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Proving theorem connecting the inverse of a holomorphic function to a contour integral of the function.

I saw this theorem in some lecture notes, but I have not been able to find a proof.

Let $f:\Omega \to \Bbb C$ be holomorphic with $f(0)=0 $ and $ f'(0)\neq0 $. Assume that $U \subset \Omega $ is a sufficiently small neighborhood of $0$ so that $f$ has a holomorphic inverse on $U$. Choose $r>0$ so small that $\bar B(0,r) \subset U$, further let $\omega \in f(B(0,r))$. Then the following formula holds:

$$ f^{-1}(\omega) = \frac{1}{2\pi i} \oint_{|z|=r}\ \frac{f'(z)z}{f(z)-\omega}\, dz $$

A proof or a link to a proof would be very much appreciated.

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The proof of $$f^{-1}(\omega) = \frac{1}{2\pi i} \oint_{|z|=r}\ \frac{z\, f'(z)}{f(z)-\omega}\, dz \tag{1}$$ is a straightforward calculation with the residue formula. The only pole of the integrand is at the point $\zeta=f^{-1}(w)$, and this pole is simple. The residue at $z=\zeta$ is $\frac{\zeta f'(\zeta)}{f'(\zeta)}=\zeta$, and the result follows.

I surely saw (1) somewhere, but don't have a reference. It can be generalized to maps in $\mathbb R^n$ using degree theory.