Manipulation of Cauchy's Integral Formula 
$\quad$ Using Cauchy's integral theorem, write down the value of a holomorphic function $f(z)$ where $|z|\lt1$ in terms of a contour integral around the unit circle, $\zeta=e^{i\theta}$.  $\quad$ By considering the point $1/\overline z$, or otherwise, show that $$f(z)=\dfrac1{2\pi}\int_0^{2\pi}f(\zeta)\dfrac{1-|z|^2}{|\zeta-z|^2}\,\mathrm d\theta.$$

Any help with the above question would be most appreciated!
It is the first part of a (so-called) long / Section II question - given this part, I can (and have) completed the rest of the question.
It is very fiddly and I just can't quite get the final answer out correctly.
(I have written out Cauchy's integral formula for the first part shown, just the final manipulation is what I need help with.)
Thanks very much for your time! :)
PS - if this is a duplicate, then I'm sorry, but this can't really be given a unique title, and I'm not going to check every single question on Cauchy's integral formula! :P
 A: Start with the usual integral formula, and expand $d\zeta$ using $\zeta = e^{i\vartheta}$:
$$f(z) = \frac{1}{2\pi i}\int_{\lvert \zeta\rvert = 1} \frac{f(\zeta)}{\zeta-z}\,d\zeta = \frac{1}{2\pi i}\int_0^{2\pi} \frac{f(\zeta)}{\zeta-z}i\zeta\,d\vartheta = \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{(\zeta-z)\overline{\zeta}}\,d\vartheta.\tag{1}$$
Now suppose $f$ is holomorphic in a neighbourhood of the closed unit disk (the general case follows by taking a limit), fix $w$ with $\lvert w\rvert < 1$, and apply $(1)$ to the function
$$z \mapsto \frac{f(z)}{1 - z\overline{w}}.$$
You obtain
$$\begin{align}
\frac{f(z)}{1-z\overline{w}} &= \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{(1-\zeta\overline{w})\overline{\zeta}(\zeta-z)}\,d\vartheta\\
&= \frac{1}{2\pi} \int_0^{2\pi} \frac{f(\zeta)}{(\overline{\zeta}-\overline{w})(\zeta-z)}\,d\vartheta.
\end{align}$$
Choosing $z = w$ gives
$$\frac{f(w)}{1-\lvert w\rvert^2} = \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{\lvert \zeta-w\rvert^2}\,d\vartheta,$$
which after multiplication with $1-\lvert w\rvert^2$ and renaming $w$ to $z$ becomes the desired
$$f(z) = \frac{1}{2\pi}\int_0^{2\pi} f(\zeta)\frac{1-\lvert z\rvert^2}{\lvert\zeta-z\rvert^2}\,d\vartheta.$$
