# How to prove that $\mathbb{C}^{\ast} \rightarrow \mathbb{C}^{\ast}, z \mapsto z^n$ is a covering map?

From a course on complex analysis. I have no idea how to start this one. I know the definition of covering maps but I am struggling to find an ansatz for a proof. Maybe someone can give me a little guidance.

EDIT: I was asked to give the definition of covering maps. Here it is:

A continuous map $p\colon X' \rightarrow X$ between manifolds is called a covering map if every $a\in X$ has a neighborhood $U$ with the following property: $p^{-1}(U)$ is a disjoint union of open sets $U'_j\subseteq X'$ such that $p|_{U'_j}$ is a homeomorphism onto $U$ for each $j$.

My problem is that I just don't know how to start. How do I choose $U$? How do I find the $U'_j$? Do I have to tinker them somehow? Is there any "standard" way to find those sets?

• Could you write down your definition of a covering map? And maybe add a detail on exactly which property you can't show that $z\mapsto z^n$ fulfills. – Arthur Oct 3 '17 at 8:53
• If $f$ is analytic and non-constant then $f(b+z) = f(b)+C z^m+o(|z^m|)$ where $C \ne 0, m \ge 1$ so $f$ sends a small neighborhood of $a$ to a small neighborhood of $f(b)$. Also if $m=1$ (ie. $C =f'(b) \ne 0$) then locally $f$ is biholomorphic. You should start with $a = f(b) =1$. – reuns Oct 3 '17 at 9:07

Hint: Think of trigonometric form and de Moivre's formula. Specifically, how do you describe solutions to $z^n=z_0$? Can you find small enough neighbourhoods around solutions that they are disjoint and all map homeomorphically to the same neighbourhood of $z_0$?
Here's how to do it for $z_0 = 1$:
• Ok I have now understood that for each of the $n$ roots of any $\alpha \in \mathbb{C}$ they have a neighborhood mapped homeomorphically to a neighborhood of $\alpha$ under $z \mapsto z^n$. How would I choose such a neighborhood wisely? Are balls of a radius small enough enough? Or are there easier sets? – Jakob Elias Oct 3 '17 at 12:34
• @Jakob Elias, not balls, observe the picture. In polar coordinates, $z\mapsto z^n$ is given by $(r,\varphi)\mapsto (r^n,n\varphi)$. – Ennar Oct 3 '17 at 12:39