Question:
Let $A_{r}$=$\lbrace z:1<|z|<r\rbrace$ , $f:A_{r_{1}}\mapsto A_{r_{2}}$ is a covering map of degree $d<\infty$. prove that $r_{2}=r_{1}^d$.
My attempt:
(We assume covering map is holomorphic.) When $d=1$ is the fact that conformal modulus is invariant under biholomorphic map. But I have no idea how to deal with $d<\infty$.
Since $f$ has finite degree, it is a proper map, and thus $|f|:A_{r_1}\to (1,r_2)$ is a proper map. I claim that this implies that both
$$\lim_{|z|\to 1}|f(z)|\ ; \lim_{|z|\to r_1}|f(z)|$$
exists and they are either $1$ or $r_2$:
Hint of the proof: consider a sequence $\{z_n\}$ which goes to either $c_1$ or $c_{r_1}$, and let $A_k=\{r_2-\frac1k\ge|z|\ge1-\frac{1}{k}\}$. Since $|f|$ is proper and $A_k$ is compact, $|f|^{-1}(A_k)$ is compact, and thus there's an $N:\forall n>N, z_n\not\in |f|^{-1}(A_k)$ ,which implies $|f|(z_n)\not\in A_k$. Thus $|f|(z_n)$ has a convergent subsequence to either $1$ or $r_2$. It is not hard to see (although it requires some calculation), that we actually have convergence.
Now, if both the limits were $1$, we would have a contradiction by the maximum modulus theorem. Let us supposte that the first limit is $1$ and the second is $r_2$.
Consider $g(z):=\ln(R_1)\ln(|f(z)|)-\ln(R_2)\ln(|z|)$. This function is zero on the boundary of $A_r$, and by the maximum principle $g(z)\equiv 0$. Thus (writing $\beta$ for $\frac{\ln(R_2)}{\ln(R_1)}$)
$$\ln(|f|)=\beta\ln(|z|)\\ |f|=|z|^{\beta}$$
Taking a local branch of the logarithm around a point $a\in A_{r_1}$, we obtain that
$$\frac{f(z)}{z^{\beta}}$$
has constant modulus $=1$, and thus $f(z)=\alpha z^{\beta}$ (where $|\alpha|=1$). Noting that $f$ is continuous and that we can apply this method to every point $a\in A_{r_1}$, we get that the equation is valid on all of $A_{r_1}$, i.e. $\frac{f(z)}{\alpha}$ is a holomorphic branch of $z^{\beta}$ defined on all of $A_{r_1}$, which is possible iff $\beta$ is an integer. It is easy to see that $\beta>0$, since $\lim_{|z|\to r_1}|f(z)|=r_2>1$, and that $\beta=d$. The conclusion follows.
If we have the opposite situation, i.e. $\lim_{|z|\to 1}|f(z)|=r_2;\lim_{|z|\to r_1}|f(z)|=1$, a composition of $f$ with $\frac{1}{z}$ brings us back to the previous case, and we are done.