Let $f:G\rightarrow \mathbb{C}$ be a holomorphic function with a double root at $z=0 \in G$.
Show that there is a open neighborhood $U\subset G$ of $0$ so that f takes every value of $f(U)\backslash \{0\}$ on U exactly two times.
My proof attempt:
Double root at $z=0$ implies : that f can be rewritten in the form $f(z) = z^{2}h(z)$, where $h \in \mathcal{O}(G)$ and $h(0) \ne 0$. $h(z)^{1/2}$ is well defined by restricting the image so that it becomes bijective. (I don't know how to do this). $h(z)^{1/2} \in \mathcal{O}(G)$ because the composition of holomorphic functions is holomorphic. Thus also $g(z):= z\sqrt{h(z)} \in \mathcal{O}(G)$. and g'(z) \ne 0 since otherwise the condition that $h(0) \ne 0$ is hurt. From this it follows that such a neighbourhood exists.
Is this correct?