$f(z) = \frac{1}{\pi} \int_E \frac{f(w)}{(1-z \bar{w})^2} dudv $ , where $w = u + iv$ 
Let $f(z)$ be holomorphic on an open region containing the closed unit disc $E = \{z \in \mathbb{C} : \mid z \mid \leq 1 \}$. Then show that 
  $$ f(z) = \frac{1}{\pi} \int_E \frac{f(w)}{(1-z \bar{w})^2} dudv , \;\; \text{where} \;\; w=u+iv$$
  for every $z$ with $\mid z \mid <1$.

My attempt
$f(z)= \frac{1}{2\pi i} \int_{C_r}\frac{f(\xi)}{\xi - z} d\xi  =  \frac{1}{2\pi}  \int_0^{2\pi} \frac{f(r e^{i \theta})} {r e^{i \theta} - z} r e^{i \theta} d\theta $  , by cauchy integral formula , where $C_r$ is a circle positively oriented contour centered at $z$.
Then, $f(z) = \int_0^1 f(z) dr= \int_0^1  \frac{1}{2\pi}  \int_0^{2\pi} \frac{f(r e^{i \theta})} {r e^{i \theta} - z} r e^{i \theta} d\theta dr  = \frac{1}{2\pi} \int_0^1    \int_0^{2\pi} \frac{f(r e^{i \theta})} {r e^{i \theta} - z} r e^{i \theta} d\theta dr $.
By polar coordinate argument, $f(z) = \frac{1}{2\pi} \int_E     \frac{f(u+iv)} {u+iv - z} \frac{u+iv}{\mid u+iv \mid}dudv = \frac{1}{2\pi} \int_E     \frac{f(w)} {w - z} \frac{w}{\mid w \mid}dudv $.
May I ask you how to proceed this problem?
 A: We have
$$\frac{1}{(1 - z\overline{w})} = \sum_{n=0}^{\infty} \overline{w}^n z^n$$
when $\vert z \vert < \frac{1}{\vert w \vert}$. In particular, this holds when $\vert z \vert, \vert w \vert < 1$. Differentiating with respect to $z$ gives
$$\frac{1}{(1 - z\overline{w})^2} = \sum_{n=0}^{\infty} (n + 1)\overline{w}^nz^n$$
Expand $f(w) = \sum_{m=0}^{\infty} a_mw^m$ as a power series about $0$. Then
$$\frac{1}{\pi} \int_E \frac{f(w)}{(1 - z\overline{w})^2} dudv = \frac{1}{\pi} \int_E\left(\sum_{m=0}^{\infty} a_mw^m \right) \left(\sum_{n=0}^{\infty} (n + 1)\overline{w}^n z^n\right) d\lambda(w),$$
where $d\lambda(w)$ denotes integration with respect to Lebesgue measure on $\mathbb{C}$. Multiplying the two series together, this becomes
$$f(z) = \frac{1}{\pi}\int_E \left(\sum_{m,n \geq 0} a_mw^m (n + 1)\overline{w}^nz^n \right) d\lambda(w)$$
Changing to polar coordinates gives us
\begin{equation*}
\begin{aligned}
&\mathrel{\phantom{=}} \frac{1}{\pi} \int_0^1\int_0^{2\pi} r\left(\sum_{m,n \geq 0} a_m(re^{i\theta})^m (n + 1)(\overline{re^{i\theta}})^nz^n \right) d\theta dr \\
&= \frac{1}{\pi} \int_0^1\int_0^{2\pi} \left(\sum_{m,n \geq 0} a_mz^n(n + 1)r^{m + n + 1}e^{i(m - n)\theta} \right) d\theta dr.
\end{aligned}
\end{equation*}
We can switch the order of integration and summation (why?), giving us
$$\frac{1}{\pi} \int_0^1 \left(\sum_{m,n \geq 0} a_mz^n(n + 1)r^{m + n + 1} \int_0^{2\pi} e^{i(m - n)\theta} d\theta\right) dr.$$
Now recall that for an integer $k$ we have $\int_0^{2\pi} e^{ik\theta} d\theta = 0$ for $k \neq 0$ and $2\pi$ if $k = 0$, so all terms of the series with $m \neq n$ vanish. Switching integration and summation again gives us
\begin{equation*}
\begin{aligned}
&\mathrel{\phantom{=}} \frac{1}{\pi} \int_0^1 \left(\sum_{m \geq 0} a_mz^m(m + 1)r^{2m + 1} \cdot 2\pi\right) dr \\
&= 2 \sum_{m \geq 0} a_mz^m(m + 1) \int r^{2m + 1} dr \\
&= 2 \sum_{m \geq 0} a_mz^m(m + 1) \cdot \frac{1}{2m + 2} \\
&= \sum_{m \geq 0}a_mz^m \\
&= f(z).
\end{aligned}
\end{equation*}
