > 0, 0 < θ < 2π by showing that it satisﬁes the polar form of Laplace’s equation, obtained in Exercise 5. Then use the technique in Example 5, Sec. 26, but involving the Cauchy–Riemann equations in polar form (Sec. 23), to derive the harmonic conjugate v(r, θ) = θ. (Compare with Exercise 6, Sec. 25.)

7. Let the function f (z) = u(x, y) + iv(x, y) be analytic in a domain D, and consider the families of level curves u(x, y) = c1 and v(x, y) = c2, where c1 and c2 are arbitrary real constants. Prove that these families are orthogonal. More precisely, show that if z0 = (x0, y0) is a point in D which is common to two particular curves u(x, y) = c1 and v(x, y) = c2 and if f (z0) = 0, then the lines tangent to those curves at (x0, y0) are perpendicular. Suggestion: Note how it follows from the pair of equations u(x, y) = c1 and v(x, y) = c2 that

∂u + ∂u dy = 0 and ∂v + ∂v dy = 0.

∂x ∂y dx

∂x ∂y dx

8. Show that when f (z) = z2, the level curves u(x, y) = c1 and v(x, y) = c2 of the component functions are the hyperbolas indicated in Fig. 32. Note the orthogonality of the two families, described in Exercise 7. Observe that the curves u(x, y) = 0 and v(x, y) = 0 intersect at the origin but are not, however, orthogonal to each other. Why
is this fact in agreement with the result in Exercise 7?

9. Sketch the families of level curves of the component functions u and v when f (z) = 1/z, and note the orthogonality described in Exercise 7.

10. Do Exercise 9 using polar coordinates.

11. Sketch the families of level curves of the component functions u and v when

f

(z)

=

z

−

1 ,

z+1

and note how the result in Exercise 7 is illustrated here.

sec. 27 c 1 =0

Uniquely Determined Analytic Functions 83

y

= < 0

= 0 c1

c1

> 0 c1

c2 > 0
x c2 < 0

FIGURE 32
27. UNIQUELY DETERMINED ANALYTIC FUNCTIONS
We conclude this chapter with two sections dealing with how the values of an analytic function in a domain D are affected by its values in a subdomain of D or on a line segment lying in D. While these sections are of considerable theoretical interest, they are not central to our development of analytic functions in later chapters. The reader may pass directly to Chap. 3 at this time and refer back when necessary.
Lemma. Suppose that (a) a function f is analytic throughout a domain D; (b) f (z) = 0 at each point z of a domain or line segment contained in D. Then f (z) ≡ 0 in D; that is, f (z) is identically equal to zero throughout D.
To prove this lemma, we let f be as stated in its hypothesis and let z0 be any point of the subdomain or line segment where f (z) = 0. Since D is a connected open set (Sec. 11), there is a polygonal line L, consisting of a ﬁnite number of line segments joined end to end and lying entirely in D, that extends from z0 to any other point P in D. We let d be the shortest distance from points on L to the boundary of D, unless D is the entire plane; in that case, d may be any positive number. We then form a ﬁnite sequence of points
z0, z1, z2, . . . , zn−1, zn
along L, where the point zn coincides with P (Fig. 33) and where each point is sufﬁciently close to adjacent ones that
|zk − zk−1| < d (k = 1, 2, . . . , n).

84 Analytic Functions

chap. 2

N0 N1 N2 z0 z1 z2

L

Nn – 1 Nn

zn – 1

P zn

FIGURE 33

Finally, we construct a ﬁnite sequence of neighborhoods
N0, N1, N2, . . . , Nn−1, Nn,
where each neighborhood Nk is centered at zk and has radius d. Note that these neighborhoods are all contained in D and that the center zk of any neighborhood Nk (k = 1, 2, . . . , n) lies in the preceding neighborhood Nk−1.
At this point, we need to use a result that is proved later on in Chap. 6. Namely, Theorem 3 in Sec. 75 tells us that since f is analytic in N0 and since f (z) = 0 in a domain or on a line segment containing z0, then f (z) ≡ 0 in N0. But the point z1 lies in N0. Hence a second application of the same theorem reveals that f (z) ≡ 0 in N1; and, by continuing in this manner, we arrive at the fact that f (z) ≡ 0 in Nn. Since Nn is centered at the point P and since P was arbitrarily selected in D, we may conclude that f (z) ≡ 0 in D. This completes the proof of the lemma.
Suppose now that two functions f and g are analytic in the same domain D and that f (z) = g(z) at each point z of some domain or line segment contained in D. The difference
h(z) = f (z) − g(z)
is also analytic in D, and h(z) = 0 throughout the subdomain or along the line segment. According to the lemma, then, h(z) ≡ 0 throught D ; that is, f (z) = g(z) at each point z in D. We thus arrive at the following important theorem.

Theorem. A function that is analytic in a domain D is uniquely determined over D by its values in a domain, or along a line segment, contained in D.

This theorem is useful in studying the question of extending the domain of
deﬁnition of an analytic function. More precisely, given two domains D1 and D2, consider the intersection D1 ∩ D2, consisting of all points that lie in both D1 and D2. If D1 and D2 have points in common (see Fig. 34) and a function f1 is analytic in D1, there may exist a function f2, which is analytic in D2, such that f2(z) = f1(z) for each z in the intersection D1 ∩ D2. If so, we call f2 an analytic continuation of f1 into the second domain D2.
Whenever that analytic continuation exists, it is unique, according to the
theorem just proved. That is, not more than one function can be analytic in D2 and assume the value f1(z) at each point z of the domain D1 ∩ D2 interior to D2. However, if there is an analytic continuation f3 of f2 from D2 into a domain D3 which

sec. 28

Reﬂection Principle 85

D1∩ D2

D1

D2

D3

FIGURE 34

intersects D1, as indicated in Fig. 34, it is not necessarily true that f3(z) = f1(z) for each z in D1 ∩ D3. Exercise 2, Sec. 28, illustrates this.
If f2 is the analytic continuation of f1 from a domain D1 into a domain D2, then the function F deﬁned by means of the equations

F (z) =

f1(z) f2(z)

when z is in D1, when z is in D2

is analytic in the union D1 ∪ D2, which is the domain consisting of all points that lie in either D1 or D2. The function F is the analytic continuation into D1 ∪ D2 of either f1 or f2; and f1 and f2 are called elements of F .

28. REFLECTION PRINCIPLE
The theorem in this section concerns the fact that some analytic functions possess the property that f (z) = f (z) for all points z in certain domains, while others do not. We note, for example, that the functions z + 1 and z2 have that property when D is the entire ﬁnite plane; but the same is not true of z + i and iz2. The theorem here, which is known as the reﬂection principle, provides a way of predicting when f (z) = f (z).
Theorem. Suppose that a function f is analytic in some domain D which contains a segment of the x axis and whose lower half is the reﬂection of the upper half with respect to that axis. Then

(1)

f (z) = f (z)

for each point z in the domain if and only if f (x) is real for each point x on the segment.
We start the proof by assuming that f (x) is real at each point x on the segment. Once we show that the function

(2)

F (z) = f (z)

86 Analytic Functions

chap. 2

is analytic in D, we shall use it to obtain equation (1). To establish the analyticity of F (z), we write

f (z) = u(x, y) + iv(x, y), F (z) = U (x, y) + iV (x, y)

and observe how it follows from equation (2) that since

(3)

f (z) = u(x, −y) − iv(x, −y),

the components of F (z) and f (z) are related by the equations

(4)

U (x, y) = u(x, t) and V (x, y) = −v(x, t),

where t = −y. Now, because f (x + it) is an analytic function of x + it, the
ﬁrst-order partial derivatives of the functions u(x, t) and v(x, t) are continuous throughout D and satisfy the Cauchy–Riemann equations∗

(5)

ux = vt , ut = −vx.

Furthermore, in view of equations (4),

dt Ux = ux , Vy = −vt dy = vt ;

and it follows from these and the ﬁrst of equations (5) that Ux = Vy. Similarly,

dt Uy = ut dy = −ut , Vx = −vx;

and the second of equations (5) tells us that Uy = −Vx. Inasmuch as the ﬁrstorder partial derivatives of U (x, y) and V (x, y) are now shown to satisfy the Cauchy–Riemann equations and since those derivatives are continuous, we ﬁnd that the function F (z) is analytic in D. Moreover, since f (x) is real on the segment of the real axis lying in D, we know that v(x, 0) = 0 on the segment; and, in view of equations (4), this means that

F (x) = U (x, 0) + iV (x, 0) = u(x, 0) − iv(x, 0) = u(x, 0).

That is,

(6)

F (z) = f (z)

at each point on the segment. According to the theorem in Sec. 27, which tells us that an analytic function deﬁned on a domain D is uniquely determined by its

∗See the paragraph immediately following Theorem 1 in Sec. 26.

sec. 28

Exercises 87

values along any line segment lying in D, it follows that equation (6) actually holds throughout D. Because of deﬁnition (2) of the function F (z), then,

(7)

f (z) = f (z) ;

and this is the same as equation (1). To prove the converse in the theorem, we assume that equation (1) holds and
note that in view of expression (3), the form (7) of equation (1) can be written

u(x, −y) − iv(x, −y) = u(x, y) + iv(x, y).

In particular, if (x, 0) is a point on the segment of the real axis that lies in D,

u(x, 0) − iv(x, 0) = u(x, 0) + iv(x, 0);

and, by equating imaginary parts here, we see that v(x, 0) = 0. Hence f (x) is real on the segment of the real axis lying in D.

EXAMPLES. Just prior to the statement of the theorem, we noted that
z + 1 = z + 1 and z2 = z2
for all z in the ﬁnite plane. The theorem tells us, of course, that this is true, since x + 1 and x2 are real when x is real. We also noted that z + i and iz2 do not have the reﬂection property throughout the plane, and we now know that this is because x + i and ix2 are not real when x is real.

EXERCISES

1. Use the theorem in Sec. 27 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant throughout any neighborhood lying in D. Suggestion: Suppose that f (z) does have a constant value w0 throughout some neighborhood in D.

2. Starting with the function

f1(z)

=

√ r

eiθ

/2

(r > 0, 0 < θ < π )

and referring to Exercise 4(b), Sec. 23, point out why

f2(z)

=

√ r

eiθ

/2

π r > 0, < θ < 2π
2

is an analytic continuation of f1 across the negative real axis into the lower half plane. Then show that the function

f3(z)

=

√ r

ei θ /2

5π r > 0, π < θ <
2

is an analytic continuation of f2 across the positive real axis into the ﬁrst quadrant but that f3(z) = −f1(z) there.