# Biholomorphically transform a polynomial into a monomial

Following up on an answer I got on a previous question, here, given a polynomial $$p(z) = \sum_{k=0}^n a_k z^k$$ over $\mathbb{C}$, how do I construct (or prove existence of) a biholomorphic map $w=\varphi(z)$, such that $$p(\varphi(z))=w^n.$$ Thanks.

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Could you precise your question ? I mean, take the polynomial $p(z)=1+z$, such a biholomorphic map $\varphi$ won't exist. Your question is maybe the following : given a complex polynomial $P\in \mathbb C[X]$ and $z_0\in \mathbb C$, then there exist a unique integer $m\in \mathbb Z_{\geq 0}$ and a biholomorphic map $\varphi:U\subset \mathbb C\rightarrow V\subset \mathbb C$ such that $z_0\in U, 0\in V, \varphi(z_0)=0$ and $P\circ \varphi^{-1}(\omega)=\omega^m$ for every $\omega\in V$. –  Bebop Sep 4 '12 at 20:34
@Bebop: $z\mapsto w-1$? –  Kevin Sep 4 '12 at 20:58
Well, the initial question was to find $\varphi$ s.t. $p(\varphi(z))=\varphi(z)^n$ where $n=deg(p)$. With my polynomial, it is certainly not possible. –  Bebop Sep 4 '12 at 21:13

This isn't true globally, since a biholomorphism will preserve the number of distinct zeros of a given function, so if $p$ has more than one zero, there is no such biholomorphism.
However, I think the person leaving that comment meant that this is true locally: in a small neighborhood of a zero of $p$, there is a biholomorphism with a small neighborhood of $0$ such that $p$ goes to $w^m$, where $m$ is the order of the zero.