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I'm reading Forster's Lectures on Riemann surfaces. The proof of theorem 2.1(you don't need to know what it is) of the book uses the following fact.

Let $g(z)$ be a holomorphic function defined on a neighborhood of $0$ in $\mathbb{C}$ such that $g(0) \neq 0$. Let $k > 0$ be an integer. Then there exists a holomorphic function defined on a neighborhood of $0$ such that $h^k = g$.

How do we prove this?

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let $g(z)=e^{f(z)}$, and $h(z)$ be something like $e^{f(z)/k}$ see math.stackexchange.com/questions/143609/… –  yoyo Sep 21 '12 at 0:24
    
And to get $f(z)$, use an antiderivative of $g'(z)/g(z)$. –  Robert Israel Sep 21 '12 at 0:29
    
Just to add to yoyo's comment, as $g(0) \neq 0$, there is some open neighbourhood of $0 \in \mathbb{C}$ on which $g$ is non-zero, so on this set $g(z)$ can be written as $e^{f(z)}$ for some $f(z)$. –  Michael Albanese Sep 21 '12 at 0:30
    
I'm expecting someone would write a detailed answer incorporating these comments. –  Makoto Kato Sep 21 '12 at 0:35

1 Answer 1

up vote 1 down vote accepted

If $\Omega \subset \mathbb{C}$ is simply connected and $g(z)$ is holomorphic and non-vanishing on $\Omega$, then there exists a holomorphic logarithm $h(z)$ which is holomorphic on $\Omega$, such that $e^{h(z)} = g(z)$. We can construct $h(z)$ as the antiderivative of $g'(z)/g(z)$, since $g'(z)/g(z)$ is holomorphic as the denominator is non-vanishing. Now take $f(z) = h(z)/k$; this is holomorphic, and hence $e^{f(z)} = e^{h(z)/k}$ is holomorphic, and exponentiating by $k$ yields $g(z)$.

(edited per comment)

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I guess there was really no need to use the argument principle; the ratio of two holomorphic functions is always holomorphic if the denominator doesn't vanish. –  Christopher A. Wong Sep 21 '12 at 1:24

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