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Math 411: Honours Complex Variables - University of Alberta

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• if w �= z:<br />

�<br />

�<br />

|g(w,z)−g(z0,z0)| = �<br />

f(w)−f(z)<br />

� −f<br />

w −z<br />

′ �<br />

�<br />

(z0) �<br />

�<br />

� �<br />

�<br />

= �<br />

1<br />

�w−z<br />

Thus, g is continuous.<br />

Next, define<br />

[f<br />

[z,w]<br />

′ (ζ)−f ′ (z0)]dζ<br />

�<br />

h0: D → C, z ↦→<br />

γ<br />

�<br />

�<br />

�<br />

�<br />

g(ζ,z)dζ.<br />

69<br />

≤ sup |f<br />

ζ∈{[z,w]}<br />

′ (ζ)−f ′ (z0)|≤ ǫ.<br />

We claim that h0 is holomorphic. It is easy to see that h0 is continuous. To see that<br />

it is indeed holomorphic, we shall show that it satisfies the Morera condition. Let<br />

∆ ⊂ D be a triangle. For fixed ζ ∈ {γ}, the function<br />

D → C, z ↦→ g(ζ,z)<br />

isholomorphicasaconsequence<strong>of</strong>Riemann’sRemovabilityCondition. Goursat’s Lemma<br />

thus yields<br />

�<br />

g(ζ,z)dz = 0<br />

∂∆<br />

for each ζ ∈ {γ}. As a consequence, we find<br />

�<br />

0 =<br />

�<br />

=<br />

�<br />

=<br />

γ<br />

∂∆<br />

∂∆<br />

so that h0 is holomorphic as claimed.<br />

Define<br />

��<br />

∂∆<br />

��<br />

γ<br />

�<br />

g(ζ,z)dz dζ<br />

�<br />

g(ζ,z)dζ dz<br />

h0(z)dz,<br />

�<br />

h1: extγ → C, z ↦→<br />

γ<br />

f(ζ)<br />

ζ −z dζ.

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