I've been stuck in this complex analysis problem since past week, and it would be really useful if someone provide me with a hint or a way to prove the next statement:
Prove using the inverse function theorem that if $f$ is holomorphic on a region $\Omega$, $f'(z_0)\neq0$, then for $\gamma$ a sufficiently small circumference centered in $z_0$ we have: $\frac{2 \pi i}{f'(z_0)}=\int_{\gamma} \frac{dz}{f(z)-f(z_0)} $
The $2\pi i$ and the difference on the integral reminds me a little bit of Cauchy's Integral Formula, but I really don't know how to apply directly to this problem, because the numerator is not in the form $z-a$. Also, I have no clue in applying inverse function theorem; the only thing I know is that $\frac{1}{f'(z_0)}=(f^{-1})'(f(z_0))$, but anyway this still makes any sense to me since I don't know who $f^{-1}$ is.
In advance, thank you for taking the time to read this question.