Path Integral using Cauchy's Integral Formula.

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Am revising Complex Analysis, and have come across this equality:

For $R\gt 1$ and $|a|\lt 1$ $$\frac{1}{2\pi i}\int_{\gamma (0;1)}\frac{\overline{f(z)}}{z-a} dz = \overline{\frac{1}{2\pi i}\int_{\gamma (0;1)}\frac{{f(z)}}{z(1-\overline{a}z)} dz}$$

And this supposedly follows from the Cauchy Integral Formula, that $$f(w)=\frac{1}{2\pi i}\int_{\gamma (w;r)}\frac{f(z)}{z-w}dz$$

But I just can't see how to apply this to get the desired result?

If someone could explain it to me, that would be really appreciated. Thanks.