Math 411: Honours Complex Variables - University of Alberta

70 CHAPTER 11. THE GENERAL CAUCHY INTEGRAL THEOREM
Then h1 is holomorphic. For z ∈ D ∩extγ, we note that
�
h0(z) = g(ζ,z)dζ
γ
�
f(ζ)−f(z)
= dζ
γ ζ −z
� �
f(ζ) 1
= dζ −f(z)
γ ζ −z γ ζ −z dζ
� �� �
=0
�
f(ζ)
=
ζ −z dζ
Define
γ
= h1(z).
h: D∪extγ, z ↦→
� h0(z), z ∈ D,
h1(z), z ∈ extγ.
Then h is holomorphic. Since γ is homologous to zero, we have C \ D ⊂ extγ.
Hence, h is entire.
For any z ∈ extγ, we have the estimate
|h(z)|= |h1(z)|≤
ℓ(γ)
dist(z,{γ}) sup|f(ζ)|.
(∗)
ζ∈{γ}
Let R > 0 be such that C \ BR(0) ⊂ extγ. Since (∗) implies that h is bounded
on C \ BR(0) and h is trivially bounded by continuity on BR[0], we see that h is
bounded on C and hence constant by Liouville’s Theorem. From (∗) again, we see
that lim |h(z)|= 0. Hence, h ≡ 0.
|z|→∞
In summary, we have for z ∈ D \{γ} that
�
0 = h(z) = h0(z) =
γ
f(ζ)−f(z)
ζ −z
�
dζ =
γ
f(ζ)
dζ −2πiν(γ,z)f(z).
ζ −z
Theorem 11.2 (Cauchy’s Integral Theorem). Let D ⊂ C be open, let f : D → C
be
�
holomorphic, and let γ be a closed curve in D that is homologous to zero. Then
f(ζ)dζ = 0.
γ
Proof. Let z0 ∈ D \{γ} be arbitrary, and define
so that
g: D → C, z ↦→ (z −z0)f(z),
�
0 = 2πiν(γ,z0)g(z0) =
γ
�
g(ζ)
dζ = f(ζ)dζ.
ζ −z0 γ