Complex Analysis - Maths KU

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88 Chapter 2 <strong>Complex</strong> Functions and Mappings Therefore, onlythe first of these two roots of i is the value of the principal square root function. Of course, we can define other “square root” functions. For example, suppose we let θ be the unique value of the argument of z in the interval π
Solve the equation z = f(w) for w to find a formula for w = f −1 (z). z 3 z 4 y z2 z 5 2π/6 Figure 2.25 Points mapped onto the same point by f(z) =z 6 z 1 z 6 x ☞ 2.4 Special Power Functions 89 EXAMPLE 6 Inverse Function of f(z) =z +3i Show that the complex function f(z) =z +3i is one-to-one on the entire complex plane and find a formula for its inverse function. Solution One wayof showing that f is one-to-one is to show that the equality f(z1) =f(z2) implies the equality z1 = z2. For the function f(z) =z +3i, this follows immediatelysince z1 +3i = z2 +3i implies that z1 = z2. As with real functions, the inverse function of f can often be found algebraicallyby solving the equation z = f(w) for the symbol w. Solving z = w +3i for w, we obtain w = z − 3i and so f −1 (z) =z − 3i. From Section 2.3 we know that the mapping f(z) =z +3i in Example 6 is a translation by3i and that f −1 (z) =z − 3i is a translation by −3i. Suppose we represent these mappings in a single copyof the complex plane. Since translating a point by3i and then translating the image by −3i moves the original point back onto itself, we see that the mappings f and f −1 “undo” each other. Inverse Functions of z n ,n > 2 We now describe how to obtain an inverse function of the power function z n , n ≥ 2. This requires some explanation since the function f(z) =z n , n ≥ 2, is not one-to-one. In order to see that this is so, consider the points z1 = re iθ and z2 = re i(θ+2π/n) with r �= 0. Because n ≥ 2, the points z1 and z2 are distinct. That is, z1 �= z2. From (5) we have that f(z1) =r n e inθ and f(z2) =r n e i(nθ+2π) = r n e inθ e i2π = r n e inθ . Therefore, f is not one-to-one since f(z1) =f(z2) but z1 �= z2. In fact, the n distinct points z1 = re iθ , z2 = re i(θ+2π/n) , z3 = re i(θ+4π/n) , ... , zn = re i(θ+2(n−1)π/n) are all mapped onto the single point w = r n e inθ by f(z) =z n . This fact is illustrated for n = 6 in Figure 2.25. The six points z1, z2, ... , z6 with equal modulus and arguments that differ by2π/6 =π/3 shown in Figure 2.25 are all mapped onto the same point by f(z) =z 6 . The preceding discussion appears to implythat the function f(z) =z n , n ≥ 2, does not have an inverse function because it is not one-to-one. You have encountered this problem before, when defining inverse functions for certain real functions in elementarycalculus. For example, the real functions f(x) =x 2 and g(x) = sin x are not one-to-one, yet we still have the inverse functions f −1 (x) = √ x and g −1 (x) = arcsin x. The keyto defining these inverse functions is to appropriatelyrestrict the domains of the functions f(x) =x 2 and g(x) = sin x to sets on which the functions are one-to-one. For example, whereas the real function f(x) =x 2 defined on the interval (−∞, ∞) is not one-to-one, the same function defined on the interval [0, ∞) is one-to-one. Similarly, g(x) = sin x is not one-to-one on the interval (−∞, ∞), but it is one-to-one on the interval [−π/2, π/2]. The function f −1 (x) = √ x is the inverse function of the function f(x) =x 2 defined on the interval [0, ∞). Since Dom(f) =[0, ∞) and Range(f) =[0, ∞), the domain and range of f −1 (x) = √ x are both [0, ∞) as well. See Figure 2.26. In a similar manner, g −1 (x) = arcsin x is the inverse function of the function g(x) = sin x defined
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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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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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-1 -0.5 1 0.5 -0.5 -1 y 0.5 Figure
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5.6 Applications 295 In Problems 5-
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C 1 y 2i C 2 -2 2 Figure 5.58 Figur
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Chapter 5 Review Quiz 299 � z 38.
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302 Chapter 6 Series and Residues y
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304 Chapter 6 Series and Residues Y
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306 Chapter 6 Series and Residues D
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308 Chapter 6 Series and Residues E
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310 Chapter 6 Series and Residues R
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312 Chapter 6 Series and Residues 3
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314 Chapter 6 Series and Residues I
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316 Chapter 6 Series and Residues D
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318 Chapter 6 Series and Residues N
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326 Chapter 6 Series and Residues a
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328 Chapter 6 Series and Residues N
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330 Chapter 6 Series and Residues T
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334 Chapter 6 Series and Residues R
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336 Chapter 6 Series and Residues i
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338 Chapter 6 Series and Residues A
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340 Chapter 6 Series and Residues S
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344 Chapter 6 Series and Residues T
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346 Chapter 6 Series and Residues (
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360 Chapter 6 Series and Residues -
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362 Chapter 6 Series and Residues z
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364 Chapter 6 Series and Residues f
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366 Chapter 6 Series and Residues v
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368 Chapter 6 Series and Residues
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378 Chapter 6 Series and Residues C
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380 Chapter 6 Series and Residues s
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386 Chapter 6 Series and Residues P
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390 Chapter 7 Conformal Mappings y
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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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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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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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