complex analysis - Derive Poisson's integral formula from Laplace's equation inside a circular disk

For Laplace's equation inside a circular disk of radius $a$:
\begin{align*}
u(r, \theta) &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\overline{\theta}) \left[ -\frac{1}{2} + \sum\limits_{n=0}^\infty \left( \frac{r}{a} \right)^n \cos n (\theta - \overline{\theta}) \right] \, d\overline{\theta} \\

\end{align*}
Using $\cos z = \mathop{Re} [e^{iz}]$, sum the geometric series to obtain Poisson's integral formula:
\begin{align*}
u(r,\theta) &= \frac{a^2 - r^2}{2\pi} \int_{-\pi}^{\pi} \frac{f(\overline{\theta}) \, d\overline{\theta}}{a^2 + r^2 - 2ar \cos (\theta - \overline{\theta})} \\
\end{align*}

My work:

\begin{align*}

\sum\limits_{n=0}^\infty \left( \frac{r}{a} \right)^n \cos n (\theta - \overline{\theta}) &= \sum\limits_{n=0}^\infty \left( \frac{r}{a} \right)^n \mathop{Re} \left[ e^{i n (\theta - \overline{\theta})} \right] \\
\end{align*}

Now for real $k \in \mathbb{R}$ and complex $z \in \mathbb{C}, z=a+ib$ we have $k \mathop{Re} [z] = \mathop{Re} [kz]$ and $k + \mathop{Re} [z] = \mathop{Re} [k+z]$ so that:

\begin{align*}
&= \mathop{Re} \left[ \sum\limits_{n=0}^\infty \left( \frac{r}{a} \right)^n e^{i n (\theta - \overline{\theta})} \right] \\
&= \mathop{Re} \left[ \sum\limits_{n=0}^\infty \left( \frac{r}{a} e^{i (\theta - \overline{\theta})} \right)^n \right] \\
\end{align*}

Converging geometric series:

\begin{align*}
&= \mathop{Re} \left[ \frac{1}{1 - \frac{r}{a} e^{i (\theta - \overline{\theta})}} \right] \\
\end{align*}

At this point, I believe I need to convert back from $\mathop{Re} [e^{iz}] = \cos z$, but I'm not so sure how to do that.

Plugging that back without converting:

\begin{align*}
u(r, \theta) &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\overline{\theta}) \left[ -\frac{1}{2} + \mathop{Re} \left[ \frac{1}{1 - \frac{r}{a} e^{i (\theta - \overline{\theta})}} \right] \right] \, d\overline{\theta} \\
u(r, \theta) &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\overline{\theta}) \mathop{Re} \left[ -\frac{1}{2} + \frac{1}{1 - \frac{r}{a} e^{i (\theta - \overline{\theta})}} \right] \, d\overline{\theta} \\
\end{align*}

... and I'm stuck.

Answer

Hint: $Re \frac 1 z=Re \frac {\overline {z}} {|z|^{2}}=\frac {Re \overline {z}} {|z|}$. When $z=1-\frac r a e^{i(\theta- \overline {\theta})}$ we have $|z|^{2}=1+\frac {r^{2}} {a^{2}}-2\frac r a \cos(\theta- \overline {\theta})$. Can you finish the computation?