# Variation of Cauchy integral formula

I am trying to show that if $D$ is the open unit disk, $f$ is holomorphic in a neighborhood of the closure of $D$, and $w$ is an arbitrary point in $D$, then $f(w)=\frac{1}{\pi}\iint_{D}\frac{f(z)}{(1-\bar{z}w)^{2}}dxdy$.

Previously I already had $f(w)=\frac{1}{2\pi}\int_{\partial D}\frac{f(z)}{1-\bar{z}w}|dz|$ so I tried applying Green's theorem but I wasn't successful.

-

There is an ad-hoc way to check the formula as follows. Expand $f$ into a power series $f(z)=\sum\limits_{n=0}^\infty a_n z^n$ and integrate in polar coordinates, using the identity $\sum\limits_{k=0}^\infty (k+1) \zeta^k = \frac{1}{(1-\zeta)^2}$ for $|\zeta|<1$, to get \begin{align*} \frac1\pi \iint_D & \frac{f(z)}{(1-\bar{z}w)^2} \, dx \, dy = \frac1\pi \int_0^1 \int_0^{2\pi} \sum_{n=0}^\infty a_n r^n e^{itn} \sum_{k=0}^\infty (k+1) r^k e^{-itk} w^k \, r \, dr \, dt \\ &=\frac{1}{\pi}\sum_{n,k=0}^\infty a_n (k+1) w^k \int_0^1 r^{n+k+1} \, dr \int_0^{2\pi} e^{it(n-k)} \, dt \end{align*} Now observe that the $t$-integral is $0$ whenever $n\ne k$, and is $2\pi$ otherwise, so you end up with $$2 \sum_{n=0}^\infty a_n (n+1) w^n \int_0^1 r^{2n+1} \, dr = \sum_{n=0}^\infty a_n w^n = f(w)$$
I guess in this setting, another way of going about it would be $f(w)=\int_{\partial D}\frac{f(z)}{z-w}dz=\int_{\partial D}\frac{\bar{z}f(z)}{1-\bar{z}w}dz$ and then apply Green's theorem. – Yong Oct 9 '12 at 20:34