I want to prove that $f: \mathbb{C}^{*} \to \mathbb{C}^{*} $ given by $f(z)=z^n$ is a covering map.
My attempt:
If $z_0 \in \mathbb{C}^{*}$ then for some $z\in \mathbb{C}^{*}$ such that $f(z)=z_0$. Implies that $z^n=z_0$. Now by changing this equation to polar form we get, $r^ne^{in\theta}=r_0e^{i\theta_1}$. Implies $r=(r_0)^{1/n}$ and $n\theta=\theta_1 +2k\pi$. Hence $n$ solution of equation $z^n=z_0$ are $z_k=r_0^{1/n}(\cos(\theta_1/n+2k\pi/n)+\sin(\theta_1/n+2k\pi/n))$ where $k=0,1,\cdot\cdot\cdot,n-1$.
I am stuck now on how to construct an open set $U$ around the point $z_0$ and how to take its inverse image and write it as a disjoint union of open sets $\{v_\alpha \}$ such that each one is mapped homeomorphic to $U$.
Thanks, for any help!!