64 CHAPTER 9. HOLOMORPHIC FUNCTIONS ON ANNULI (a) On A0,1(0): For |z|< 1, we have and, for |z|< 3, 1 3−z = We thus have for z ∈ A0,1(0) that f(z) = (b) On A1,3(0): For |z|> 1, we have 1 1−z so that, for z ∈ A1,3(0): 1 1−z = n� z n n=0 1 3 � 1− z � = 3 1 3 ∞� n=0 ∞� n=0 � z �n . 3 � 1− 1 3n+1 � z n . 1 1 = − = − z −1 z � 1− 1 ∞� 1 � = − z z n+1, f(z) = − (c) On A3,∞(0): For |z|> 3, we have − 1 3−z and thus, for z ∈ A3,∞(0): � ∞� n=1 = 1 z −3 = f(z) = 1 + zn ∞� n=0 1 z � 1− 3 � = z n=0 zn 3n+1 � . ∞� (3 n−1 −1) 1 zn. n=1 ∞� n=0 3 n z n+1
Chapter 10 The Winding Number <strong>of</strong> a Curve Definition. Let γ be a closed curve in C, and let z ∈ C \ {γ}. Then the winding number <strong>of</strong> γ with respect to z is defined as ν(γ,z) := 1 � 1 2πi ζ −z dζ. Remark. Geometrically, ν(γ,z) is the number <strong>of</strong> times γ winds around z in the counterclockwise direction. Lemma 10.1. Let γ: [0,1] → C be a curve, and let z ∈ C \{γ}. Then there exist open discs D1,...,Dn ⊂ C\{z} and a partition 0 = t0 < t1 < ··· < tn = 1 such that γ([tj−1,tj]) ⊂ Dj for j = 1,...,n. Pro<strong>of</strong>. Let ǫ := dist(z,{γ}) > 0. Since γ is uniformly continuous, there exists δ > 0 such that |γ(t)−γ(t ′ )|< ǫ for all t,t ′ ∈ [0,1] such that |t−t ′ |< δ. Choose 0 = t0 < t1 < ··· < tn = 1 such that |tj−1−tj|< δ for j = 1,...,n, and set Dj := Bǫ(γ(tj)) for j = 1,...,n. Bythechoice <strong>of</strong>ǫ, itisclearthatD1,...,Dn ⊂ C\{z}. Forj = 1,...,n, let t ∈ [tj−1,tj], and note that |t−tj|≤ |tj−1 −tj|< δ, so that |γ(t)−γ(tj)|< ǫ, i.e. γ(t) ∈ Dj; consequently, γ([tj−1,tj]) ⊂ Dj holds. Proposition 10.1. Let γ be a closed curve in C, and let z ∈ C \ {γ}. Then ν(γ,z) ∈ Z. Pro<strong>of</strong>. Choose a partition 0 = t0 < t1 < ··· < tn = 1 and open discs D1,...,Dn such that γ([tj−1,tj]) ⊂ Dj for j = 1,...,n. Let j ∈ {1,...,n}. Since z /∈ Dj, there exists a holomorphic function Lj: Dj → C such that e Lj(w) = w−z for w ∈ Dj. On noting that γ(tn) = γ(t0), it is convenient to denote Dn+1 := D1 and Ln+1 := L1. For j = 1,...,n we then see that γ(tj) ∈ Dj ∩Dj+1 and hence exp(Lj(γ(tj))) = γ(tj)−z = exp(Lj+1(γ(tj))), 65 γ
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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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- Page 15 and 16: holds over C. We obtain: � ∞�
- Page 17 and 18: Figure 3.2: Surface plot of cos(z)
- Page 19 and 20: It follows that {an(z−z0) n } ∞
- Page 21 and 22: Problem 3.1. Show in Theorem 3.2 th
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- Page 25 and 26: 1. Given a < b < c and two curves
- Page 27 and 28: That is, F ′ (z) = f(z). Theorem
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- Page 35 and 36: γ1 Then it is clear that � � 1
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- Page 39 and 40: Proof. Apply Theorem 5.3 and 5.4. E
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- Page 43 and 44: (ii) for each compact K ⊂ D, the
- Page 45 and 46: Proof. Let ζ ∈ ∂Br(z0), and no
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- 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
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- Page 73 and 74: Examples. 1. Let f(z) = eiz z 2 +1
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- 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 )
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By the Open Mapping Theorem, there
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Proof. (i) =⇒ (ii) is Corollary 1
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Index C is a Field, 5 Biholomorphis
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INDEX 121 straight line segment, 25