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Why is it that $$ f(z)=\frac{1}{2\pi i}\oint_{\partial D(0,1)}\frac{\overline\zeta}{\zeta-z}d\zeta=0,\forall z\in D(0,1) ? $$

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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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  • $\begingroup$ Great! This is a very smart solution. Aleks, thank you very much. Intelligence is unlimited! $\endgroup$ – Sam Oct 19 '12 at 7:10
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We may parametrize the circle as $t\mapsto e^{it}$, $0<t<2\pi$, then $$f(z)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{e^{-it}}{e^{it}-z}ie^{it}dt=\frac{1}{2\pi }\int_0^{2\pi}\frac{1}{e^{it}-z}dt =\frac{1}{2\pi }\int_0^{2\pi}\sum_{n\geq0} z^ne^{-(n+1)it}dt=0$$ since $|z|<1$ (which by the M-test ensures that the last sum converges uniformly, and we may switch the order of integration and summation).

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  • $\begingroup$ Thank you very much, AD. This is a very nice and clear proof. Especially I like the series expansion. $\endgroup$ – Sam Oct 19 '12 at 6:46
  • $\begingroup$ I am happy to help. $\endgroup$ – AD. Oct 19 '12 at 6:47

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