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 $z \in \Omega$.
Existence of holomorphic n-th root and non-vanishing
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complex-analysis
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0Note that, if $\exists z_0 \in \Omega$ s.t. $f(z_0)=0$ with multiplicity $N \geq 1,$ then locally, $f(z)=z^N.$ – 2012-01-26
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1Consider 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 $z$... – 2012-01-26
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0The 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. – 2012-01-27
1 Answers
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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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0I 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. – 2012-01-27
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0@Georges Elencwajg: Dear Georges, $f=e^h.$ – 2012-01-27
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0Dear @ehsanmo: corrected , thanks. – 2012-01-27