Q. If f: $\Omega \rightarrow \Bbb C $ is a holomorphic function, $ 0 \in \Omega $ and $r > 0$ chosen such that $B_r(0) \subset \Omega $, show that for all $z\in B_r(0)$ that,
$$ f(z) = {1\over 2 \pi i} \int_{|w|=1} {f(w)\over w} {r^2-|z|^2 \over |w-z|^2} dw.$$
By starting with Cauchy's integral formula, $$ f(z) = {1\over 2\pi i} \int {f(w)\over (w-z)}dw $$ and subtracting an integral of the form, $$ {1\over 2 \pi i } \int {f(w) \bar z \over w\bar z-r^2}dw =0 $$ (which is zero by Cauchy's theorem), I get $$ {1\over 2\pi i} \int f(w) {r^2-|z|^2 \over(z-w)(w\bar z -r^2)}dw. $$ This appears to be along the right lines, however any advice on if this is the correct approach and/or on how to proceed is appreciated.
I think there's a typo in the question: it should be $\lvert w \rvert = r$ (otherwise the integral has a nasty singularity if $\lvert z \rvert = 1$, for one thing). Then, you have $r^2 = w\bar{w}$, so $w\bar{z}-r^2 = w(\bar{z}-\bar{w})$. Hence $$ \frac{r^2-\lvert z \rvert^2}{(z-w)(w\bar{z}-r^2)} = \frac{r^2-\lvert z \rvert^2}{w(z-w)(\bar{z}-\bar{w})} = \frac{r^2-\lvert z \rvert^2}{w\lvert z-w\rvert^2}, $$ as you want.