Deriving an alternate form of the Cauchy Integral Formula There is a well known formula in complex analysis called the Cauchy Integral Formula:  
$$f(z) = \frac{1}{2\pi i} \int_{C} \frac{f(p)}{p-z} \, dp$$
which holds for the circle of integration $C$ when $f$ is holomorphic in an open region containing the disk defined by $C$ and any $z$ strictly inside the disk outlined by $C$.  
There is an alternate formula which I am trying to derive for the special case when $C$ is a circle of radius $R$ around the origin:  
$$ f(z) = \frac{1}{2 \pi} \int_{0}^{2\pi} f(Re^{i \theta}) \mathbb{R}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta$$ where the symbol $\mathbb{R}(\cdot)$ is meant to mean the real part of some complex number.
Additional Information:
Here is a hint from the text: note that if $w = \frac{R^{2}}{\overline{z}}$, then the integral of $\frac{f(p)}{p -w} $ around $C$ is $0$, which I'm pretty sure follows from holomorphicity of said integrand. However, this has not been much help to me.  
What I've tried:
It is necessary and sufficient to show:  
$$ f(z) + \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)i \, d\theta$$
$$ = \frac{1}{2 \pi} \int_{0}^{2\pi} f(Re^{i \theta}) \mathbb{R}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta + \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)i \, d\theta$$
*Here I've only added the "imaginary" version of the formula to both sides. Now working with the right hand side:
$$
\begin{align}
& = \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta})\frac{Re^{i\theta} + z}{Re^{i\theta} - z}d\theta \\[8pt]
& = \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta})\frac{Re^{i\theta} + z}{Re^{i\theta} - z}d\theta \\[8pt]
& = \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}(Re^{i\theta} + z) \, d\theta \\[8pt]
& = \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}Re^{i\theta}d\theta + \frac{1}{2 \pi} \int_{0}^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt]
& = \frac{1}{2 \pi i} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}Rie^{i\theta}d\theta + \frac{1}{2 \pi} \int_{0}^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt]
& = \frac{1}{2\pi i} \int_C \frac{f(p)}{p-z}dp + \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt]
& = f(z) + \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta
\end{align}
$$
And canceling $f(z)$ from both sides yields:
$$ \int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta =\int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta$$
But I don't know where to take it from here.
 A: This is called the Poisson integral formula, by the way.
$$\dfrac{1}{2\pi}\int_0^{2 \pi} f(Re^{i\theta})\ \text{Re}\left(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\right)\ d\theta = \frac{1}{4\pi} \int_0^{2\pi} f(Re^{i\theta}) \left( \frac{Re^{i\theta}+z}{Re^{i\theta}-z} + \frac{Re^{-i\theta} + \overline{z}}{Re^{-i\theta}-\overline{z}}\right)\ d\theta $$
Break this up into four terms. Taking $\zeta = R e^{i\theta}$ on the circle,
$$\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{Re^{i\theta}}{Re^{i\theta} - z}\ d\theta = \frac{1}{4\pi i} \oint_C f(\zeta) \dfrac{d\zeta}{\zeta - z}  = \frac{f(z)}{2}$$
$$ \eqalign{\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{z}{Re^{i\theta} - z}\ d\theta &=
\frac{1}{4\pi i} \oint_C f(\zeta) \frac{z}{\zeta-z} \frac{d\zeta}{\zeta}\cr
&= \frac{1}{4\pi i} \oint_C \left( \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta} \right) \ d\zeta = \frac{f(z)}{2} - \frac{f(0)}{2}\cr}$$
$$\eqalign{\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{Re^{-i\theta}}{Re^{-i\theta} - \overline{z}}\ d\theta &= \frac{1}{4 \pi i} \oint_C f(\zeta) \frac{R^2 \zeta^{-1}}{R^2 \zeta^{-1} - \overline{z}} \frac{d\zeta}{\zeta}\cr
&= \frac{1}{4\pi i} \oint_C \left( \frac{f(\zeta)}{\zeta} + \frac{\overline{z} f(\zeta)}{R^2 - \overline{z} \zeta}\right) d\zeta = \frac{f(0)}{2}
}$$
(note that if $\overline{z} \ne 0$, $|R^2/\overline{z}| > R$).  I'll let you do the last term.
