# Holomorphic n-th covering map between annulus must be $z^n$, essentially.

Let $$A_R$$ denotes the annulus $$\{ z\in \mathbb{C}:1<|z|. I guess and tend to prove the following(may not true):

If $$f:A_r\mapsto A_R$$ is a holomorphic covering map with degree $$n$$, then $$R=r^n$$ and $$f$$ must has the form $$f=\phi\circ z^n\circ \psi.$$ Where $$\psi\in \text{Aut}A_r$$, and $$\phi \in \text{Aut}{A_R}$$.

Is this true? If not, does $$R=r^n$$ still holds(I strongly believe this is true)?

This question rises when I look this post. Which says about $$A_r$$ and $$A_R$$ are conformal equivalent iff $$R=r$$ (they have the same moduli). So how about the covering map? I think may it should first prove $$\text{Mod}(A_R)=n\text{Mod}(A_r).$$ Here $$\text{mod}$$ means the moduli. I want use $$\text{Mod}(A_r)=\frac{1}{\lambda(\Gamma_r)},$$ where $$\lambda$$ is the extremal length. Give a metric on $$A_R$$, pull back by $$f$$ to obtain a metric on $$A_r$$ I can prove $$\lambda(\Gamma_r)\geq n \lambda(\Gamma_R).$$ Hence $$\text{Mod}(A_R)\geq n \text{Mod}(A_r).$$ But for another direction, I don’t know how to continue.

• Here $\Gamma_R$ denotes the family of all finite length circ with winding number 1 who lies in $A_R$. The same notation work for $r$. Jun 12, 2021 at 15:57