TITLE: Is my proof correct? (Conformal equivalence of two circular annuli)
QUESTION [9 upvotes]: I want to show that the two annuli $$A=\{r<|z-z_0|<R\} $$
$$A'=\{r'<|z-z_0'|<R'\} $$
are conformally equivalent (i.e. there exists a biholomorphic map between the two) iff $$\frac{R}{r}=\frac{R'}{r'}. $$
Sufficiency:
Suppose the ratios of the radii are the same. A suitable scaling (by $\frac{r'}{r}$) will make $A$ congruent to $A'$, and a translation will map the scaled $A$ onto $A'$. This is clearly a conformal homeomorphism.
Necessity:
The annulus $A$ has the following set of periods
$$\left\{ \pm \frac{2 \pi}{\log (R/r)} \right\} $$
(this step required some computations which I will omit)
Similarly $A'$ has 
$$\left\{ \pm \frac{2 \pi}{ \log(R'/r')} \right\}$$
as its set of periods. Since the set of periods is a conformal invariant. The ratios of the radii must indeed by the same.
Is this proof correct? 
Thanks!

REPLY [5 votes]: You did not present much of actual proof here, but yes, the logic is correct. The periods of a domain are determined by the set of harmonic functions it supports, and conformal maps preserve harmonicity. Hence, the periods are conformal invariants of a domain. 
For a completely different proof, see When can we find holomorphic bijections between annuli? 
And for further results, Conformal maps of doubly connected regions to annuli.