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Let $A$ be a region in complex plane $\Bbb C$, $f\colon A\to f(A)$ is continuous on $A$. $f^n$ is holomorphic on $A$. How to prove that $f$ is holomorphic on $A$?

Partial proof. I have proved the theorem when $0$ doesn't belong to $A$. (by composition of functions)

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Is $n$ an integer? What are the condition about it? –  Davide Giraudo Sep 9 '12 at 8:20
    
The problem should be easier in the case when $0 \not \in f(A)$, rather than $A$. –  ronno Sep 9 '12 at 8:26

1 Answer 1

If $f^n$ is holomorphic in $A$, it is either identically $0$ or its zeros are isolated. As an isolated singularity of $f$, an isolated zero is removable.

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