Let $U$ be an open connected subset in $\mathbb C$ and $f:U\rightarrow \mathbb C$ be a continuous map such that the map $z\mapsto (f(z))^n$ is analytic. Show that $f(z)$ is analytic. Now if we take $g(z)=(f(z))^n$ then $g$ is analytic. Then derivative of $g$ of any order is also analytic. But I am not sure whether it will help in proving analyticity of $f$. Another thought that I am having is since $f$ is given to be continuous we may use Morera'st theorem. Any help?
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No.The $n-$th derivative is not $n! f(z)$. – Kavi Rama Murthy Jul 15 '21 at 05:00
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Yes sir. Silly mistake. – Shreya Chauhan Jul 15 '21 at 06:24
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See a more general result here: https://math.stackexchange.com/questions/556080/composition-of-continious-and-analytic-map – Kavi Rama Murthy Jul 15 '21 at 06:29