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Let $n$ be a fixed positive integer. Show that there exists a holomorphic function $f : \mathbb{C} \setminus \mathbb{R}_{\leq 0}$ such that $[f(z)]^n = z$.

This question is related, but the domain differs from mine.

My attempt is as follows: It's clear the idea is to show that "$z^{\frac{1}{n}}$" is holomorphic in $\mathbb{C} \setminus \mathbb{R}_{\leq 0}$. Alternatively, we can write $z^{\frac{1}{n}} = e^{\frac{1}{n}\log{z}}$. Now since $\log{z}$ is holomorphic on $\mathbb{C} \setminus \mathbb{R}_{\leq 0}$ with branch cut $|\theta| < \pi$, we have that $e^{\frac{1}{n}\log{z}} = z^{\frac{1}{n}}$ is also holomorphic.

However, this attempt feels odd to me. I always have the feeling that in complex analysis, we try to avoid the $\log$ function (if we even treat it as a function). I would like to know if my attempted solution is complete, and if I can improve it any further.

Clement Yung
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1 Answers1

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Your proof is correct. In general, I refer to the following answer from J. Loreaux, which in particular shows that on any simply connected domain excluding the origin we have roots of all orders:

Given a holomorphic function $f:\Omega\to\mathbb{C}\setminus\{0\}$, we say that a holomorphic function $g:\Omega\to\mathbb{C}\setminus\{0\}$ is a logarithm of $f$ if $f(z)=e^{g(z)}$ for $z\in\Omega$. Similarly, we say that a holomorphic function $h_n:\Omega\to\mathbb{C}\setminus\{0\}$ is an $n$-th root of $f$ if $f(z)=h_n(z)^n$ for $z\in\Omega$. The question is, when do such functions exist?

The quick answer is that for an $n$-th root to exist, we must have, for any closed curve $\gamma$ in $\Omega$, $$ \frac{1}{2\pi i}\int_\gamma \frac{f'}{f}\ dz \in n\mathbb{Z}. $$ Furthermore, a branch of the logarithm of $f$ exists if and only if for every $n\in\mathbb{N}$, an $n$-th root of $f$ exists. The reason for this is that a logarithm of $f$ exists if and only if for every closed curve $\gamma$ in $\Omega$, $$ \frac{1}{2\pi i}\int_\gamma \frac{f'}{f}\ dz = 0. $$ Notice that if we have a logarithm of $f$, say $g$, then we can construct an $n$-th root of $f$ for any $n$ by letting $h_n(z):=e^{\frac{g(z)}{n}}$. For the other direction, notice that if the aforementioned integral is an element of $n\mathbb{Z}$ for every $n$, then it must be zero, and thus the logarithm of $f$ must exist.

A full proof can be found in the linked answer.

Brevan Ellefsen
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