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Let $U$ be an open connected set in $\Bbb C$ and $f:U\to \Bbb C$ be a continuous map such that $z\mapsto f(z)^n$ is analytic for some positive integer $n$. Prove that $f$ is analytic.

I think the statement is FALSE. Consider the function $f(z)=\sqrt z$ in any open connected set $U\subset \mathbb C$ containing $0$. Then $f$ is continuous and $f(z)^2$ is analytic , but $f$ is not analytic.

Is my argument correct or there are some misunderstanding ?

If the statement is TRUE then how I can proceed to prove it ?

Empty
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    you're right, you need the additional assumption $f(z) \neq 0$, –  Mar 14 '17 at 19:41
  • @james.nixon Ok. According to you suppose that $f(z)\not= 0$. Then how I proceed to prove the same ? – Empty Mar 14 '17 at 19:44
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    Wait wait wait wait. I'm all mixed up. The theorem as stated is true. Your counter example is false because $\sqrt{z}$ is NOT continuous on a domain $U$ containing zero. Continuity is in fact enough. I was looking at the problem backwards, trying to construct an $n$'th root with only assumptions on the function itself. –  Mar 14 '17 at 19:59
  • Generalization answered here. – Jonas Meyer Mar 15 '17 at 14:09

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Near the points $z_0$ with $f(z_0)\ne 0$, $f$ is a composition of holomorphic functions, namely $f =\root n\of{f^n}$ where $\root n\of\cdot$ is some branch (what branch?) of the $n$-th root.

And the points with $f(z_0) = 0$? As the zeros of holomorphic functions are isolated, $f$ will be holomorphic in a perforated nhood of $z_0$. But as $f$ is continuous in $z_0$, then also will be holomorphic in $z_0$ (see the Riemann's theorem on removable singularities).

Martín-Blas Pérez Pinilla
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