Show that $f(z)=\frac{1}{2\pi}\int\limits_0^{2\pi}f\left(\frac{e^{i\theta}+z}{1+\overline{z}e^{i\theta}}\right)d\theta$ 
Let $f$ be analytic on domain $\Omega$ which contains the closed unit
  disk $\overline{\mathbb{D}}$. Show that
   (a)
  $$f(0)=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta})d\theta$$
  (b) Use part (a) to show that whenever $z\in\mathbb{D}$,
  $$f(z)=\frac{1}{2\pi}\int_0^{2\pi}f\left(\frac{e^{i\theta}+z}{1+\overline{z}e^{i\theta}}\right)d\theta$$
Hint: Consider the conformal self maps of the unit disk  

For part (a) I used the Cauchy integral formula on the unit disk and then substituted the parameterization $z(\theta)=e^{i\theta}$ where $0<\theta<2\pi$ 
But for part (b) I don't see how it is related with part (a) and how the automorphisms on the unit disk comes into play.
Anyway I know that $\Phi_{\alpha}:\mathbb{D}\rightarrow\mathbb{D}$ defined as $\Phi_\alpha(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$ is an automorphism when $|\alpha|<1$ 
Appreciate your help

Consider the function $g_z(w) = \frac{z+w}{1+\overline{z}w}$ for $|z|<1$ and $|w|\leq 1$. When $|w| = 1$, observe that 
$$|g_z(w)| = \left| \frac{z+w}{1+\overline{z}w} \right| = \left| \frac{z+w}{1+\overline{z}w} \right| \cdot \left| \frac{1}{\overline{w}} \right| = \left| \frac{z+w}{\overline{w} + \overline{z}} \right| = \left| \frac{z+w}{\overline{z+w}} \right| = 1.$$ 
When $|w| < 1$ observe that, for example, $|g_z(0)| = |z| < 1$. By the maximum principle, $|g_z(w)| < 1$ in this case. 
Therefore, we see that $f \circ g_z$ is analytic on a domain containing the closed unit disc. 
So by part (a), we have 
$$(f\circ g_z)(0) = \frac{1}{2 \pi} \int_{0}^{2 \pi} (f \circ g_z)(e^{i \theta}) d\theta,$$ 
and the result follows. 
