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I'm trying to prove the follwing:

For a holomorphic function $f$ with a zero $z_0$ of multiplicity $n$ the $k$-th root of $f$ exists near $z_0$ if and only if $k\mid n$.

I thought it might be a good idea to prove the fact for $f=z^n$ first. If $k|n$ I can define $h(z):=z^{n/k}$ which is clearly holomorphic and satisfies $h(z)^k=f(z)$. Converserly if $k$ does not divide $n$ I tried to compute $(z^n)^{1/k}=\exp(\frac{1}{k}\log|z^n|+\frac{1}{k}\arg(z^n))=\exp(\frac{n}{k}(\log |z|+\arg z))$. How can I continue from here? How to prove the result for general $f$?

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2 Answers

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We may suppose $z_0=0$ and then write $f(z)=z^nu(z)$ with $u(0)\neq 0$.
Since $u$ has a logarithm, it has a $k$-th root for any $k$; this means that we can reduce the problem to the case $f(z)=z^n$.

a) if $k$ divides $n$ we have of course $f(z)=z^n=(z^{n/k})^k$

b) If conversely $f=g^k$, write $g=z^lv(z)$ with $v(0)\neq 0$. This implies $$f(z)=z^n=g(z)^k=z^{lk} v (z)^k$$
which in turn implies $n=lk$ (and $v (z)^k=1)$ and so $k$ divides $l$, as desired.

The proof summed up in one sentence:
The ring $\mathbb C\lbrace z\rbrace$ of germs of holomorphic functions is a DVR in which units are $k$ powers for all $k$.

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Assume that the $k$-th root of $f$ exists. We can write $f(z)=(z-z_0)^nf_0(z)$ where $f_0(z_0)\neq 0$ and $f_0$ is holomorphic. $f_0$ has a $k$-th root, hence $(z-z_0)^n$ has a $k$-th root near $z_0$. We can find an holomorphic function $g$ such that near $z_0$, $g(z)^k=(z-z_0)^n$. Since $g(z_0)=0$, we can write $g(z)=(z-z_0)^jh(z)$, where $h$ is holomorphic and $h(z_0)\neq 0$. We get $$(z-z_0)^{kj}h(z)=(z-z_0)^n.$$ This implies $kj=n$ hence $k\mid n$.

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