# 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!

• Why is the set of periods a conformal invariant? It seems like the whole proof is being hidden under this fact. Nov 19, 2014 at 2:10