# Existence of holomorphic n-th root and non-vanishing

Suppose $f \in H(\Omega)$, $\Omega =$ arbitrary region. Suppose $f$ has a holomorphic $n-$th root in $\Omega$ for every positive integer $n$. Then I need to show that $f(z)\neq 0$ for all $z \in \Omega$.

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Note that, if $\exists z_0 \in \Omega$ s.t. $f(z_0)=0$ with multiplicity $N \geq 1,$ then locally, $f(z)=z^N.$ – Ehsan M. Kermani Jan 26 '12 at 23:40
Consider the contrapositive: if $f(z) = 0$, show that $f$ cannot have a holomorphic $n$-th root for some $n > 1$. Hint: any such will have a branch point at $z$... – Zhen Lin Jan 26 '12 at 23:41
The étiquette on this site is not to ask questions by giving orders ("show that"). However, since you are new here, I'll answer your question all the same, since I'm sure you'll follow that rule now that you know it. – Georges Elencwajg Jan 27 '12 at 0:05

## 1 Answer

Let $z_0\in G$ and let $ord_{z_0}(f)\in \mathbb N$ be the order of vanishing of $f$ at $z_0$. If $f=g^n$ we have $ord_{z_0}(f)=n\cdot ord_{z_0}(g)$. Hence $ord_{z_0}(f)$ is being divisible by all integers must be zero; in other words $f$ does not vanish at $z_o$.

But this is small beer.

The much stronger conclusion of your hypothesis is that actually $f$ is an exponential: there exists $h\in H(\Omega)$ with $f=e^h$.
This is an old chestnut that you can find for example in Remmert's Theory of Complex Functions, Chapter 9.

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I don't know what I had in mind when I wrote my comment but you're right, I can't think of a proof that doesn't use this fact. I'll be deleting the comment to avoid confusion. – Jose27 Jan 27 '12 at 2:29
@Georges Elencwajg: Dear Georges, $f=e^h.$ – Ehsan M. Kermani Jan 27 '12 at 2:54
Dear @ehsanmo: corrected , thanks. – Georges Elencwajg Jan 27 '12 at 9:55
Thanks..! This really helps. – facebook-100000771761284 Jan 27 '12 at 12:01