Math 411: Honours Complex Variables - University of Alberta

γ1
Then it is clear that
� �
1
dζ +
ζ −z
γ1
γ2
z0
z
�
1
dζ =
ζ −z ∂Br(z0)
z0
z
�
1
dζ +
ζ −z ∂Bǫ(z) −
γ2
1
ζ −z dζ.
The sketches also show that there exist star-shaped open set Dj ⊂ C \ {z} with
{γj} ⊂ Dj for j = 1,2. Cauchy’s integral theorem for star-shaped domains thus
yields that
for j = 1,2, which proves the claim.
�
γj
1
dζ = 0
ζ −z
Theorem 5.3 (Cauchy’s Integral Formula for Circles). Let D ⊂ C be open, let
f: D → C be holomorphic, and let z0 ∈ D and r > 0 be such that Br[z0] ⊂ D. Then
we have
for all z ∈ Br(z0).
f(z) = 1
�
f(ζ)
2πi ∂Br(z0) ζ −z dζ
Remark. A consequence of Theorem 5.3 is that the values of f on all of Br[z0] are
pre-determined by those on ∂Br(z0)!
Proof of Theorem 5.3. Since Br[z0] is compact, we may choose R > 0 be such that
Br[z0] ⊂ BR(z0) ⊂ D. Let z ∈ Br(z0), and note that z is a center for the star-shaped
domain BR(z0).
Define
g: D → C, ζ ↦→
� f(ζ)−f(z)
ζ−z , ζ �= z,
f ′ (z), ζ = z.
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