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 $$\frac{R}{r}=\frac{R'}{r'}. $$


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.


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?


  • 6
    $\begingroup$ Why is the set of periods a conformal invariant? It seems like the whole proof is being hidden under this fact. $\endgroup$
    – fhyve
    Nov 19, 2014 at 2:10

1 Answer 1


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.


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