Let $f$ be analytic in an open set containing $\bar{\Bbb{D}}$. If $w\in \Bbb{D}$ then prove that $$f(w)=\frac{1}{2\pi}\int_{\partial\Bbb{D}}\frac{f(z)}{1-w\bar{z}}|dz|$$
First of all,since $\frac{1}{1-u}=\sum_{k=0}^{\infty}u^k$ for all $u\in \mathbb{D}$, I can let $u=w\bar{z}$ and we have $\frac{1}{1-w\bar{z}}=\sum_{k=0}^{\infty}(w\bar{z})^k$. Also, since $f(z)$ analytic, it has a power series expansion $\sum_{n=0}^{\infty}a_nz^n$. If $w\in \mathbb{D}$ is fixed, then for any $z\in \mathbb{D}$, $\frac{1}{1-w\bar{z}}$ converges uniformly, as well as $f(z)$. Consequently, I can deduce that it's product $\frac{f(z)}{1-w\bar{z}}$ is uniformly convergent.
Can anyone help me finish the proof?
On $\partial \Bbb D$, $\bar{z} = z^{-1}$ and $|dz| = \dfrac{dz}{iz}$, making
$$\frac{1}{2\pi}\int_{\partial \Bbb D} \frac{f(z)}{1 - w\bar{z}}\lvert dz\rvert = \frac{1}{2\pi}\int_{\partial \Bbb D} \frac{f(z)}{1 - wz^{-1}} \frac{dz}{iz} = \frac{1}{2\pi i}\int_{\partial \Bbb D} \frac{f(z)}{z - w}\, dz$$ The last expression is equal to $f(w)$ by the Cauchy integral formula.