Question about univalent branch Problem: Let the holomorphic function $f$ have a zero of order $m$ at the point $z_0$. Prove that there is an open disk centered at $z_0$ in which there is a univalent branch of $f^{1/m}$. I suppose I can find a disk by local mapping theorem. But I don't know how to show that $f^{1/m}(z_0)=0$ is of order 1, because this implies that is injective (univalent branch). Thanks in advance.

By Taylor's theorem $$f(z)=a_m(z-z_0)^m +a_{m+1}(z-z_0)^{m+1}+\cdots$$ in a neighbourhood of $z_0$ where $a_m\
e0$. Then $$f(z)=c^m (z-a_0)^mg(z)$$ for some $c$ where $g$ is holomorphic and $g(z_0)=1$. Then $g$ has a holomorphic $m$-th root near $z_0$. (Compose $g$ with a holomorphic branch of $w\mapsto w^{1/m}$ near $w=1$).