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 14 Harmonic Functions Definition. Let D ⊂ R N be open, and let u: D → R be twice continuously partially differentiable. Then u is called harmonic if ∆u = ∂2 u ∂x 2 1 +···+ ∂2 u ∂x 2 N ≡ 0. We will only be concerned with harmonic functions on R 2 , i.e. on C. Proposition 14.1 (Harmonic Components). Let D ⊂ C be open, and let f: D → C be holomorphic. Then Ref and Imf are harmonic. Proof. Clearly, Ref and Imf are twice continuously differentiable. We have ∂2 (Ref) ∂ ∂ = ∂x2 ∂x ∂x Ref = ∂ ∂ Imf, by Cauchy–Riemann, ∂x∂y = ∂ ∂ ∂y ∂x Imf = − ∂2 (Ref) ∂y2 , by Cauchy–Riemann again, so that ∆Ref ≡ 0, i.e. Ref is harmonic. Similarly, one sees that Imf is harmonic. Remark. The converse of Proposition 14.1 is not true: a harmonic function need not be the real part of some holomorphic function. Consider so that u: C\{0} → R, z ↦→ log|z|, u(x,y) = log � x 2 +y 2 = 1 2 log(x2 +y 2 ) 94
for (x,y) ∈ R 2 \{(0,0)}. The partial derivatives of u with respect to x are and ∂u ∂x = x x 2 +y 2. ∂2u ∂x2 = (x2 +y2 )−2x 2 (x2 +y2 ) 2 Moreover, since u is symmetric in x and y, we find ∂ 2 u ∂y 2 = −y2 +x 2 (x 2 +y 2 ) 2. = −x2 +y 2 (x 2 +y 2 ) 2 Consequently, u is harmonic. Now suppose that there is a holomorphic function f : C \ {0} → C such that Ref = u. On C−, we then have that Ref = log|z|= ReLog. The Cauchy–Riemann Equations thus yield so that ∂(Imf) (z) = − ∂x ∂Ref ∂y (z) = −∂(ReLog) ∂y (z) = ∂(ImLog) (z), ∂x f ′ (z) = ∂Ref ∂x (z)+i∂(Imf) (z) = ∂x ∂ReLog (z)+i ∂x ∂(ImLog) (z) = Log ∂x ′ z = 1 z for z ∈ C−. By continuity, it follows that f ′ (z) = 1 for all z ∈ C\{0}, so that f is z an antiderivative of z ↦→ 1 on C\{0}. This is impossible (cf. page 24). z Definition. Let D ⊂ C be open, and let u: D → R be harmonic. We call a harmonic function v: D → R a harmonic conjugate of u if u+iv is holomorphic. Theorem 14.1 (Harmonic Conjugates). Let D ⊂ C be open and suppose that there exists (x0,y0) ∈ D with the following property: for each (x,y) ∈ D, we have • (x,t) ∈ D for each t between y and y0 and • (s,y0) ∈ D for each s between x and x0. Then every harmonic function on D has a harmonic conjugate. Proof. Let u: D → R be harmonic. We will find a harmonic v: D → R such that For (x,y) ∈ D, define ∂u ∂x = ∂v ∂y and ∂v ∂x � y ∂u v(x,y) = ∂x (x,t)dt+φ(x), y0 95 = −∂u. (∗) ∂y
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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
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 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: Proof. Let p(z) = anz n +···+a1z
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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