Math 411: Honours Complex Variables - University of Alberta

four subtriangles, denoted by ∆(z0,z1,z2), ∆(z1,z3,z2), ∆(z1,w,z3), and ∆(z2,z3,z).
As in the proof of Goursat’s Lemma, we have
� � � � �
f = f + f + f + f.
∂∆
∂∆(z0,z1,z2)
∂∆(z1,z3,z2)
∂∆(z1,w,z3)
∂∆(z2,z3,z)
Since ∆(z1,z3,z2),∆(z1,w,z3),∆(z2,z3,z) ⊂ D\{z0}, and since f is holomorphic on
D \{z0}, Goursat’s Lemma yields
� � �
f = f = f = 0,
∂∆(z1,z3,z2)
so that �
It follows that
��
�
�
�
∂∆
�
�
f�
� =
��
�
�
�
∂∆(z0,z1,z2)
∂∆
∂∆(z1,w,z3)
�
f =
∂∆(z0,z1,z2)
∂∆(z2,z3,z)
f.
�
�
f�
� ≤ ℓ(∂∆(z0,z1,z2)) sup |f(ζ)|.
ζ∈∂∆(z0,z1,z2)
Since|f|iscontinuouson∆, itisboundedabovebysomeM > 0. Byplacingz1 andz2
sufficiently close to z0, we see that ℓ(∂∆(z0,z1,z2)) can be made smaller than every
ǫ/M > 0. We deduce that �
f = 0. The result then follows from Theorem 5.2.
∂∆
Let z0 ∈ C, and let r > 0. Slightly abusing notation, we use ∂Br(z0) to denote
the boundary of Br(z0) oriented counterclockwise.
Lemma 5.2. Let D ⊂ C be open, let z0 ∈ D, and let r > 0 be such that Br[z0] ⊂ D.
Then �
1
dζ = 2πi
ζ −z
for all z ∈ Br(z0).
∂Br(z0)
Proof. Through direct computation, we saw on pg. 24 that
�
1
dζ = 2πi.
ζ −z0
∂Br(z0)
Let z ∈ Br(z0), and choose ǫ > 0 such that Bǫ[z] ⊂ Br(z0), so that
�
1
dζ = 2πi.
ζ −z
We need to show that
� �
1
dζ =
ζ −z
∂Br(z0)
∂Bǫ(z)
∂Bǫ(z)
�
1
dζ = −
ζ −z ∂Bǫ(z) −
1
ζ −z dζ.
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