\fCOMPLEX 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. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOC/DOC 0 9 8 ISBN 978-0-07-305194-9 MHID 0-07-305194-2 Editorial Director: Stewart K. Mattson Director of Development: Kristine Tibbetts Senior Sponsoring Editor: Elizabeth Covello Developmental Editor: Michelle Driscoll Editorial Coordinator: Adam Fischer Senior Marketing Manager: Eric Gates Project Manager: April R. Southwood Senior Production Supervisor: Kara Kudronowicz Associate Design Coordinator: Brenda A. Rolwes Cover Designer: Studio Montage, St. Louis, Missouri Project Coordinator: Melissa M. Leick Compositor: Laserwords Private Limited Typeface: 10.25/12 Times Roman Printer: R. R. Donnelly Crawfordsville, IN Library of Congress Cataloging-in-Publication Data Brown, James Ward. Complex variables and applications / James Ward Brown, Ruel V. Churchill.8th ed. p. cm. Includes bibliographical references and index. ISBN 978-0-07-305194-9ISBN 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 .9dc22 2007043490 www.mhhe.com 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 rst wrote almost 70 years ago. He was also the author of Operational Mathematics. Dr. Churchill held various ofces 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 Inuenced The Careers of Many People, Including Myself. JWB CONTENTS Preface 1 x 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 2 31 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 Innity 50 Continuity 53 Derivatives 56 Differentiation Formulas 60 Cauchy-Riemann Equations 63 Sufcient Conditions for Differentiability 66 Polar Coordinates 68 Analytic Functions 73 Examples 75 Harmonic Functions 78 Uniquely Determined Analytic Functions 83 Reection Principle 85 3 Elementary Functions 89 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 117 Derivatives of Functions w(t) 117 Denite 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 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 181 Convergence of Sequences Convergence of Series 181 184 Taylor Series 189 Proof of Taylor's Theorem Examples 190 192 Laurent Series 197 Proof of Laurent's Theorem Examples 199 202 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 6 222 Residues and Poles 229 Isolated Singular Points 229 Residues 231 Cauchy's Residue Theorem 234 Residue at Innity 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 viii 7 contents Applications of Residues 261 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 Denite Integrals Involving Sines and Cosines 288 Argument Principle 291 Rouch 's Theorem e 294 Inverse Laplace Transforms 298 Examples 8 301 Mapping by Elementary Functions 311 Linear Transformations 311 The Transformation w = 1/z Mappings by 1/z 313 315 Linear Fractional Transformations 319 An Implicit Form 322 Mappings of the Upper Half Plane The Transformation w = sin z 325 330 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 367 355 contents 10 Applications of Conformal Mapping ix 373 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 11 The Schwarz-Christoffel Transformation 403 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 12 Integral Formulas of the Poisson Type 429 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 Appendixes 449 Bibliography 449 Table of Transformations of Regions 452 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 rst two of which were written by the late Ruel V. Churchill alone. The rst 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, nding 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 uid ow. 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 rst 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 rst 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 rst 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 identied. 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 signicant 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 innity, 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 denition 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 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 dened 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 3:32pm 1 Brown-chap01-v3 2 Complex Numbers chap. 1 y z = (x, y) i = (0, 1) O x x = (x, 0) FIGURE 1 The sum z1 + z2 and product z1 z2 of two complex numbers z1 = (x1 , y1 ) and z2 = (x2 , y2 ) are dened as follows: (3) (4) (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ), (x1 , y1 )(x2 , y2 ) = (x1 x2 y1 y2 , y1 x2 + x1 y2 ). Note that the operations dened 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) = (x1 x2 , 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 = z2 z, etc., we have i 2 = (0, 1)(0, 1) = (1, 0), or (6) In i 2 = 1. electrical engineering, the letter j is used instead of i. 10/29/07 3:32pm 2 Brown-chap01-v3 sec. 2 Basic Algebraic Properties 10/29/07 3 Because (x, y) = x + iy, denitions (3) and (4) become (7) (8) (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 ), (x1 + iy1 )(x2 + iy2 ) = (x1 x2 y1 y2 ) + i(y1 x2 + x1 y2 ). 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 i 2 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 veried in the exercises. The commutative laws z1 + z2 = z2 + z1 , (1) z1 z2 = z2 z1 and the associative laws (2) (z1 + z2 ) + z3 = z1 + (z2 + z3 ), (z1 z2 )z3 = z1 (z2 z3 ) follow easily from the denitions 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 . Verication 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 z1 z2 z3 is well dened without parentheses, as is the case with real numbers. 3:32pm 3 Brown-chap01-v3 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, z+0=z (4) 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 z = (x, y), (5) 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 z1 such that zz = 1. This multiplicative inverse is less obvious than the additive one. To nd it, we seek real numbers u and v, expressed in terms of x and y, such that 1 (x, y)(u, v) = (1, 0). According to equation (4), Sec. 1, which denes 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 . + y2 x2 So the multiplicative inverse of z = (x, y) is (6) z1 = x y , x2 + y2 x2 + y2 (z = 0). The inverse z1 is not dened when z = 0. In fact, z = 0 means that x 2 + y 2 = 0 ; and this is not permitted in expression (6). 10/29/07 3:32pm 4 Brown-chap01-v3 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 z1 z2 is zero, then so is at least one of the factors 1 z1 and z2 . For suppose that z1 z2 = 0 and z1 = 0. The inverse z1 exists; and any complex number times zero is zero (Sec. 1). Hence 1 1 1 1 z2 = z2 1 = z2 (z1 z1 ) = (z1 z1 )z2 = z1 (z1 z2 ) = z1 0 = 0. That is, if z1 z2 = 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 z1 z2 . Subtraction and division are dened in terms of additive and multiplicative inverses: (1) z1 z2 = z1 + (z2 ), (2) z1 1 = z1 z2 z2 (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 x2 y2 = (x1 , y1 ) , 2 2 2 2 z2 x2 + y2 x2 + y2 = x1 x2 + y1 y2 y1 x2 x1 y2 , 2 2 2 2 x2 + y2 x2 + y2 (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 x1 x2 + y1 y2 y1 x2 x1 y2 = +i 2 2 2 2 z2 x2 + y2 x2 + y2 (z2 = 0). Although expression (6) is not easy to remember, it can be obtained by writing (see Exercise 7) (7) (x1 + iy1 )(x2 iy2 ) z1 , = z2 (x2 + iy2 )(x2 iy2 ) 10/29/07 3:32pm 6 Brown-chap01-v3 sec. 3 Further Properties 10/29/07 5 EXERCISES 1. Verify that (a) ( 2 i) i(1 2i) = 2i; (c) (3, 1)(3, 1) 1 1 , 5 10 (b) (2, 3)(2, 1) = (1, 8); = (2, 1). 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 satises 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 satises the given equation to show that y = 0. 1 3 Ans. z = , . 2 2 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 3:32pm 5 Brown-chap01-v3 sec. 3 Further Properties 10/29/07 7 multiplying out the products in the numerator and denominator on the right, and then using the property z1 + z2 z1 z2 1 1 1 = (z1 + z2 )z3 = z1 z3 + z2 z3 = + z3 z3 z3 (8) (z3 = 0). The motivation for starting with equation (7) appears in Sec. 5. EXAMPLE. The method is illustrated below: 4+i (4 + i)(2 + 3i) 5 + 14i 5 14 = = = + i. 2 3i (2 3i)(2 + 3i) 13 13 13 There are some expected properties involving quotients that follow from the relation 1 1 (9) = z2 (z2 = 0), z2 which is equation (2) when z1 = 1. Relation (9) enables us, for instance, to write equation (2) in the form z1 1 = z1 z2 z2 (10) (z2 = 0). Also, by observing that (see Exercise 3) 1 1 1 1 (z1 z2 )(z1 z2 ) = (z1 z1 )(z2 z2 ) = 1 (z1 = 0, z2 = 0), 1 1 and hence that z1 z2 = (z1 z2 )1 , one can use relation (9) to show that (11) 1 z1 1 z2 1 1 = z1 z2 = (z1 z2 )1 = 1 z1 z2 (z1 = 0, z2 = 0). Another useful property, to be derived in the exercises, is (12) z1 z3 z2 z4 = z1 z2 z3 z4 (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 n (13) (z1 + z2 )n = k=0 where n k nk z z k 1 2 n n! = k k!(n k)! (n = 1, 2, . . .) (k = 0, 1, 2, . . . , n) and where it is agreed that 0! = 1. The proof is left as an exercise. 3:32pm 7 Brown-chap01-v3 8 Complex Numbers chap. 1 EXERCISES 1. Reduce each of these quantities to a real number: (a) 2i 1 + 2i + ; 3 4i 5i (b) Ans. (a) 2/5; 5i ; (1 i)(2 i)(3 i) (b) 1/2; (c) (1 i)4 . (c) 4. 2. Show that 1 =z 1/z (z = 0). 3. Use the associative and commutative laws for multiplication to show that (z1 z2 )(z3 z4 ) = (z1 z3 )(z2 z4 ). 4. Prove that if z1 z2 z3 = 0, then at least one of the three factors is zero. Suggestion: Write (z1 z2 )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 z3 z2 z4 = z1 z2 z3 z4 (z3 = 0, z4 = 0). 7. Use the identity obtained in Exercise 6 to derive the cancellation law z1 z1 z = z2 z z2 (z2 = 0, z = 0). 8. Use mathematical induction to verify the binomial formula (13) in Sec. 3. More precisely, 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 ) k=0 m = k=0 m k m+1k z z + k 1 2 m k=0 m k mk z z k 1 2 m k+1 mk z z k 1 2 and replace k by k 1 in the last sum here to obtain m m+1 (z1 + z2 )m+1 = z2 + k=1 m m + k k1 m+1 k m+1k z1 z2 + z1 . 10/29/07 3:32pm 8 Brown-chap01-v3 sec. 4 Vectors and Moduli 10/29/07 9 Finally, show how the right-hand side here becomes m m + 1 k m+1k m+1 z 1 z2 + z1 = k m+1 z2 + k=1 m+1 k=0 m + 1 k m+1k z1 z2 . 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. y (-2, 1) 1 -2 +i z = (x, y) z= x+ iy x O -2 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 z1+ z2 z2 z1 O 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. 3:32pm 9 Brown-chap01-v3 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 dened as the nonnegative real number x 2 + y 2 and is denoted by |z|; that is, |z| = (1) 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 denition (1) that |z1 z2 | = (x1 x2 )2 + (y1 y2 )2 . y (x2, y2) z2 |z1 - z2 | (x1, y1) z1 O z1 - z 2 -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. 10/29/07 3:32pm 10 Brown-chap01-v3 sec. 4 Vectors and Moduli 10/29/07 11 It also follows from denition (1) that the real numbers |z|, Re z = x, and Im z = y are related by the equation |z|2 = (Re z)2 + (Im z)2 . (2) Thus Re z |Re z| |z| (3) 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 : |z1 + z2 | |z1 | + |z2 |. (4) 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 |z1 + z2 | ||z1 | |z2 ||. (5) 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 |. 3:32pm 11 Brown-chap01-v3 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 nite 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 2 (a) z1 = 2i, z2 = i; (b) z1 = ( 3, 1), z2 = ( 3, 0); 3 (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 |, |z1 | + |z2 | Re(z1 + z2 ) . |z3 + z4 | ||z3 | |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. 10/29/07 3:32pm 12 Brown-chap01-v3 sec. 5 Complex Conjugates 10/29/07 13 5. COMPLEX CONJUGATES The complex conjugate, or simply the conjugate, of a complex number z = x + iy is dened as the complex number x iy and is denoted by z ; that is, z = x iy. (1) The number z is represented by the point (x, y), which is the reection 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 (x, -y) 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: z1 + z2 = z1 + z2 . (2) In like manner, it is easy to show that (3) z1 z2 = z1 z2 , (4) z1 z2 = z1 z2 , and (5) z1 z2 = z1 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 2 and Im z = zz . 2i 3:32pm 13 Brown-chap01-v3 14 Complex Numbers chap. 1 An important identity relating the conjugate of a complex number z = x + iy to its modulus is z z = |z|2 , (7) where each side is equal to x 2 + y 2 . 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, 5 + 5i (1 + 3i)(2 + i) 5 + 5i 1 + 3i = = = = 1 + i. 2 2i (2 i)(2 + i) |2 i| 5 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 |z1 z2 | = |z1 ||z2 | (8) and z1 |z1 | = z2 |z2 | (9) (z2 = 0). Property (8) can be established by writing |z1 z2 |2 = (z1 z2 )(z1 z2 ) = (z1 z2 )(z1 z2 ) = (z1 z1 )(z2 z2 ) = |z1 |2 |z2 |2 = (|z1 ||z2 |)2 and recalling that a modulus is never negative. Property (9) can be veried 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. 10/29/07 3:32pm 14 Brown-chap01-v3 sec. 5 Exercises 10/29/07 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) z1 z2 z3 = 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 z2 z3 = z1 ; z2 z3 (b) |z1 | z1 = . z2 z3 |z2 ||z3 | 7. Show that |Re(2 + z + z3 )| 4 when |z| 1. 8. It is shown in Sec. 3 that if z1 z2 = 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 1 . z4 4z2 + 3 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) z1 z2 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 a 0 + a1 z + a2 z 2 + + an z n = a0 + a 1 z + a 2 z 2 + + a n z 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 = R 2 . 14. Using expressions (6), Sec. 5, for Re z and Im z, show that the hyperbola x 2 y 2 = 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 ) = z1 z1 + (z1 z2 + z1 z2 ) + z2 z2 . 3:32pm 15 Brown-chap01-v3 16 Complex Numbers (b) Point out why chap. 1 z1 z2 + z1 z2 = 2 Re(z1 z2 ) 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 be polar coordinates of the point (x, y) that corresponds to a nonzero complex number z = x + iy. Since x = r cos and y = r sin , the number z can be written in polar form as z = r(cos + i sin ). (1) If z = 0, the coordinate is undened; 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 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, has an innite number of possible values, including negative ones, that differ by integral multiples of 2. Those values can be determined from the equation tan = y/x, where the quadrant containing the point corresponding to z must be specied. Each value of 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 < . Evidently, then, arg z = Arg z + 2n (2) (n = 0, 1, 2, . . .). Also, when z is a negative real number, Arg z has value , not . y z = x + iy r x FIGURE 6 EXAMPLE 1. The complex number 1 i, which lies in the third quadrant, has principal argument 3/4. That is, Arg(1 i) = 3 . 4 10/29/07 3:32pm 16 Brown-chap01-v3 sec. 6 Exponential Form It must be emphasized that because of the restriction < argument , it is not true that Arg(1 i) = 5/4. According to equation (2), arg(1 i) = 3 + 2n 4 10/29/07 17 of the principal (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) = 5 + 2n 4 (n = 0, 1, 2, . . .). The symbol ei , or exp(i ), is dened by means of Euler's formula as ei = cos + i sin , (3) where is to be measured in radians. It enables one to write the polar form (1) more compactly in exponential form as z = rei . (4) The choice of the symbol ei 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 1 i = (5) 2 exp i 3 4 . With the agreement that ei = ei() , this can also be written 1 i = 2 ei3/4 . Expression (5) is, of course, only one of an innite number of possibilities for the exponential form of 1 i: (6) 1 i = 2 exp i 3 + 2n 4 (n = 0, 1, 2, . . .). Note how expression (4) with r = 1 tells us that the numbers ei lie on the circle centered at the origin with radius unity, as shown in Fig. 7. Values of ei are, then, immediate from that gure, without reference to Euler's formula. It is, for instance, geometrically obvious that ei = 1, ei/2 = i, and ei4 = 1. 3:32pm 17 Brown-chap01-v3 18 Complex Numbers chap. 1 y 1 O x FIGURE 7 Note, too, that the equation z = Rei (7) (0 2) is a parametric representation of the circle |z| = R, centered at the origin with radius R. As the parameter increases from = 0 to = 2, 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 z = z0 + Rei (8) (0 2). 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 xed vector z0 and a vector of length R whose angle of inclination varies from = 0 to = 2. y z z0 O x FIGURE 8 7. PRODUCTS AND POWERS IN EXPONENTIAL FORM Simple trigonometry tells us that ei has the familiar additive property of the exponential function in calculus: ei1 ei2 = (cos 1 + i sin 1 )(cos 2 + i sin 2 ) = (cos 1 cos 2 sin 1 sin 2 ) + i(sin 1 cos 2 + cos 1 sin 2 ) = cos(1 + 2 ) + i sin(1 + 2 ) = ei(1 +2 ) . 10/29/07 3:32pm 18 Brown-chap01-v3 sec. 7 Products and Powers in Exponential Form 10/29/07 19 Thus, if z1 = r1 ei1 and z2 = r2 ei2 , the product z1 z2 has exponential form z1 z2 = r1 ei1 r2 ei2 = r1 r2 ei1 ei2 = (r1 r2 )ei(1 +2 ) . (1) Furthermore, (2) z1 r1 ei1 r1 ei1 ei2 r1 ei(1 2 ) r1 = = i i = = ei(1 2 ) . z2 r2 ei2 r2 e 2 e 2 r2 ei0 r2 Note how it follows from expression (2) that the inverse of any nonzero complex number z = rei is z1 = (3) 1ei0 1 1 1 = i = ei(0) = ei . z re r 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 = rei is zn = r n ein (4) (n = 0, 1, 2, . . .). It is easily veried for positive values of n by mathematical induction. To be specic, we rst note that it becomes z = rei 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 = zm z = r m eim rei = (r m r)ei(m+) = r m+1 ei(m+1) . Expression (4) is thus veried 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 dene zn in terms of the multiplicative inverse of z by writing zn = (z1 )m where m = n = 1, 2, . . . . Then, since equation (4) is valid for positive integers, it follows from the exponential form (3) of z1 that zn = 1 i() e r m = 1 r m eim() = 1 r n ei(n)() = r n ein (n = 1, 2, . . .). Expression (4) is now established for all integral powers. Expression (4) can be useful in nding powers of complex numbers even when they are given in rectangular form and the result is desired in that form. 3:32pm 19 Brown-chap01-v3 20 Complex Numbers chap. 1 EXAMPLE 1. In order to put ( 3 + 1)7 in rectangular form, one need only write ( 3 + i)7 = (2ei/6 )7 = 27 ei7/6 = (26 ei )(2ei/6 ) = 64( 3 + i). Finally, we observe that if r = 1, equation (4) becomes (ei )n = ein (5) (n = 0, 1, 2, . . .). When written in the form (6) (cos + i sin )n = cos n + i sin n (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 + i sin )2 = cos 2 + i sin 2, or cos2 sin2 + i2 sin cos = cos 2 + i sin 2. By equating real parts and then imaginary parts here, we have the familiar trigonometric identities cos 2 = cos2 sin2 , sin 2 = 2 sin cos . (See also Exercises 10 and 11, Sec. 8.) 8. ARGUMENTS OF PRODUCTS AND QUOTIENTS If z1 = r1 ei1 and z2 = r2 ei2 , the expression (1) z1 z2 = (r1 r2 )ei(1 +2 ) in Sec. 7 can be used to obtain an important identity involving arguments: (2) arg(z1 z2 ) = arg z1 + arg z2 . This result is to be interpreted as saying that if values of two of the three (multiplevalued) arguments are specied, then there is a value of the third such that the equation holds. We start the verication of statement (2) by letting 1 and 2 denote any values of arg z1 and arg z2 , respectively. Expression (1) then tells us that 1 + 2 is a value of arg(z1 z2 ). (See Fig. 9.) If, on the other hand, values of arg(z1 z2 ) and 10/29/07 3:32pm 20 Brown-chap01-v3 sec. 8 Arguments of Products and Quotients 10/29/07 21 z1z2 y z2 z1 x O FIGURE 9 arg z1 are specied, those values correspond to particular choices of n and n1 in the expressions arg(z1 z2 ) = (1 + 2 ) + 2n (n = 0, 1, 2, . . .) and arg z1 = 1 + 2n1 (n1 = 0, 1, 2, . . .). Since (1 + 2 ) + 2n = (1 + 2n1 ) + [2 + 2(n n1 )], equation (2) is evidently satised when the value arg z2 = 2 + 2(n n1 ) is chosen. Verication when values of arg(z1 z2 ) and arg z2 are specied 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(z1 z2 ) = Arg(i) = 2 but Arg z1 + Arg z2 = + 3 = . 2 2 If, however, we take the values of arg z1 and arg z2 just used and select the value Arg(z1 z2 ) + 2 = 3 + 2 = 2 2 of arg(z1 z2 ), we nd that equation (2) is satised. Statement (2) tells us that arg z1 z2 1 1 = arg(z1 z2 ) = arg z1 + arg(z2 ); 3:32pm 21 Brown-chap01-v3 22 Complex Numbers chap. 1 and, since (Sec. 7) 1 z2 = 1 i2 e , r2 one can see that 1 arg(z2 ) = arg z2 . (3) 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 nd the principal argument Arg z when z= 2 , 1 + 3i observe that arg z = arg(2) arg(1 + Since Arg(2) = and Arg(1 + 3i). 3i) = , 3 one value of arg z is 2/3; and, because 2/3 is between and , we nd that Arg z = 2/3. EXERCISES 1. Find the principal argument Arg z when i ; (b) z = ( 3 i)6 . (a) z = 2 2i Ans. (a) 3/4; (b) . 2. Show that (a) |ei | = 1; (b) ei = ei . 3. Use mathematical induction to show that ei1 ei2 ein = ei(1 +2 ++n ) (n = 2, 3, . . .). 4. Using the fact that the modulus |ei 1| is the distance between the points ei and 1 (see Sec. 4), give a geometric argument to nd a value of in the interval 0 < 2 that satises the equation |ei 1| = 2. Ans. . 10/29/07 3:32pm 22 Brown-chap01-v3 sec. 8 10/29/07 Exercises 23 5. By writing the individual factors on the left in exponential form, performing the needed operations, and nally changing back to rectangular coordinates, show that (a) i(1 3i)( 3 + i) = 2(1 + 3i); (b) 5i/(2 + i) = 1 + 2i; (d) (1 + 3i)10 = 211 (1 + 3i). (c) (1 + i)7 = 8(1 + i); 6. Show that if Re z1 > 0 and Re z2 > 0, then Arg(z1 z2 ) = 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 = rei and m = n = 1, 2, . . . . Using the expressions zm = r m eim z1 = and 1 i() , e r verify that (zm )1 = (z1 )m and hence that the denition zn = (z1 )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 = c1 c2 and z2 = c1 c2 . Suggestion: Note that exp i 1 + 2 2 exp i 1 2 2 = exp(i1 ) 1 + 2 2 exp i 1 2 2 = exp(i2 ). and [see Exercise 2(b)] exp i 9. Establish the identity 1 + z + z2 + + zn = 1 zn+1 1z (z = 1) and then use it to derive Lagrange's trigonometric identity: 1 + cos + cos 2 + + cos n = 1 sin[(2n + 1)/2] + 2 2 sin(/2) (0 < < 2 ). Suggestion: As for the rst identity, write S = 1 + z + z2 + + zn and consider the difference S zS. To derive the second identity, write z = ei in the rst one. 10. Use de Moivre's formula (Sec. 7) to derive the following trigonometric identities: (a) cos 3 = cos3 3 cos sin2 ; (b) sin 3 = 3 cos2 sin sin3 . 3:32pm 23 Brown-chap01-v3 24 Complex Numbers chap. 1 11. (a) Use the binomial formula (Sec. 3) and de Moivre's formula (Sec. 7) to write n n cosnk (i sin )k k cos n + i sin n = k=0 (n = 0, 1, 2, . . .). Then dene 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 n = k=0 n (1)k cosn2k sin2k 2k (n = 0, 1, 2, . . .). (b) Write x = cos in the nal summation in part (a) to show that it becomes a polynomial m Tn (x) = k=0 n (1)k x n2k (1 x 2 )k 2k of degree n (n = 0, 1, 2, . . .) in the variable x. 9. ROOTS OF COMPLEX NUMBERS Consider now a point z = rei , lying on a circle centered at the origin with radius r (Fig. 10). As is increased, z moves around the circle in the counterclockwise direction. In particular, when is increased by 2, we arrive at the original point; and the same is true when is decreased by 2. It is, therefore, evident from Fig. 10 that two nonzero complex numbers z1 = r1 ei1 and z2 = r2 ei2 y r O x FIGURE 10 These are called Chebyshev polynomials and are prominent in approximation theory. 10/29/07 3:32pm 24 Brown-chap01-v3 sec. 9 Roots of Complex Numbers 10/29/07 25 are equal if and only if r1 = r2 and 1 = 2 + 2k, where k is some integer (k = 0, 1, 2, . . .). This observation, together with the expression zn = r n ein in Sec. 7 for integral powers of complex numbers z = rei , is useful in nding the nth roots of any nonzero complex number z0 = r0 ei0 , 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 = rei such that zn = z0 , or r n ein = r0 ei0 . According to the statement in italics just above, then, r n = r0 and n = 0 + 2k, 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 0 2k 0 + 2k = + n n n Consequently, the complex numbers = z= n r0 exp i 2k 0 + n n (k = 0, 1, 2, . . .). (k = 0, 1, 2, . . .) are the nth roots of z0 . We are able to see immediately from this exponential form of the roots that they all lie on the circle |z| = n r0 about the origin and are equally spaced every 2/n radians, starting with argument 0 /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 ck = (1) n r0 exp i 2k 0 + n n (See Fig. 11.) ck-1 ck y n O n r0 x FIGURE 11 (k = 0, 1, 2, . . . , n 1). 3:32pm 25 Brown-chap01-v3 26 Complex Numbers chap. 1 The number n r0 is the length of each of the radius vectors representing the n roots. The rst root c0 has argument 0 /n; and the two roots when n = 2 lie at the opposite ends of a diameter of the circle |z| = n r0 , the second root being c0 . When n 3, the roots lie at the vertices of a regular polygon of n sides inscribed in that circle. 1/n We shall let z0 denote the set of nth roots of z0 . If, in particular, z0 is a 1/n positive real number r0 , the symbol r0 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 0 that is used in expression (1) is the principal value of arg z0 ( < 0 ), 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 0 n exp i 2k n (k = 0, 1, 2, . . . , n 1), and also write n = exp i (2) 2 n , it follows from property (5), Sec. 7. of ei that k n = exp i (3) 2k n (k = 0, 1, 2, . . . , n 1) and hence that k ck = c0 n (4) (k = 0, 1, 2, . . . , n 1). The number c0 here can, of course, be replaced by any particular nth root of z0 , since n represents a counterclockwise rotation through 2/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) z0 = r0 ei(0 +2k) (5) (k = 0, 1, 2, . . .) and to formally apply laws of fractional exponents involving real numbers, keeping in mind that there are precisely n roots: 1/n z0 = r0 ei(0 +2k) 1/n = n r0 exp 2k i(0 + 2k) 0 = n r0 exp i + n n n (k = 0, 1, 2, . . . , n 1). 10/29/07 3:32pm 26 Brown-chap01-v3 sec. 10 Examples 10/29/07 27 The examples in the next section serve to illustrate this method for nding 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 nd all values of (8i)1/3 , or the three cube roots of the number 8i. One need only write 8i = 8 exp i + 2k (k = 0, 1, 2, . . .) 2 to see that the desired roots are ck = 2 exp i (1) 2k + 6 3 (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 2/3 radians, starting with the principal root (Fig. 12) c0 = 2 exp i = 2 cos i sin = 3 i. 6 6 6 Without any further calculations, it is then evident that c1 = 2i; and, since c2 is symmetric 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 2 c0 , c0 3 , c0 3 where 3 = exp i y c1 2 c2 x c0 FIGURE 12 2 3 . 3:32pm 27 Brown-chap01-v3 28 Complex Numbers EXAMPLE 2. chap. 1 In order to determine the nth roots of unity, we start with 1 = 1 exp[i(0 + 2k)] and nd that 0 2k n (2) 11/n = 1 exp i + n n = exp i (k = 0, 1, 2 . . .) 2k 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 2 n1 1, n , n , . . . , n where n = exp i 2 n . n See Fig. 13, where the cases n = 3, 4, and 6 are illustrated. Note that n = 1. y y 1x y 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 + 2k (k = 0, 1, 2, . . .) 6 and (see Fig. 14) (3) ck = + k 2 exp i 12 (k = 0, 1). Euler's formula tells us that = 2 cos + i sin , c0 = 2 exp i 12 12 12 and the trigonometric identities 10/29/07 3:32pm 28 Brown-chap01-v3 sec. 10 Exercises 10/29/07 29 y c0 2 c1 = - c0 x FIGURE 14 cos2 (4) 1 + cos = , 2 2 enable us to write 1 cos = 2 2 3 1 = 1+ 2 2 3 1 = 1 2 2 1 cos = 1 + cos 12 2 6 2 sin2 sin2 1 = 1 cos 12 2 6 2+ 3 = , 4 2 3 = . 4 Consequently, 2+ 3 +i c0 = 2 4 2 3 1 = 4 2 Since c1 = c0 , the two square roots of (5) EXERCISES 1 2 2+ 2+ 3+i 2 3 . 3 + i are, then, 3+i 2 3 . 1. Find the square roots of (a) 2i; (b) 1 3i and express them in rectangular coordinates. 3i Ans. (a) (1 + i); (b) . 2 2. In each case, nd 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). 3:32pm 29 Brown-chap01-v3 30 Complex Numbers chap. 1 3. In each case, nd 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 . 1 3i 1 + 3i , . Ans. (b) 2, 2 2 4. According to Sec. 9, the three cube roots of a nonzero complex number z0 can be 2 written c0 , c0 3 , c0 3 where c0 is the principal cube root of z0 and 2 1 + 3i = . 3 = exp i 3 2 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 ( 3 1) ( 3 + 1)i 2 , c0 3 = . c0 3 = 2 2 5. (a) Let a denote any xed real number and show that the two square roots of a + i are A exp i 2 where A = a 2 + 1 and = 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 Aa . 2 (Note that this becomes the nal 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 ei/4 = 1 + i. Then use those zeros to factor z2 + 4 into quadratic factors with real coefcients. 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 + + cn1 = 0. Suggestion: Use the rst 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 coefcients a, b, and c are complex numbers. Specically, by completing the square on the left-hand side, derive the quadratic formula b + (b2 4ac)1/2 , 2a where both square roots are to be considered when b2 4ac = 0, z= 10/29/07 3:32pm 30 Brown-chap01-v3 sec. 11 Regions in the Complex Plane 10/29/07 31 (b) Use the result in part (a) to nd the roots of the equation z2 + 2z + (1 i) = 0. i 1 i 1 + , . 1 Ans. (b) 1 + 2 2 2 2 9. Let z = rei be a nonzero complex number and n a negative integer (n = 1, 2, . . .). Then dene z1/n by means of the equation z1/n = (z1 )1/m where m = n. By showing that the m values of (z1/m )1 and (z1 )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 neighborhood |z z0 | < (1) of a given point z0 . It consists of all points z lying inside but not on a circle centered at z0 and with a specied positive radius (Fig. 15). When the value of 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, 0 < |z z0 | < (2) consisting of all points z in an neighborhood of z0 except for the point z0 itself. y |z - z0 | z O z0 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. 3:32pm 31 Brown-chap01-v3 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 rst 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 nite 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 z2 z1 O 1 2 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, . . .). 10/29/07 3:32pm 32 Brown-chap01-v3 sec. 11 Exercises 10/29/07 33 EXERCISES 1. Sketch the following sets and determine which are domains: (a) |z 2 + i| 1; (b) |2z + 3| > 4; (c) Im z > 1; (d) Im z = 1; (e) 0 arg z /4 (z = 0); (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) < arg z < (z = 0); (b) |Re z| < |z|; (c) Re 1 z 1 ; 2 (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 = i n (n = 1, 2, . . .); (b) zn = i n /n (n = 1, 2, . . .); n1 (c) 0 arg z < /2 (z = 0); (d) zn = (1)n (1 + i) (n = 1, 2, . . .). n 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 nite set of points z1 , z2 , . . . , zn cannot have any accumulation points. 3:32pm 33 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 dened 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 denition of f . It must be emphasized that both a domain of denition and a rule are needed in order for a function to be well dened. When the domain of denition 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 dened 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 denition is often a domain as dened 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: f (z) = u(x, y) + iv(x, y). (1) If the polar coordinates r and , instead of x and y, are used, then u + iv = f (rei ) where w = u + iv and z = rei . In that case, we may write f (z) = u(r, ) + iv(r, ). (2) EXAMPLE 2. If f (z) = z2 , then f (x + iy) = (x + iy)2 = x 2 y 2 + i2xy. Hence u(x, y) = x 2 y 2 and v(x, y) = 2xy. When polar coordinates are used, f (rei ) = (rei )2 = r 2 ei2 = r 2 cos 2 + ir 2 sin 2. Consequently, u(r, ) = r 2 cos 2 and v(r, ) = r 2 sin 2. 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 = x 2 + y 2 + 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 + a1 z + a2 z2 + + an zn is a polynomial of degree n. Note that the sum here has a nite number of terms and that the domain of denition is the entire z plane. Quotients P (z)/Q(z) of sec. 12 Exercises 37 polynomials are called rational functions and are dened 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 denition. 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 ( < ) 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 2 (r > 0, < ), the (single-valued) function (3) is well dened 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 dened on the entire plane. EXERCISES 1. For each of the functions below, describe the domain of denition that is understood: 1 1 ; (b) f (z) = Arg ; (a) f (z) = 2 z +1 z z 1 (c) f (z) = . ; (d) f (z) = z+z 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) = (x 3 3xy 2 + x + 1) + i(3x 2 y y 3 + y). 3. Suppose that f (z) = x 2 y 2 2y + i(2x 2xy), where z = x + iy. Use the expressions (see Sec. 5) zz z+z and y = x= 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, ) + iv(r, ). Ans. f (z) = r + 1 1 cos + i r sin . 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 denition 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 denition 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 denition 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 reection 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 = ei/2 , the mapping w = iz = r exp i + 2 , where z = re