Let $f$ be an entire function, $z_{1}$, $z_{2}$ $\in$ $C$, with $z_{1} \neq z_{2}$ and $R>\max{(|z_{1}|,|z_{2}|)}$. Prove that $$2\pi i\dfrac{f(z_{1})-f(z_{2})}{z_{1}-z_{2}} = \oint_{\gamma}{\dfrac{f(z)}{(z-z_{1})(z-z_{2})}},$$ where $\gamma$ is the circle of center $0$ and radius $R$ positively oriented. Deduce that every bounded entire function is constant.
Progress
By Partial fractions, we have $\dfrac{1}{(z-z_{1})(z-z_{2})} = \dfrac{1}{z_{2}-z_{1}}\left( \dfrac{1}{z-z_{2}} - \dfrac{1}{z-z_{1}}\right)$
$$\oint_{\gamma}{\dfrac{f(z)}{(z-z_{1})(z-z_{2})}}dz=\dfrac{1}{z_{2}-z_{1}}\oint{\left( \dfrac{f(z)}{z-z_{2}}dz - \dfrac{f(z)}{z-z_{1}}dz\right)}$$
Now, applying Cauchy's Theorem $$\oint_{\gamma}{\dfrac{f(z)}{(z-z_{1})(z-z_{2})}}dz=\dfrac{1}{z_{2}-z_{1}}(2\pi if(z_{2})-2\pi if(z_{1}))$$
Then, $$2\pi i \left(\dfrac{f(z_{2})-f(z_{1})}{z_{2}-z_{1}}\right)=\oint_{\gamma}{\dfrac{f(z)}{(z-z_{1})(z-z_{2})}}dz$$
But, I don't know how determinated the last part of problem (or maybe, my progress isn't than better for the problem. And finally, who knows where is this problem in some book, greetings
Note that equality holds for sufficiently large $R>0$.
Consider $|f(z)|\le M$
$\displaystyle \left|\int_{\gamma}\frac{f(z)}{(z-z_1)(z-z_2)}dz\right|\le \frac{M}{(R-|z_1|)(R-|z_2|)}\cdot 2\pi R\longrightarrow 0$ , $R\longrightarrow \infty$ (ML-Inequality)
Therefore $\;2\pi i\dfrac{f(z_{1})-f(z_{2})}{z_{1}-z_{2}}=0$.