greene and krantz, function of one complex variable, Theorem 3.1.3 Why is it that
$$
f(z)=\frac{1}{2\pi i}\oint_{\partial D(0,1)}\frac{\overline\zeta}{\zeta-z}d\zeta=0,\forall z\in D(0,1) ?
$$
 A: 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).
A: 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.
