Math 411: Honours Complex Variables - University of Alberta

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$Math 527 A1 Homework 5 (Due Nov. 26 in Class) Proof. i. Set u = v(x ...$

82 CHAPTER 12. THE RESIDUE THEOREM AND APPLICATIONS In particular, we see for α = 1/2 that Γ2 (1/2) = 2 � π/2 dθ = π and 0 � ∞ e −∞ −x2 � ∞ dx = 2 e 0 −x2 � � 1 dx = Γ = 2 √ π. A substitution then leads to the important result � ∞ a > 0. −∞ e−ax2 For arbitrary α ∈ (0,1), we find, on substituting z = tan 2 θ, � π/2 I(α) := Γ(α)Γ(1−α) = 2 tan 2α−1 θdθ = 0 + � ∞ 0 + dx = � π/a for z α−1 1+z dz. The integral here can be evaluated by a contour integration in the complex plane, noting that the function z α−1 = e (α−1)logz is holomorphic on the starshaped domain obtained by slicing the complex plane along the positive real axis. This branch cut is shown in red in the following figure. In other words we choose the antiderivative logz = log|z|+iargz of the function z ↦→ 1/z, where argz ∈ [0,2π). CR −1 Imz ir Cr iR Rez Here the large circular contour CR is chosen to have radius R ≥ 2, so that |1+z| ≥ R/2 on CR, and the small semicircular contour Cr is chosen to have radius r ≤ 1/2, so that |1+z| ≥ 1/2 on Cr. On denoting f(z) := zα−1 1+z e(α−1)logz = , 1+z we then see, accounting for the residue from the pole of f at z = −1, that 2πie (α−1)iπ � R+ir � = f + ir CR f + � −ir R−ir � f + Cr f.
12.2. THE GAMMA FUNCTION 83 Since α < 1, we see that the contribution from the circular arc CR is �� f� � CR ≤ Rα−1 R 2 ·2πR = 4πR α−1 → R→∞ 0. Likewise, since α > 0, the contribution from the semicircular contour Cr is �� f� � We thus deduce that Thus Cr 2πie (α−1)iπ = lim r→0 ≤ rα−1 1 2 R→∞ ir � ∞ e (α−1)log|z| ·πr = 2πr α → r→0 0. �� R+ir � R−ir f − = 0 + dz − 1+z = I(α) � 1−e (α−1)2πi� . � f −ir � ∞ e (α−1)(log|z|+i2π) 0 + 1+z π = I(α)· e−(α−1)πi −e (α−1)πi = I(α)· 2i −e−απi +eαπi , 2i from which we see that I(α) = Γ(α)Γ(1−α) = π sinπα , On extending this result by analytic continuation, one finds for all z ∈ C \ Z that Γ(z)Γ(1−z) = π sinπz . • For α ≥ 1 and positive x and λ, another frequently encountered integral can be expressed in terms of Γ using the substitution u = xtλ : � � ∞ 0 e −xtλ t α−1 dt = 1 λx α λ 0 � ∞ 0 e −u u α λ−1 du = Γ� α λ λx α λ For the special case α = x = 1, this result simplifies to � ∞ e −tλ dt = 1 λ Γ � � � 1 = Γ 1+ λ 1 � . λ For 0 < α < 1 and x > 0, a related integral is � ∞ 0 e ixt t α−1 dt = iα Γ(α) x α . dz . (12.3) (12.4) To establish this result, it is convenient to introduce a branch cut, shown in red, along the negative real axis:
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Math 411: Honours Complex Variables
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Contents 1 The Complex Numbers 5 2
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Chapter 1 The Complex Numbers Defin
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Proof. (i): If z = x+iy, then ¯z =
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Chapter 2 Complex Differentiation D
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(i) there exists c ∈ C such that
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Example. Let Then f is totally diff
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holds over C. We obtain: � ∞�
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Figure 3.2: Surface plot of cos(z)
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It follows that {an(z−z0) n } ∞
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Problem 3.1. Show in Theorem 3.2 th
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Definition. The length of a piecewi
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1. Given a < b < c and two curves
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That is, F ′ (z) = f(z). Theorem
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∆ (4) ∆ (1) ∆ (2) As the line
Page 31 and 32: Proof. Let z0 ∈ D be a center for
Page 33 and 34: four subtriangles, denoted by ∆(z
Page 35 and 36: γ1 Then it is clear that � � 1
Page 37 and 38: Proof. There is no loss of generali
Page 39 and 40: Proof. Apply Theorem 5.3 and 5.4. E
Page 41 and 42: is entire. Since p is a nonconstant
Page 43 and 44: (ii) for each compact K ⊂ D, the
Page 45 and 46: Proof. Let ζ ∈ ∂Br(z0), and no
Page 47 and 48: holds for some a0,a1,a2,... ∈ C a
Page 49 and 50: Proof. Let z0 ∈ D be such that |f
Page 51 and 52: Theorem 7.5 (Biholomorphisms of D).
Page 53 and 54: Define ⎧ ⎪⎨ g: D ∪{z0} →
Page 55 and 56: Define g: C → C, z ↦→ ∞�
Page 57 and 58: Chapter 9 Holomorphic Functions on
Page 59 and 60: It follows that � � f(ζ)dζ +
Page 61 and 62: Definition. The function h in Theor
Page 63 and 64: For the converse, let g: Br(z0) →
Page 65 and 66: Chapter 10 The Winding Number of a
Page 67 and 68: Proposition 10.2 (Winding Numbers A
Page 69 and 70: • if w �= z: � � |g(w,z)−
Page 71 and 72: Corollary 11.2.1. Let D be an open,
Page 73 and 74: Examples. 1. Let f(z) = eiz z 2 +1
Page 75 and 76: 12.1. APPLICATIONS OF THE RESIDUE T
Page 81: 12.2. THE GAMMA FUNCTION 81 12.2 Th
Page 85 and 86: 12.2. THE GAMMA FUNCTION 85 6 5 4 3
Page 87 and 88: Chapter 13 Function Theoretic Conse
Page 89 and 90: Lemma 13.1. Let D ⊂ C be open and
Page 91 and 92: Definition. Let D ⊂ C be open, an
Page 93 and 94: Proof. Let p(z) = anz n +···+a1z
Page 95 and 96: for (x,y) ∈ R 2 \{(0,0)}. The par
Page 97 and 98: Proof. Of course, only (ii) =⇒ (i
Page 99 and 100: so that u(z) = � 2π u(re 0 iθ )
Page 101 and 102: Set K := supζ∈∂D|f(ζ)|. For z
Page 103 and 104: Chapter 15 Analytic Continuation al
Page 105 and 106: z0 ∈ D, the germ of f at z0—den
Page 107 and 108: Chapter 16 Montel’s Theorem Defin
Page 109 and 110: Define g: D → RM as follows: for
Page 111 and 112: and note that z−i 1+ z+i g(f(z))
Page 113 and 114: Example. Let z be a complex number.
Page 115 and 116: By the Open Mapping Theorem, there
Page 117 and 118: Proof. (i) =⇒ (ii) is Corollary 1
Page 119 and 120: Index C is a Field, 5 Biholomorphis
Page 121: INDEX 121 straight line segment, 25
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Math 411: Honours Complex Variables - University of Alberta

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