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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?

Makoto Kato
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  • let $g(z)=e^{f(z)}$, and $h(z)$ be something like $e^{f(z)/k}$ see http://math.stackexchange.com/questions/143609/representation-of-holomorphic-functions-by-exponential – yoyo Sep 21 '12 at 00:24
  • And to get $f(z)$, use an antiderivative of $g'(z)/g(z)$. – Robert Israel Sep 21 '12 at 00:29
  • 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 00:30
  • I'm expecting someone would write a detailed answer incorporating these comments. – Makoto Kato Sep 21 '12 at 00:35

1 Answers1

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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)

Christopher A. Wong
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