Math 411: Honours Complex Variables - University of Alberta

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

12 CHAPTER 2. COMPLEX DIFFERENTIATION (ii) f is totally differentiable at z0 (in the sense of multivariable calculus), and the Cauchy–Riemann differential equations hold. Proof. (i) =⇒ (ii): Define ∂u ∂x (z0) = ∂v ∂y (z0) and T: C → C, z ↦→ f ′ (z0)z, ∂u ∂y (z0) = − ∂v ∂x (z0) and note that � |f(z)−f(z0)−T(z −z0)| � = � f(z)−f(z0) |z −z0| � −f z −z0 ′ � � (z0) � � → 0 as z → z0. Therefore, f is totally differentiable at z0. From multivariable calculus, it follows that the matrix representation of T with respect to the standard basis of R is the Jacobian of f, i.e. � � ux(z0) uy(z0) Jf(z0) = . vx(z0) vy(z0) Since T is C-linear, Lemma 2.1 yields that ux(z0) = vy(z0) and uy(z0) = −vx(z0). (ii) =⇒ (i): Since f is totally differentiable at z0, we have a unique R-linear map T : C → C such that |f(z)−f(z0)−T(z −z0)| lim = 0. z→z0 |z −z0| Asweknowfrommultivariablecalculus, T isrepresented byJf(z0)withrespect tothe standard basis of R2 . Since the Cauchy–Riemann differential equations are supposed to hold, Jf(z0) is of the form described in Lemma 2.1(iv). By Lemma 2.1, there thus exists c ∈ C such that T(z) = cz for all z ∈ C. It follows that � � � � f(z)−f(z0) � � −c� z � −z0 = |f(z)−f(z0)−c(z −z0)| → 0 |z −z0| as z → z0. Hence, f is complex differentiable at z0. Remark. In the situation of Theorem 2.1, we have f ′ � �� ux(z0) uy(z0) 1 (z0)1 = = ux(z0)+ivx(z0) vx(z0) vy(z0) 0 as well as so that f ′ (z0)i = � ux(z0) uy(z0) vx(z0) vy(z0) �� 0 1 � = uy(z0)+ivy(z0), f ′ (z0) = ux(z0)+ivx(z0) = vy(z0)−iuy(z0).
Example. Let Then f is totally differentiable, with f: C → C, z ↦→ |z| 2 . ux = 2x, uy = 2y, vx = vy = 0, noting that v = 0. The Cauchy–Riemann equations ux(z0) = vy(z0) and uy(z0) = −vx(z0) thus hold if and only if z0 = 0. By Theorem 2.1, this means that f is complex differentiable at z0 if and only if z0 = 0. Corollary 2.1.1. Let D ⊂ C be open and connected, and let f: D → C be complex differentiable. Then f is constant on D if and only if f ′ ≡ 0. Proof. Suppose that f ′ ≡ 0. From the remark after Theorem 2.1, it follows that ux = vx = uy = vy ≡ 0. Multivariable calculus then yields that f is constant. 13
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: (i) there exists c ∈ C such that
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 and 32: Proof. Let z0 ∈ D be a center for
Page 33 and 34: four subtriangles, denoted by ∆(z
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
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• if w �= z: � � |g(w,z)−
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Corollary 11.2.1. Let D be an open,
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Examples. 1. Let f(z) = eiz z 2 +1
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12.1. APPLICATIONS OF THE RESIDUE T
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Page 79 and 80:
Page 81 and 82:
12.2. THE GAMMA FUNCTION 81 12.2 Th
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12.2. THE GAMMA FUNCTION 83 Since
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12.2. THE GAMMA FUNCTION 85 6 5 4 3
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Chapter 13 Function Theoretic Conse
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Lemma 13.1. Let D ⊂ C be open and
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Definition. Let D ⊂ C be open, an
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Proof. Let p(z) = anz n +···+a1z
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for (x,y) ∈ R 2 \{(0,0)}. The par
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Proof. Of course, only (ii) =⇒ (i
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so that u(z) = � 2π u(re 0 iθ )
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Set K := supζ∈∂D|f(ζ)|. For z
Page 103 and 104:
Chapter 15 Analytic Continuation al
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z0 ∈ D, the germ of f at z0—den
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Chapter 16 Montel’s Theorem Defin
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Define g: D → RM as follows: for
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and note that z−i 1+ z+i g(f(z))
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Example. Let z be a complex number.
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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
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Math 411: Honours Complex Variables - University of Alberta

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