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?
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*}