Complex Variables Churchill 8ed Solution

Student's Solutions Manual to accompany Complex Variables and Applications Eighth Edition James Ward Brown University of Michigan - Dearborn Ruel V. Churchill Late Professor of Mathematics University of Michigan Prepared by James Ward Brown University of Michigan - Dearborn McGraw-Hill Higher Education Boston Burr Ridge, IL Dubuque, IA New York San Francisco St. Louis Bangkok Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto Contents 1 Complex Numbers 1 2 Basic Algebraic Properties 1 3 Further Properties 2 4 Vectors and Moduli 3 5 Complex Conjugates 5 8 Arguments of Products and Quotients 8 10 Examples 12 11 Regions in the Complex Plane 18 2 Analytic Functions 22 12 Functions of a Complex Variable 22 18 Continuity 22 20 Differentiation Formulas 24 23 Polar Coordinates 25 25 Examples 31 26 Harmonic Functions 32 3 Elementary Functions 35 29 The Exponential Function 35 31 Branches and Derivatives of Logarithms 39 32 Some Identities Involving Logarithms 41 33 Complex Exponents 43 34 Trigonometric Functions 45 35 Hyperbolic Functions 49 iii 4 Integrals 53 38 Definite Integrals of Functions w(t) 53 39 Contours 54 42 Examples with Branch Cuts 56 43 Upper Bounds for Moduli of Contour Integrals 61 45 Proof of the Theorem 65 49 Multiply Connected Domains 66 52 Some Consequences of the Extension 69 5 Series 75 56 Convergence of Series 75 59 Examples 77 62 Examples 80 66 Uniqueness of Series Representations 86 67 Multiplication and Division of Power Series 89 6 Residues and Poles 94 71 Residue at Infinity 94 72 The Three Types of Isolated Singular Points 99 74 Examples 103 76 Zeros and Poles 109 7 Applications of Residues 118 79 Example 118 81 Jordan's Lemma 129 iv 84 Integration Along a Branch Cut ................................................ 138 85 Definite Integrals Involving Sines and Cosines ....................... 151 87 Rouche's Theorem ..................................................................... 153 89 Examples .................................................................................. 155 Note to the reader: The numbering system used here to identify chapters, sections, and exercises is consistent with that used in the text. For instance, according to the table of contents just above, solutions of exercises following Section 10 in Chapter 1 of the text start on page 12 of this solutions manual. Chapter 1 SECTION 2 1. (a) ( √2−í)−í(1−√2í) = √2−í−í−√2 = −2í; (b) (2−3X)−2(1) = (−4+3.6+2) = (−1,8); (c) (3,1)(3,−1)( 1 , 1 ) = (10,0)( 1 , 1 ) = (2,1). 5 10 5 10 2. (a) Re(íx) = Re[{(x+íy)] = Re(−y+íx) = −y = −Im z; (b) Im(íx) = Im[{(x+íy)] = Im(−y+íx) = x = Re z. 3. (1+z)² = (1+2)(1+z) = (1+z)•1+1(1+z)z=1•1(1+z)+z(1+z) = 1+z+z+z² = 1+2z+z². 4. If z = 1±í, then z²−2z+2 = (1±í)²−2(1±í)+2 = ± 2í −2±2í+2 = 0. 5. To prove that multiplication is commutative, write z1z2 = (x1, y1)(x2, y2) = (x1x2−y1y2, y1x2+x1y2) = (x1x2−y1y2, y2x1+x1y2) = (x2, y2)(x1, y1) = z2z1. 6. (a) To verify the associative law for addition, write (z4 + z2) + z3 = [(x1, y1)+(x2, y2)] + (x3, y3) = (x1 + x2, y1 + y2) + (x3, y3) = ((x1 + x2) + x3, (y1 + y2) + y3) = (x1 + (x2 + x3), y1 + (y2 + y3)) = (x1, y1) + (x2 + x3, y2 + y3) = (x1, y1) + [(x2, y2) + (x3, y3)] = z4 + (z2 + z3). 5. (a) Rewrite |z - 1 + i| = 1 as |z - (1 - i)| = 1. This is the circle centered at 1 - i with radius 1. It is shown below. 6. (a) Write |z - 4i + 1/z + 4i| = 10 as |z - 4i + 1/z - (-4i)| = 10 to see that this is the locus of all points z such that the sum of the distances from z to 4i and -4i is a constant. Such a curve is an ellipse with foci ± 4i. (b) Write |z - 1| = |1/z - i| as |z - 1| = |1/z - (-i)| to see that this is the locus of all points z such that the distance from z to 1 is always the same as the distance to -i. The curve is, then, the perpendicular bisector of the line segment from 1 to -i. SECTION 5 1. (a) \(\bar{z} + 3i = \bar{z} + 3i = z - 3i\; (b) \(i\bar{z} = i\bar{z} = -iz\; (c) \((2 + i)^2 = (2 + i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i\; (d) \(|(2z + 5\sqrt{2} - i)| = |2z + 5|\sqrt{2} - i| = |2z + 5|\sqrt{2} + 1 = \sqrt{3} |2z + 5|.\) 2. (a) Rewrite Re(z - i) = 2 as Re[x + i(-y - 1)] = 2, or x = 2. This is the vertical line through the point z = 2, shown below.