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Chapter 15 / Exercise 23
Calculus
Stewart
Expert Verified
See discussions, stats, and author profiles for this publication at: Complex Analysis: Problems with solutions Book · August 2015 DOI: 10.13140/RG.2.1.1305.7362 READS 2,403 1 author: Juan Carlos Ponce-Campuzano University of Queensland 30 PUBLICATIONS 11 CITATIONS SEE PROFILE All in-text references underlined in blue are linked to publications on ResearchGate, letting you access and read them immediately. Available from: Juan Carlos Ponce-Campuzano Retrieved on: 18 June 2016
##### We have textbook solutions for you!
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Chapter 15 / Exercise 23
Calculus
Stewart
Expert Verified
e π i + 1 = 0 C omplex A nalysis Problems with solutions Juan Carlos Ponce Campuzano
Contents Proem 1 1 Complex Numbers 3 1.1 Basic algebraic properties . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Exponential and Polar Form, Complex roots . . . . . . . . . . . . . 10 2 Functions 15 2.1 Basic notions and limits . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Continuity and Differentiation . . . . . . . . . . . . . . . . . . . . . 25 2.3 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3 Complex Integrals 35 3.1 Contour integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.2 Cauchy Integral Theorem and Cauchy Integral Formula . . . . . . 40 3.3 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4 Series 59 4.1 Taylor and Laurent series . . . . . . . . . . . . . . . . . . . . . . . . 59 4.2 Classification of singularities . . . . . . . . . . . . . . . . . . . . . . 67 iii
iv CONTENTS 4.3 Applications of residues . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.3.1 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . 80 Bibliography 83
Proem This text constitutes a collection of problems for using as an additional learning resource for those who are taking an introductory course in complex analysis. The problems are numbered and allocated in four chapters corresponding to different subject areas: Complex Numbers , Functions , Complex Integrals and Series . The majority of problems are provided with answers, detailed procedures and hints (sometimes incomplete solutions). Of course, no project such as this can be free from errors and incompleteness. I will be grateful to everyone who points out any typos, incorrect solutions, or sends any other suggestion for improving this manuscript. Contact: [email protected] 2015
Chapter 1 Complex Numbers 1.1 Basic algebraic properties 1. Verify that (a) 2 - i - i 1 - 2 i = - 2 i (b) ( 2 - 3 i ) ( - 2 + i ) = - 1 + 8 i Solution. We have 2 - i - i 1 - 2 i = 2 - i - i + 2 = - 2 i , and ( 2 - 3 i ) ( - 2 + i ) = - 4 + 2 i + 6 i - 3 i 2 = - 4 + 3 + 8 i = - 1 + 8 i . 2. Reduce the quantity 5 i ( 1 - i )( 2 - i )( 3 - i ) to a real number. Solution. We have 5 i ( 1 - i )( 2 - i )( 3 - i ) = 5 i ( 1 - i )( 5 - 5 i ) = i ( 1 - i ) 2 = i - 2 i = 1 2 3
4 1. Complex Numbers 3. Show that (a) Re ( iz ) = - Im ( z ) ; (b) Im ( iz ) = Re ( z ) . Proof. Let z = x + yi with x = Re ( z ) and y = Im ( z ) . Then Re ( iz ) = Re ( - y + xi ) = - y = - Im ( z ) and Im ( iz ) = Im ( - y + xi ) = x = Re ( z ) . 4. Verify the associative law for multiplication of complex numbers. That is, show that ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 ) for all z 1 , z 2 , z 3 C . Proof. Let z k = x k + iy k for k = 1, 2, 3. Then ( z 1 z 2 ) z 3 = (( x 1 + y 1 i )( x 2 + y 2 i ))( x 3 + y 3 i ) = (( x 1 x 2 - y 1 y 2 ) + i ( x 2 y 1 + x 1 y 2 ))( x 3 + y 3 i ) = ( x 1 x 2 x 3 - x 3 y 1 y 2 - x 2 y 1 y 3 - x 1 y 2 y 3 ) + i ( x 2 x 3 y 1 + x 1 x 3 y 2 + x 1 x 2 y 3 - y 1 y 2 y 3 ) and z 1 ( z 2 z 3 ) = ( x 1 + y 1 i )(( x 2 + y 2 i ))( x 3 + y 3 i )) = ( x 1 + y 1 i )(( x 2 x 3 - y 2 y 3 ) + i ( x 2 y 3 + x 3 y 2 )) = ( x 1 x 2 x 3 - x 3 y 1 y 2 - x 2 y 1 y 3 - x 1 y 2 y 3 ) + i ( x 2 x 3 y 1 + x 1 x 3 y 2 + x 1 x 2 y 3 - y 1 y 2 y 3 ) Therefore, ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 )
1.1. Basic algebraic properties 5 5. Compute (a) 2 + i 2 - i ; (b) ( 1 - 2 i ) 4 .