Math 411: Honours Complex Variables - University of Alberta

Proof. Apply Theorem 5.3 and 5.4.
Example. We shall use Cauchy’s integral theorem and formula to evaluate the line
integral
�
eζ ζ(ζ −1) dζ
for various curves γ:
(a) γ = ∂Bπ(−3): The function
γ
f: C\{1} → C, z ↦→ ez
z −1
is holomorphic, and we have Bπ[−3] ⊂ C\{1}. Cauchy’s integral formula thus
yields:
�
(b) γ = ∂B1
2
∂Bπ(−3)
eζ �
f(ζ)
dζ = dζ = 2πif(0) = −2πi.
ζ(ζ −1) ∂Bπ(−3) ζ
(i): As the integrand is holomorphic in the star-shaped domain B3(i),
4
Cauchy’s integral theorem yields that
�
∂B1 (i)
2
eζ dζ = 0.
ζ(ζ −1)
(c) γ = ∂B2(0): the method of partial fractions yields
∂B2(0)
1 1 1
= −
z(z −1) z −1 z .
Since 0,1 ∈ B2(0), we obtain with the help of Cauchy’s integral formula:
�
eζ �
dζ =
ζ(ζ −1)
eζ �
dζ −
ζ −1
eζ dζ = 2πi(e−1).
ζ
∂B2(0)
∂B2(0)
Theorem 5.5 (Characterizations of Holomorphic Functions). Let D ⊂ C be open,
and let f: D → C be continuous. Then the following are equivalent:
(i) f is holomorphic;
(ii) the Morera condition holds, i.e. �
(iii) for each z0 ∈ D and r > 0 with Br[z0] ⊂ D, we have
f(z) = 1
�
f(ζ)
2πi ζ −z dζ
for z ∈ Br(z0);
∂∆
f(ζ)dζ = 0 for each triangle ∆ ⊂ D;
∂Br(z0)
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