Is my proof correct? (Conformal equivalence of two circular annuli) 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!

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.
