Prove Bergman's kernel formula by theorem of residues I am having trouble with exercise $5$, $4.5.3$ in Ahlfors's complex analysis. I was asked to prove the Bergman's kernel formula:
\begin{equation}
f(\zeta)=\frac{1}{\pi}\iint\limits_{|z|<1}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\bar{z}\zeta)^{2}}
\end{equation}

under the conditions f(z) is bounded and analytic in the unit disk, moreover $\zeta$ lies in the disk. Two solutions of this problem are available on this website by using series expansion or Green's formula. However Ahlhors suggested another approach by using polar coordinates first, then transforming the inside integral into a line integral which can be evaluated by theorem of residues. I really don't know how to proceed in this way. I haved tried as following:
\begin{align}
\iint\limits_{D}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\bar{z}\zeta)^{2}}&=\int_{0}^{1}\mathrm{d}r\int_{0}^{2\pi}\frac{f(z)r}{(1-\bar{z}\zeta)^{2}}\mathrm{d}\theta\\
&=\int_{0}^{1}r\mathrm{d}r\oint\limits_{|z|=r}\frac{f(z)}{(1-\bar{z}\zeta)^{2}z}\mathrm{d}z\\
\text{Since $z \bar{z}=r^{2}$}\\ &= \int_{0}^{1}r\mathrm{d}r\oint\limits_{|z|=r}\frac{f(z)z}{(z-r^{2}\zeta)^{2}}\mathrm{d}z
\end{align}
At this stage, I have observed that if $\zeta=0$, then by Cauchy's theorem, the equality holds. If not, I use Residues to evaluate the contour integral and get the value $f(r^{2}\zeta)+f'(r^{2}\zeta)r^{2}\zeta$. But I stacked here because it seems impossible to get an explicit result of the integral with respect to r. Then I spent a huge amount of time trying to construct a special change of variable for the inside contour integral by a fractional linear transformation from a unit disk onto the smaller disk. But I just can not find the right linear transformation. Maybe I am on the wrong track.
 A: Using
$$ z=re^{i\theta} $$
one has
$$ \iint\limits_{D}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\bar{z}\zeta)^{2}}=\int_{0}^{1}\mathrm{d}r\int_{0}^{2\pi}\frac{f(re^{i\theta})r}{(1-re^{-i\theta}\zeta)^{2}}\mathrm{d}\theta=\int_{0}^{1}r\mathrm{d}r\int_{0}^{2\pi}\frac{e^{2i\theta}f(re^{i\theta})}{(e^{i\theta}-r\zeta)^{2}}\mathrm{d}\theta.$$
Then using
$$z=e^{i\theta}, dz=iz\mathrm{d}\theta$$
one has
\begin{eqnarray}
&&\int_{0}^{1}r\mathrm{d}r\int_{0}^{2\pi}\frac{e^{2i\theta}f(re^{i\theta})}{(e^{i\theta}-r\zeta)^{2}}\mathrm{d}\theta= \int_{0}^{1}r\mathrm{d}r\oint\limits_{|z|=1}\frac{z^2f(rz)}{(z-r\zeta)^{2}}\frac{\mathrm{d}z}{iz}\\
&=& -i\int_{0}^{1}r\mathrm{d}r\oint\limits_{|z|=1}\frac{zf(rz)}{(z-r\zeta)^{2}}\mathrm{d}z=-i\int_{0}^{1}r\cdot2\pi i[f(r^2\zeta)+r^2\zeta f'(r^2\zeta)]\mathrm{d}r\tag1\\
&=&2\pi\bigg[\int_{0}^{1}rf(r^2\zeta)\mathrm{d}r+\int_{0}^{1}r^3\zeta f'(r^2\zeta)\mathrm{d}r\bigg]\\
&=&2\pi\bigg[\frac12\int_{0}^{1}f(r^2\zeta)\mathrm{d}(r^2)+\int_{0}^{1}r^3\zeta f'(r^2\zeta)\mathrm{d}r\bigg]\tag2\\
&=&2\pi\cdot\frac12f(r^2\zeta)r^2\bigg|_0^1\\
&=&\pi f(\zeta).
\end{eqnarray}
So
$$ f(\zeta)=\frac{1}{\pi}\iint\limits_{|z|<1}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\bar{z}\zeta)^{2}}. $$
Here in (1),
$$ zf(rz)=r\zeta f(r^2\zeta)+[f(r^2\zeta)+r^2\zeta f'(r^2\zeta)](z-r\zeta)+O((z-r\zeta)^2) $$
and in (2) the integration by parts is used.
