Why is it that $ f(z)=\frac{1}{2\pi i}\oint_{\partial D(0,1)}\frac{\overline\zeta}{\zeta-z}d\zeta=0,\forall z\in D(0,1) ? $
greene and krantz, function of one complex variable, Theorem 3.1.3
2 Answers
Here is a different approach, circumventing the use of infinite series. We know that
$\frac{\zeta^*}{\zeta-z} = \frac{\zeta\zeta^*}{\zeta(\zeta-z)} = \frac{1}{\zeta(\zeta-z)}$
for $\zeta \in \partial D(0,1)$. If $z = 0$, we have $z^{-2}$ and when we integrate via A.D.'s parametrization, we immediately obtain 0. Otherwise, by partial fractions we have
$\frac{1}{\zeta(\zeta-z)} = \frac{1}{z (\zeta -z)}-\frac{1}{\zeta z}$
And the integral becomes
$\frac{1}{2\pi iz}\int_{\partial D}\frac{1}{\zeta -z}d\zeta - \frac{1}{2\pi iz}\int_{\partial D}\frac{1}{\zeta}d \zeta$
Each of these integrals gives you $2\pi i$ and after subtracting, we have 0.
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0Great! This is a very smart solution. Aleks, thank you very much. Intelligence is unlimited! – 2012-10-19
We may parametrize the circle as $t\mapsto e^{it}$, $0
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0I am happy to help. – 2012-10-19