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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?

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HINT: Look at the power series of $f(z)$ which can be written as $z^2 g(z)$ where $g(z)$ is holomorphic and $g(0) = a_0 \neq 0$. – user17762 Nov 6 '11 at 18:28
By local normal form, (for example see,…) $f$ can be considered as $z^2.$ – Ehsan M. Kermani Nov 6 '11 at 18:35
@VVV: what is $g'$? – Ehsan M. Kermani Nov 6 '11 at 19:37
@VVV: By long-standing internet convention, all caps is considered yelling/shouting. Please avoid all caps. – Arturo Magidin Nov 6 '11 at 23:24

Forming a holomorphic square root of the function as you suggest does indeed give a correct proof.

However, it is much easier to use the argument principle. Indeed, let $C$ be a small circle around zero, so that the disk $D$ bounded by $C$ does not contain any other roots or critical points.

Then the argument principle tells you that the number of $a$-values taken by $f$ in $D$ is given by a certain integral. This integral depends continuously on $a$, where $a$ does not belong to $f(C)$, and hence is constant on the component of $\mathbb{C}\setminus f(C)$ containing $0$. Since $f$ has exactly two roots in $D$ (counting multiplicities), the claim follows. I will leave it to you to flesh out the details.

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