Conformal equivalence of annuli

For $0 < r < R$, let $A(r, R) = \{z \in \mathbb{C}: r < |z| < R\}$, the annulus centered at the origin with inner radius $r$ and outer radius $R$. We are going to show that the conformal equivalence classes of such anuuli in the complex plane are parametrized by the ratio $R/r$.

Theorem. $A(r_1, R_1)$ is conformally equivalent to $A(r_2, R_2)$ if and only if $R_1/r_1 = R_2/r_2$.
Proof: Suppose $R_1/r_1 = R_2/r_2$. Then there exists $k > 0$ such that $R_2 = k R_1$ and $r_2 = k r_1$. The map $f(z) = kz$ gives a conformal equivalence from $A(r_1, R_1)$ onto $A(r_2, R_2)$.
Conversely, let $f$ be a biholomorphic map from $A(r_1, R_1)$ onto $A(r_2, R_2)$. By scaling if necessary,  we may assume $r_1 = r_2 = 1$. Let $A_1 = A(1, R_1)$ and $A_2 = A(1, R_2)$. Fix some $1 < r < R_2$.
Let $C = \{z \in \mathbb{C}: |z| = r\}$.
Since $f^{-1}$ is continuous, $f^{-1}(C)$ is compact, so we can find some $\delta > 0$ such that $A(1, 1 + \delta) \cap f^{-1}(C) = \emptyset$. Let $V = f(A(1, 1 + \delta))$. Since $f$ is continuous, $V$ is connected, so either $V \subseteq A(1, r)$ or $V \subseteq A(r, R_2)$. By replacing $f$ with $R_2/f$, we may assume the first case holds.
Claim: $|f(z_n)| \rightarrow 1$ whenever $|z_n| \rightarrow 1$.
Proof of claim: Let $\{z_n\} \subseteq A(1, 1 + \delta)$. Note that $\{f(z_n)\}$ does not have a limit point in $A_2$ since otherwise $\{z_n\}$ would have a limit point in $A_1$ by continuity of $f^{-1}$, contradicting $|z_n| \rightarrow 1$. Hence $|f(z_n)| \rightarrow 1$ or $|f(z_n)| \rightarrow R_2$ (it must converge by continuity). The latter case is ruled out as $f(z_n) \in V \subseteq A(1, r)$.
Similarly, we also have:
Claim:  $|f(z_n)| \rightarrow R_2$ whenever $|z_n| \rightarrow  R_1$.
Now set $\alpha = \log R_2/ \log R_1$. Define $g: A_1 \rightarrow \mathbb{R}$ by $$g(z)  = \log |f(z)|^2 - \alpha \log |z|^2 = 2(\log |f(z)| - \alpha \log |z|).$$ We know that $\log |h|$ is harmonic whenever $h$ is holomorphic and nonzero, so $g$ is a harmonic function. By the two claims, $g$ extends to a continuous function on $\overline{A_1}$ vanishing on $\partial A_1$. This forces $g$ to vanish identically on $A_1$. In particular, $$0 = \frac{\partial g}{\partial z} = \frac{f'(z)}{f(z)} - \frac{\alpha}{z}.$$ Take some $1 < c < R_1$ and let $\gamma(t) = ce^{it}, t \in [0, 2\pi]$. We have $$\alpha = \frac{1}{2 \pi i}\int_\gamma \frac{\alpha}{z} dz = \frac{1}{2 \pi i}\int_\gamma \frac{f'(z)}{f(z)} dz = \mathrm{Ind}_{f \circ \gamma}(0),$$ so $\alpha > 0$ is an integer. Observe that $$\frac{d}{dz}(z^{-\alpha}f(z)) = z^{-\alpha - 1}(-\alpha f(z) + zf'(z)) = 0,$$ thus $f(z) = Kz^{\alpha}$ for some nonzero constant $K$. Since $f$ is injective, $\alpha = 1$. Therefore $R_2 = R_1$.

Reference:
Rudin - Real and Complex Analysis (3rd Ed.)