$f$ continuous, $f^N$ analytic on a domain D implies $f$ analytic on D

Solution 1:

Strictly speaking, one should dispose of $f\equiv 0$ first. After that, we know that the zeros of $f$ are isolated, because they are precisely the zeros of $f^N$. Let $Z$ be the set of zeros of $f$. The answer by Harald Hanche-Olsen explains how to show that $f$ is holomorphic in $D\setminus Z$. To finish off the problem, it is not necessary to follow the hint about $N$ dividing $m$. Instead one can use the following

Fact. A function that is continuous in $D$ and holomorphic in $D\setminus Z$ (where $Z$ is a discrete set) is holomorphic in $D$.

Sketch of proof: for any $a\in Z$ the Laurent series of $f$ around $a$ does not contain any negative powers, since $f$ is bounded in a neighborhood of $a$. (Use the Cauchy estimate for coefficients, for example).

Solution 2:

Write $g(z)=f(z)^N$. Away from the zeros, $f(z)$ is locally of the form $g(z)^{1/N}$, where the appropriate branch of the $N$th root is (locally) consistent. This should give you analyticity away from the zeros.

For the last part, a knowledge of winding numbers is certainly helpful. As for the conclusion from knowing that $N\mid m$, think back to those $N$th roots. You will find that, as you go around the zero, you can choose the $N$th root in a continuous manner all the way, getting back to the same branch after a full circle.

At least this is how I would think about it. I can't say if it is appropriate given what you do and don't know at that stage in the book.