Prove $f|_{U_0}$ is $m$-to-$1$ except at $z_0$.

Question

Let $f$ be analytic on a domain $U$, $z_0\in U$, and $w_0=f(z_0)$. Suppose that $\mbox{ord}_{z_0}(f-w_0)=m\in\mathbb N$. Prove that there is an open set $U_0$ with $z_0\in U_0\subset U$ such that $f^{-1}(w_0)\cap U_0=\{z_0\}$ and $f^{-1}(w)\cap U_0$ contains exactly $m$ elements (without repetition) for $w\in f(U_0)\setminus\{w_0\}$. This means that $f|_{U_0}$ is $m$-to-$1$ except at $z_0$.

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Proof: First, since $\mbox{ord}_{z_0}(f-w_0)=m$, when writing $f-w_0$ as its Laurent series, then $a_m \ne 0$, and $a_n=0$ for all $n<m$ (this is the definition of $\mbox{ord}_{z_0}(f-w_0)=m$, which essentially says $m$ is the minimum power actually appearing in the Laurent series with a nonzero coefficient.) Then we can write \begin{align*} f(z)-w_0&=\sum_{n=m}^\infty a_n(z-z_0)^n \\ &=a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots \end{align*} This also means $z_0$ is either a removable singularity or a pole of $f-w_0$. Since $m\in \mathbb{N}$, we know $z_0$ is removable and after removing the singularity, we say that $z_0$ is a zero of $f-w_0$ (of order $m$). Also, it means there exists a holomorphic function $g$ in the disk $D(z_0,r)$ with $g(z_0) \ne 0$ such that $f(z)-w_0=(z-z_0)^m g(z)$. Would this be correct so far?

We have just finished talking about the Open Mapping Theorem and Inverse Mapping Theorem, so it's could be possible this question is an application of one of those.

Answer 1

Actually this claim is included in some proof of the open mapping theorem, for example, see W. Rudin's Real and Complex Analysis, Thm 10.23.

Your discussion is correct. Let's start from $f(z)-w_0=(z-z_0)^m g(z)$, where $g(z)$ is holomorphic in $D(z_0,r)$ and never vanishes.

Claim: There exists a holomorphic function $h(z)$, such that $g(z)=(h(z))^m$ in $D(z_0,r)$.
Proof: Consider $g'/g$ which is holomorphic in $D(z_0,r)$. Hence we can find some $G$ holomorphic in $D(z_0,r)$, such that $G'=g'/g$. Hence, we can choose one $G$ such that $\exp G=g$. Thus, $h(z)=G(z)/m$ is what we want.

Now we have $f(z)-w_0=((z-z_0)h(z))^m$. Choose a small neighborhood $V$ of $z_0$ such that $(z-z_0)h(z)$ is one-to-one in $V$, by the open mapping theorem, since the derivative of that is not zero at $z_0$. Now the conclusion follows from the fact that $z\mapsto z^m$ is $m$-to-$1$.

Answer 2

As you said, you can write $f(z)-f(c)=(z-c)^ng(z)$ where $g(c)\neq 0$ is holomorphic where $f$ is. Since $f$ is holomorphic, we can restrict ourselves to a nbhd $N$ where $f$ attains the value $f(c)$ only once, by the identity theorem.

Since $g(c)\neq 0$ there is a neighborhood $V\subseteq N$ of $c$ where $g$ is nonzero. In such neighborhood you can pick an $n$-th root $q$ of $g$; i.e. $q$ holomorphic with $q^n=g$.

If you let $h=(z-c)q$ then $f(z)-f(c)=q(z)^n$, and $h'(c)=q(c)$ and since $q(c)^n=g(c)\neq 0$; $h'(c)\neq 0$. This means there is further a nbhd $U$ where $h:U\to h(U)$ is biholomorphic, i.e. we have written $f(z)=f(c)+h(z)^n$ with $h$ biholomorphic, all this in $U$. Since $h(c)=0$, the open mapping theorem guarantees there is an open ball $B=B(0,r)$ inside $h(U)$. Let $W=h^{-1}(B)$, so that $\tilde h=r^{-1}h:W\to B(0,1)$. Note then that we can write $f$ has a series of compositions $t\circ p \circ \tilde h$ where $p(z)=z^n$ and $t(z)=r^nz+f(c)$.

The end result is we have factored $f$ into a biholomorphism $\tilde h:W\simeq B(0,1)$, a linear biholomorphism $t:B(0,1)\to B(f(c),r^n)$ and the usual $n$-fold cover of the unit disk (minus the origin!) by itself $z\mapsto z^n$.

This is known as the local normal form of a holomorphic mapping.