Showing $f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)\,dx\,dy}{(1-\bar{z}\zeta)^2}$ I'm doing practice physics qualifying exam problems and came across this one I didn't know how to solve:
Show that if $f(x)$ is bounded and analytic for $|z|=|x+iy|<1$, then
$$f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)\,dx\,dy}{(1-\bar{z}\zeta)^2}$$
Hint: First express the area integral in polar coordinates, then transform one of the integrals to
a suitable line integral of a rational function that can be evaluated using the calculus of residues.
I tried using $z=re^{i\theta}$ and messing around with the integral, but after a long writeout I am left with a puddle of muddled thoughts. Could someone explain the next steps?
 A: You can prove this by Green formula:
$$\int_{|z|=1}F(z)dz=2i\int_{|z|<1}\frac {\partial F}{\partial \bar{z}}dxdy.$$
By the Cauchy Int Formula $$f(\zeta)=\frac{1}{2\pi i }\int_{|z|=1} \frac{f(z)}{z-\zeta}dz.$$
Let $F(z)=\frac{f(z)}{z-\zeta}$, on the circle $|z|=1$, we have $z=\frac{1}{\bar z}$, so 
$F(z)=\frac{f(z)}{z-\zeta}=\frac{\bar zf(z)}{1-\bar z\zeta}$, by  easy computation , we can get $\frac {\partial F}{\partial \bar{z}}=\frac{f(z)}{(1-\bar z\zeta)^2}$. Finally, using the Green Formula, we get $$f(\zeta)=\frac{1}{\pi }\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar z\zeta)^2}.$$
A: The function $$K(\zeta,z)=\frac{1}{\pi}\frac{1}{(1-\bar{z}\zeta)^2}$$ is known as the Bergman reproducing kernel. 
Hint:


*

*Compute the series expansion of $g(x)=\frac{1}{(1-x)^2}$, e.g. by using that $$\frac{1}{(1-x)^2} =\frac{d}{dx}\frac{1}{1-x}$$

*First prove the statement for $f_n(z)=z^n$, e.g. by using the fact
$$\int_0^{2\pi}e^{ikt}dt=0,\text{  for all integers $k\ne0$}$$

Hopefully leaving the fun parts to you...
