104 CHAPTER 15. ANALYTIC CONTINUATION ALONG A CURVE Since fj is an antiderivative <strong>of</strong> g on Dj for j = 1,2,3, we obtain � γz 1 ,z 2 It follows that g = f1(z2)−f1(z1), � c = f3(z1)−f1(z1) γz 2 ,z 3 g = f2(z3)−f2(z2), and � γz 3 ,z 1 = f3(z1)−f3(z3)+f2(z3)−f2(z2)+f1(z2)−f1(z1) � � � = g + g + g � = γz 3 ,z 1 � 1 = ∂D ζ dζ = 2πi. z3 γz 2 ,z 3 γz 1 ,z 2 ⊕γz 2 ,z 3 ⊕γz 3 ,z 1 g D2 D3 y γz 1 ,z 2 z2 z1 D1 g = f3(z1)−f3(z3). Definition. A function elementis a pair (D,f), where D ⊂ C is open and connected, and f : D → C is a holomorphic function. For a given function element (D,f) and x
z0 ∈ D, the germ <strong>of</strong> f at z0—denoted by 〈f〉z0—is the collection <strong>of</strong> all function elements (E,g) such that z0 ∈ E and there is a neighbourhood U ⊂ D∩E <strong>of</strong> z0 such that f(z) = g(z) for all z ∈ U. Definition. Let γ: [0,1] → C be a path, and suppose that, for each t ∈ [0,1], there is a function element (Dt,ft) such that: (a) γ(t) ∈ Dt; (b) there exists δ > 0 such that, whenever s ∈ [0,1] is such that |s − t|< δ, then γ(s) ∈ Dt and 〈fs〉γ(s) = 〈ft〉γ(s). Then we call {(Dt,ft) : t ∈ [0,1]} an analytic continuation along γ and say that (D1,f1) is obtained by analytic continuation <strong>of</strong> (D0,f0) along γ. Remark. Since γ is continuous and Dt is open for each t ∈ Dt, it is clear that there exists δ > 0 such that γ(s) ∈ Dt for all s ∈ [0,1] such that |s − t|< δ. What is important about part (b) <strong>of</strong> the definition is that 〈fs〉γ(s) = 〈ft〉γ(s), i.e. there is a neighbourhood Us ⊂ Ds ∩Dt <strong>of</strong> γ(s) such that fs(z) = ft(z) for z ∈ Us. γ(0) D0 γ Theorem 15.1 (Monodromy Theorem). Let γ : [0,1] → C be a path, and let {(Dt,ft) : t ∈ [0,1]} and {(Et,gt) : t ∈ [0,1]} be analytic continuations along γ such that 〈f0〉γ(0) = 〈g0〉γ(0). Then we have 〈f1〉γ(1) = 〈g1〉γ(1). D1 γ(1) 105
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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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- Page 55 and 56: Define g: C → C, z ↦→ ∞�
- Page 57 and 58: Chapter 9 Holomorphic Functions on
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- Page 61 and 62: Definition. The function h in Theor
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- Page 65 and 66: Chapter 10 The Winding Number of a
- Page 67 and 68: Proposition 10.2 (Winding Numbers A
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- Page 73 and 74: Examples. 1. Let f(z) = eiz z 2 +1
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- 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: Chapter 15 Analytic Continuation al
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- Page 109 and 110: Define g: D → RM as follows: for
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- Page 117 and 118: Proof. (i) =⇒ (ii) is Corollary 1
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