Conformality of two concentric annuli I'm trying to prove that two annuli, $A_1:=\{z:r_1<\left| z \right| < R_1  \}$ and $A_2:=\{z:r_2<\left| z \right| < R_2  \}$ are conformally equivalent if and only if $\frac{R_1}{r_1}=\frac{R_2}{r_2}$. I'd need to prove this using relatively simply concepts, i.e. the proof should probably be quite short. I think that in this case some other theorems, like Riemann's mapping theorem, should be used, but I don't know how Riemann's mapping theorem might apply in this case.
One possibility of such a proof could be via the proof that if $A:=\{z:1<\left| z \right| < R_1  \}$ and $A':=\{z:1<\left| z \right| < R_2  \}$ if and only if $R_1=R_2$, by deducing that if we divide the set $A_1$ by $r_1$, and set $A_2$ by $r_2$, then $\frac{R_1}{r_1}=\frac{R_2}{r_2}$ (or is this too lame?). Unfortunately, the proof I could find, of this theorem, for the case of $A$ and $A'$, uses concepts which I don't yet know.
I'd appreciate some help with this problem.
 A: By scaling, we can assume wlog that $r_1 = r_2 = 1$.
Suppose $f: A \to A'$ is analytic, one-to-one and onto.  As $|z| \to 1$, either $|f(z)| \to 1$ or $|f(z)| \to R_2$.  If the latter, replace $f$ by
$R_2/f$, so we may assume it is the former.  As $|z| \to R_1$ we have $|f(z)| \to R_2$.  Now $u(z) = \log |f(z)| - \dfrac{\log|R_2|}{\log|R_1|} \log |z|$ is a harmonic function on $A$, extending continuously to its closure with boundary values $0$ on the circles $|z|=1$ and $|z|=R_1$.  Since harmonic functions have no local max or min, $u = 0$.  Thus
$$|f(z)| = \exp\left( \dfrac{\log|R_1|}{\log|R_2|} \log |z|\right) = |z|^\alpha$$ where $\alpha = \log|R_1|/\log|R_2| > 0$.  Now locally $z = \exp(w)$ and $f(z) = \exp(g(w))$ where $\text{Re}(g(w)) = \log |f(z)| = \alpha \log |z| = \alpha \text{Re}(w)$, implying that $g(w) = \alpha w + ic$ for real constant $c$.  We get $$\dfrac{f'(z)}{f(z)} = \dfrac{\alpha}{z}$$
This has a continuous nonzero solution on the annulus iff $\alpha$ is an integer, and the solutions are one-to-one only for $\alpha = \pm 1$.
