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.