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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.$2012-01-26
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    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 $z$...2012-01-26
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    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.2012-01-27

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