Math 411: Honours Complex Variables - University of Alberta

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

32 CHAPTER 5. CAUCHY’S INTEGRAL THEOREM AND FORMULA has the initial point |z| and the endpoint z. It follows that [1,|z|]⊕γ is curve with initial point 1 and endpoint z as shown: It follows that � Logz := [1,|z|] y θ [1,z] 1 [1,|z|] |z| � � θ 1 1 |z|e dζ + dζ = log|z|+i ζ γ ζ 0 it dt = log|z|+iArgz. |z|eit Lemma 5.1. Let D ⊂ C be open and star shaped with center z0, and let f : D → C be continuous such that f|D\{z0} is holomorphic. Then f has an antiderivative on D. Proof. Let ∆ be a triangle in D having a vertex z0: z2 z0 z1 w Let z1 be an interior point of [z0,w], let z2 be an interior point of [z0,z], and let z3 be an interior point of [w,z]. As shown above, we use these points to split ∆ into z z3 γ x z
four subtriangles, denoted by ∆(z0,z1,z2), ∆(z1,z3,z2), ∆(z1,w,z3), and ∆(z2,z3,z). As in the proof of Goursat’s Lemma, we have � � � � � f = f + f + f + f. ∂∆ ∂∆(z0,z1,z2) ∂∆(z1,z3,z2) ∂∆(z1,w,z3) ∂∆(z2,z3,z) Since ∆(z1,z3,z2),∆(z1,w,z3),∆(z2,z3,z) ⊂ D\{z0}, and since f is holomorphic on D \{z0}, Goursat’s Lemma yields � � � f = f = f = 0, ∂∆(z1,z3,z2) so that � It follows that �� ∂∆ � � f� � = �� ∂∆(z0,z1,z2) ∂∆ ∂∆(z1,w,z3) � f = ∂∆(z0,z1,z2) ∂∆(z2,z3,z) f. � � f� � ≤ ℓ(∂∆(z0,z1,z2)) sup |f(ζ)|. ζ∈∂∆(z0,z1,z2) Since|f|iscontinuouson∆, itisboundedabovebysomeM > 0. Byplacingz1 andz2 sufficiently close to z0, we see that ℓ(∂∆(z0,z1,z2)) can be made smaller than every ǫ/M > 0. We deduce that � f = 0. The result then follows from Theorem 5.2. ∂∆ Let z0 ∈ C, and let r > 0. Slightly abusing notation, we use ∂Br(z0) to denote the boundary of Br(z0) oriented counterclockwise. Lemma 5.2. Let D ⊂ C be open, let z0 ∈ D, and let r > 0 be such that Br[z0] ⊂ D. Then � 1 dζ = 2πi ζ −z for all z ∈ Br(z0). ∂Br(z0) Proof. Through direct computation, we saw on pg. 24 that � 1 dζ = 2πi. ζ −z0 ∂Br(z0) Let z ∈ Br(z0), and choose ǫ > 0 such that Bǫ[z] ⊂ Br(z0), so that � 1 dζ = 2πi. ζ −z We need to show that � � 1 dζ = ζ −z ∂Br(z0) ∂Bǫ(z) ∂Bǫ(z) � 1 dζ = − ζ −z ∂Bǫ(z) − 1 ζ −z dζ. 33
Page 1 and 2: Math 411: Honours Complex Variables
Page 3 and 4: Contents 1 The Complex Numbers 5 2
Page 5 and 6: Chapter 1 The Complex Numbers Defin
Page 7 and 8: Proof. (i): If z = x+iy, then ¯z =
Page 9 and 10: Chapter 2 Complex Differentiation D
Page 11 and 12: (i) there exists c ∈ C such that
Page 13 and 14: Example. Let Then f is totally diff
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
Page 23 and 24: Definition. The length of a piecewi
Page 25 and 26: 1. Given a < b < c and two curves
Page 27 and 28: That is, F ′ (z) = f(z). Theorem
Page 29 and 30: ∆ (4) ∆ (1) ∆ (2) As the line
Page 31: Proof. Let z0 ∈ D be a center for
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 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 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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