# If $f$ has a zero of order m at $z_0$, then there is a g such that $g^m = f$

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?

• I think the inverse function theorem can guarantee a bijection.link – 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 $$D_r(0)$$.