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 ?
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).