Math 411: Honours Complex Variables - University of Alberta

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

Chapter 17 The Riemann Mapping Theorem Definition. Let D1,D2 ⊂ C be open and connected. We say that D1 and D2 are biholomorphically equivalent if there is a biholomorphic map from D1 onto D2. Examples. 1. Let z1,z2 ∈ C, and let r1,r2 > 0. Then Br1(z1) and Br2(z2) are biholomorphically equivalent because is biholomorphic. 2. Consider the Cayley transform Br1(z1) → Br2(z2), z ↦→ r2 (z −z1)+z2 f: H → C, z ↦→ r1 z −i z +i . Let x,y ∈ R with y > 0, and let z = x+iy. Then |z −i| 2 = |x+i(y −1)| 2 = x 2 +y 2 −2y +1 < x 2 +y 2 +2y +1 = |x+i(y +1)| 2 = |z +i| 2 holds, so that |f(z)|< 1. Consequently, we have f(H) ⊂ D. Consider g: D → C, z ↦→ i 1+z 1−z , 110
and note that z−i 1+ z+i g(f(z)) = i 1− z−i z+i z +i+z −i = i z +i−z +i = i 2z 2i = z for z ∈ H. Hence, f is injective. Let x 2 +y 2 < 1, and note that g(x+iy) = i (1+x)+iy (1−x)−iy For z ∈ D, we can thus evaluate = i [(1+x)+iy][(1−x)+iy] (1−x) 2 +y2 2y = − (1−x) 2 +y2 +i 1−(x2 +y2 ) (1−x) 2 +y2 ∈ H. � �� >0 f(g(z)) = i1+z 1−z −i i 1+z 1−z +i = 1+z −1+z = 2z 2 = z. 1+z +1−z 111 Hence, f is also surjective and thus bijective with inverse g. Since f and g are obviously holomorphic, this means that D and H are biholomorphically equivalent. 3. Thereisnobiholomorphicmapf: C → Dbecause anyholomorphicmapfromC to D is bounded and thus constant by Liouville’s theorem. Hence, C and D are not biholomorphically equivalent.
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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
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Proof. Let z0 ∈ D be a center for
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four subtriangles, denoted by ∆(z
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γ1 Then it is clear that � � 1
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Proof. There is no loss of generali
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Proof. Apply Theorem 5.3 and 5.4. E
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is entire. Since p is a nonconstant
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(ii) for each compact K ⊂ D, the
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Proof. Let ζ ∈ ∂Br(z0), and no
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holds for some a0,a1,a2,... ∈ C a
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Proof. Let z0 ∈ D be such that |f
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Theorem 7.5 (Biholomorphisms of D).
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Define ⎧ ⎪⎨ g: D ∪{z0} →
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Define g: C → C, z ↦→ ∞�
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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 and 82: 12.2. THE GAMMA FUNCTION 81 12.2 Th
Page 83 and 84: 12.2. THE GAMMA FUNCTION 83 Since
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: Define g: D → RM as follows: for
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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