Complex Analysis - Maths KU

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344 Chapter 6 Series and Residues Theorem 6.15 Residue at a Pole of Order n If f has a pole of order n at z = z0, then Res(f(z),z0) = 1 (n − 1)! lim z→z0 d n−1 dz n−1 (z − z0) n f(z). (2) Proof Because f is assumed to have pole of order n at z = z0, its Laurent expansion convergent on a punctured disk 0 < |z − z0|
An alternative method for computing a residue at a simple pole ☞ 6.5 Residues and Residue Theorem 345 Now at the pole of order 2, the result in (2) gives Res(f(z), 1) = 1 1! lim d z→1dz (z − 1)2f(z) d 1 = lim z→1dz z − 3 −1 = lim = −1 z→1(z − 3) 2 4 . When f is not a rational function, calculating residues by means of (1) or (2) can sometimes be tedious.It is possible to devise alternative residue formulas.In particular, suppose a function f can be written as a quotient f(z) =g(z)/h(z), where g and h are analytic at z = z0.If g(z0) �= 0 and if the function h has a zero of order 1 at z0, then f has a simple pole at z = z0 and Res(f(z),z0) = g(z0) h ′ . (4) (z0) To derive this result we shall use the definition of a zero of order 1, the definition of a derivative, and then (1).First, since the function h has a zero of order 1 at z0, we must have h(z0) = 0 and h ′ (z0) �= 0.Second, by definition of the derivative given in (12) of Section 3.1, h ′ (z0) = lim z→z0 0 � �� h(z) − h(z0) h(z) = lim . z − z0 z→z0 z − z0 We then combine the preceding two facts in the following manner in (1): Res(f(z),z0) = lim (z − z0) z→z0 g(z) = lim h(z) z→z0 g(z) h(z) z − z0 = g(z0) h ′ (z0) . There are several alternative ways of arriving at formula (4).For instance, it can be obtained by a single application of L’Hôpital’s rule (page 147), but you are asked in Problem 40 in Exercises 6.5 to derive (4) using (5) of Section 6.4. Residue formulas for poles of order greater than 1 are far more complicated than (4) and will not be presented here.Practicality aside, a derivation of one of these higher-order formulas provides an opportunity to review and use important concepts.See Problem 41 in Exercises 6.5. EXAMPLE 3 Using (4) to Compute Residues The polynomial z 4 + 1 can be factored as (z − z1)(z − z2)(z − z3)(z − z4), where z1, z2, z3, and z4 are the four distinct roots of the equation z 4 +1=0
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A First Course in Complex Analysis
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For Dana, Kasey, and Cody
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vi Contents Chapter 4. Elementary F
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Preface 7.2 Preface Philosophy This
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Preface xi to, but not formally cov
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3π 2π π 0 -π -2π -3π -1 0 1 -
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1.1 Complex Numbers and Their Prope
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y z 2 z 2 - z 1 or (x 2 - x 1 , y 2
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1.2 Complex Plane 13 From (8) with
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1.2 Complex Plane 15 30. Find an up
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O y θ x = r cos θ (r, θ) or (x,
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1.3 Polar Form of Complex Numbers 1
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1.3 Polar Form of Complex Numbers 2
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1.4 Powers and Roots 23 arg(z) =π/
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√ 4 = 2 and 3 √ 27 = 3 are the
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1.4 Powers and Roots 27 16. Rework
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z 0 ρ ρ |z - z 0 |= Figure 1.15 C
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y Interior Exterior Boundary Figure
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1.5 Sets of Points in the Complex P
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1.5 Sets of Points in the Complex P
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Note: The roots z1 and z2 are conju
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1.6 Applications 39 c = 0.The latte
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1.6 Applications 41 Electrical engi
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1.6 Applications 43 In Problems 25
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Chapter 1 Review Quiz 45 Here the s
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Chapter 1 Review Quiz 47 27. The pr
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3i 2i i S 1 2 3 Image of a square u
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2.1 Complex Functions 51 interchang
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2.1 Complex Functions 53 Definition
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2.1 Complex Functions 55 respective
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2.1 Complex Functions 57 Focus on C
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2.2 Complex Functions as Mappings 5
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-4 -4 -3 C 2 3 4 (a) The vertical l
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4 2 -6 -4 -2 2 4 6 -2 -4 v Figure 2
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3i 2i i S z 1 2 3 S′ T(z) Figure
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ar Figure 2.12 Magnification C C′
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Note: The order in which you perfor
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2.3 Linear Mappings 75 triangle S4
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2.3 Linear Mappings 77 In Problems
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2.3 Linear Mappings 79 36. (a) Give
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2θ r θ r 2 Figure 2.17 The mappin
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-3 y 3 2 1 -2 -1 1 2 3 -1 -2 -3 (a)
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y 2 1.5 1 0.5 -2 -1.5 -1 -0.5 -0.5
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Note ☞ 2.4 Special Power Function
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Solve the equation z = f(w) for w t
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Remember: Arg(z) is in the interval
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S y (a) A circular sector 3π/8 v 3
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B y y A w = z 2 x (a) A maps onto A
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3π 2π π 0 -π -2π -3π -1 0 1 -
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2.4 Special Power Functions 99 43.
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z c(z) Figure 2.41 Complex conjugat
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w = 1/z Figure 2.43 The reciprocal
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2.5 Reciprocal Function 105 of the
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2.5 Reciprocal Function 107 underst
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2.5 Reciprocal Function 109 24. Acc
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L y y = L + ε y = L - ε y = f(x)
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2.6 Limits and Continuity 113 Crite
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2.6 Limits and Continuity 115 Befor
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2.6 Limits and Continuity 117 Theor
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2.6 Limits and Continuity 119 See (
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-1 y z = e iθ Figure 2.54 Figure f
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2.6 Limits and Continuity 123 It th
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2.6 Limits and Continuity 125 While
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y Values of arg decreasing Values o
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2.6 Limits and Continuity 129 In Pr
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2.6 Limits and Continuity 131 In Pr
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-4 -4 -3 -2 (-2, -1) 4 3 2 1 y -1 1
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Note: Normalized vector fields shou
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4 2 y -4 -2 2 4 -2 -4 Figure 2.63 S
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Chapter 2 Review Quiz 139 10. The l
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3 2 1 0 -1 -2 -3 -3 -2 -1 0 1 2 3 L
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3.1 Differentiability and Analytici
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☞ 3.1 Differentiability and Analy
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3.2 Cauchy-Riemann Equations 153 We
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3.2 Cauchy-Riemann Equations 155 EX
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3.2 Cauchy-Riemann Equations 157 EX
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3.3 Harmonic Functions 159 then sol
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3.3 Harmonic Functions 161 EXAMPLE
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3.3 Harmonic Functions 163 In Probl
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3.4 Applications 165 tangent is the
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Lines of force ψ = c2 Lower potent
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In Section 4.5, for purposes that w
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y a b φ = k 1 x φ = k 0 Figure 3.
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Chapter 3 Review Quiz 173 11. The C
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176 Chapter 4 Elementary Functions
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Normalized velocity vector field fo
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5.1 Real Integrals 237 3. Let �P
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y π t = gives 2 (0,4) C t = 0 give
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(-1, -1) y C (2, 8) x Figure 5.5 Gr
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5.1 Real Integrals 243 denotes the
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5.2 Complex Integrals 245 In Proble
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z 0 C z k * z 1 * z 2 * z 2 z 1 z k
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These properties are important in t
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5.2 Complex Integrals 251 Now in vi
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5.2 Complex Integrals 253 Proof It
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y 1 + i 1 Figure 5.21 Figure for Pr
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D C Figure 5.26 Simply connected do
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D D A C C (a) (b) B C 1 C 1 Figure
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C 1 C 1 C (a) C (b) Figure 5.31 Tri
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C C y Figure 5.34 Figure for Proble
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z 0 C 1 D z 1 C Figure 5.38 If f is
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5.4 Independence of Path 267 and z(
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Be careful when using Ln z as an an
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5.4 Independence of Path 271 (ii) I
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5.5 Cauchy’s Integral Formulas an
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3i -3i y Figure 5.44 Contour for Ex
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y i 0 C 2 C 1 Figure 5.45 Contour f
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5.6 Applications 285 A B A B A B (a
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5.6 Applications 287 Indeed, by rew
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Note ☞ 5.6 Applications 289 If yo
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-2 -1 2 1 -1 -2 y Figure 5.51 Zero
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394 Chapter 7 Conformal Mappings π
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396 Chapter 7 Conformal Mappings Re
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398 Chapter 7 Conformal Mappings 15
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400 Chapter 7 Conformal Mappings De
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402 Chapter 7 Conformal Mappings Th
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404 Chapter 7 Conformal Mappings v
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406 Chapter 7 Conformal Mappings Re
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408 Chapter 7 Conformal Mappings No
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410 Chapter 7 Conformal Mappings y
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412 Chapter 7 Conformal Mappings x
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414 Chapter 7 Conformal Mappings A
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416 Chapter 7 Conformal Mappings fo
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418 Chapter 7 Conformal Mappings EX
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420 Chapter 7 Conformal Mappings
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428 Chapter 7 Conformal Mappings (a
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436 Chapter 7 Conformal Mappings φ
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438 Chapter 7 Conformal Mappings ψ
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444 Chapter 7 Conformal Mappings In
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448 Chapter 7 Conformal Mappings In
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Appendixes 1
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Appendix II Proof of the Cauchy-Gou
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Integrals � � � FE , GF , EG
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Appendix I Proof of Theorem 2.1 APP
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Appendix III Table of Conformal Map
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Answers to Selected Odd-Numbered Pr
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Indexes 1
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Word Index IND-7 Convergence of an
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Word Index IND-5 Cauchy-Riemann equ
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Symbol Index IND-3 z α , 194 sin z
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Word Index IND-9 principal square r
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IND-14 Word Index partial sums of,
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IND-12 Word Index Partial fractions
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Word Index IND-15 mapping by, 206-2
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Complex Analysis - Maths KU

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