Problem in adjustment for solving problem 
Let $f$ be an holomorphic function on disc $D_{R_0}$ centered at origin and radius $R_0$.
Prove that whenever $0<R<R_0$ and $|z|<R$ then:
$$f(z)=\frac{1}{2\pi}\int^{2\pi}_0f(Re^{i\theta})\Re\left(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\right)d\theta$$

My attempt:
I had considered disk of radius $R$. Now By
Cauchy Integral Formula 
$$f(z)=\frac{1}{2\pi i}\int_\mathcal{C}\frac{f(w)}{w-z}dw$$ 
where $\mathcal{C}$ is the circle with radius $R$.
Now $w=Re^{i\theta}$ by inserting and solving I get

$$f(z)=\frac{1}{2\pi }\int_c\frac{f(Re^{i\theta})Re^{i\theta}}{Re^{i\theta}-z}d\theta$$
I used $\Re(t)=\frac{t+\bar t}{2}$ for $t=\frac{Re^{i\theta}+z}{Re^{i\theta}-z}$ but I got 
$$\Re(t)=\frac{R^2-z\bar z}{R^2+z\bar z-2R\cos(ze^{-i\theta})}$$ where I got stuck.
Any help will be appreciated
 A: This is the right direction. For simplicity let us consider the case $R=1$. So $|z| <1$.
Using:
$$\Re \left(\frac{e^{i \theta}+z}{e^{i \theta}-z} \right)=\frac{e^{i \theta}}{e^{i \theta}-z}-\frac{e^{i \theta}}{e^{i \theta}-\frac{1}{\bar z}}$$
 we obtain:
$$\frac{1}{2\pi}\int^{2\pi}_0f(e^{i\theta})\Re\left(\frac{e^{i\theta}+z}{e^{i\theta}-z}\right)d\theta=\underbrace{\frac{1}{2\pi}\int^{2\pi}_0f(e^{i\theta})\left(\frac{e^{i\theta}}{e^{i\theta}-z}\right)d\theta}_{(1)}-\underbrace{\frac{1}{2\pi}\int^{2\pi}_0f(e^{i\theta})\left(\frac{e^{i\theta}}{e^{i\theta}-\bar z^{-1}}\right)d\theta}_{(2)}$$
and:


*

*The term $(1)$ is indeed $f(z)$.

*As $\left| \bar z^{-1} \right|>1$ i.e $\bar z^{-1}$ is outside the disc, the term $(2)$ is $0$.

To prove the first formula notice that, with $\zeta=z e^{- i \theta}$ it is the same as showing:
$$\Re \left(\frac{1+\zeta}{1-\zeta} \right)=\frac{1}{1-\zeta}-\frac{1}{1-\bar \zeta^{-1}}$$
and:
\begin{align}\Re \left(\frac{1+\zeta}{1-\zeta} \right)&=\frac{(1+\zeta)(1-\bar \zeta)+(1+\bar \zeta)(1-\zeta)}{2(1-\zeta)(1-\bar \zeta)}=\frac{2(1-\zeta \bar \zeta)}{2(1-\zeta)(1-\bar \zeta)}\\
&=\frac{(1-\bar \zeta)+\bar \zeta (1-\zeta)}{(1-\zeta)(1-\bar \zeta)}=\frac{1}{1-\zeta}-\frac{\bar \zeta}{\bar \zeta -1}=\frac{1}{1-\zeta}-\frac{1}{1-\bar \zeta^{-1}}\end{align}
