I'm trying to show that given a non-constant holomorphic $f$ in a region $\Omega \ni z_0$, then there exists a holomorphic $g$ such that $g^m = f$ for all $z$ in some open set $V \subseteq \Omega$ containing $z_0$. Furthermore, $g$ is a bijection between $V$ and $D_r(0)$ for some $r > 0$.

So far, I've written $f(z) = (z-z_0)^mh(z)$ and then used the fact that $h$ doesn't have a zero to take the log, and hence m-th root. Then my $g(z) = (z-z_0)e^{\frac{1}{m}\log(h(z))}$. I've managed to show that $g(z_0) = 0$ and $g'(z) \neq 0$ for all $z \in V$, and I want to try and use Bloch's Theorem to find some $S$ such that $g$ is injective on $S$ and the image contains a disk, however, I have no guarantee that the disk in the image contains 0. (If it did, I could just take a smaller disk thats then centred at 0, and find a new V thats the preimage of this new disk).

Edit: Another idea I had is to use the open mapping theorem to conclude that there is some $D_r(0)$ contained in the image, $g(V)$, and then take the preimage of $D_r(0)$. This gives us a surjection from some smaller $U \subseteq V$ to $D_r(0)$. I don't know how to show injectivity then.

Any help with fixing either of these approaches?

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    $\begingroup$ I think the inverse function theorem can guarantee a bijection.link $\endgroup$ – Apocalypse Nov 16 '18 at 3:58

Heres an attempt at the proof: By the inverse function theorem, we can find a small enough $V \subseteq \Omega$ such that $g$ is invertible on $V$, which shows that it is a bijection. Then by the open mapping theorem, $g(V)$ is open and contains $0$, so there is some $D_r(0) \subseteq g(V)$. Then by continuity of $g$, we know that the preimage of $D_r(0)$ is open, so we let $V' \subseteq V$ be the preimage of $D_r(0)$. Then we have $g \mid_{V'}$ is a bijection from $V'$ to $D_r(0)$.


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