Math 411: Honours Complex Variables - University of Alberta

Proof. Let z0 ∈ D be a center for D. Define
�
F : D → C, z ↦→
Let z ∈ D, and let δ > 0 be such that Bδ(z) ⊂ D. Let w ∈ Bδ(z). Since [z0,z] ⊕
[z,w]⊕[w,z0] is the boundary of a triangle ∆ ⊂ D,
�
f = 0,
so that
�
F(w) =
[z0,w]
[z0,z]⊕[z,w]⊕[w,z0]
[z0,z]
� � � �
f = − f = f + f = F(z)+ f.
[w,z0] [z0,z] [z,w] [z,w]
Lemma 4.1 then implies that F is an antiderivative for f.
Corollary 5.2.1. Let D ⊂ C be open and star shaped, and let f : D → C be
holomorphic. Then f has an antiderivative.
Proof. Apply Goursat’s Lemma and Theorem 5.2.
Corollary 5.2.2. Let D ⊂ C be open, and let f : D → C be holomorphic. Then,
for each z0 ∈ D, there exists a neighbourhood U ⊂ D of z0 such that f|U has an
antiderivative.
Corollary 5.2.3 (Cauchy’s Integral Theorem for Star-Shaped Domains). Let D ⊂ C
be open and star shaped, and let f: D → C be holomorphic. Then �
f(ζ)dζ = 0 for
γ
each closed curve γ in D.
Proof. This follows from Corollary 5.2.1 and Theorem 4.1.
Example. The sliced plane is defined as
f.
C− := {z ∈ C : z /∈ (−∞,0]}.
Then C− is star shaped (1 is a center, for instance). As seen in the proof of Theorem
5.2, the function
�
Log: C− → C, z ↦→
1
ζ dζ
is an antiderivative of z ↦→ 1
z on C−; it is called the principal branch of the logarithm.
Let z ∈ C−, and let γz be any curve in C− with initial point 1 and endpoint z.
From Theorem 4.1, we conclude that �
1
dζ = Logz. γz ζ
For any z ∈ C−, there exists a unique θ ∈ (−π,π)—the principal argument Argz
of z—such that z = |z|eiθ . For z ∈ C−, the curve
[1,z]
γ: [0,θ] → C, t ↦→ |z|e it .
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