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

Question

Let $A_R$ denotes the annulus $\{ z\in \mathbb{C}:1<|z|<R \}$. 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)?

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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.

Answer 1

The conjecture is true. See Shoshichi Kobayashi's book Hyperbolic Manifolds And Holomorphic Mappings: An Introduction Theorem 6.1.

The proof is not trivial, and it relies on some tools on Differential Geometry. It would be interesting if there is a simpler proof which only based on complex analysis.