Suppose is the annulus for . Then

is not biholomorphic to whenever .

The proof goes like following: Suppose there exists a biholomorphic map between them, then this map can be extended to the boundary by the so called Kellog theorem, then one can reflect the annulus the inner part and outside part. Near the origin, one can use the removable singularity theorem to prove the extended map is holomophic. Thus we can get a biholomorphic map between and which must have the form . First , because the map keeps the origin. Second must have the unit length to keep the inner boundary to be . Thus the map can only be rotation around origin.

This actually tells us the automorphism of an annulus can only be the rotation. Let us verify this fact by another perspective.

As we all know, is biholomorphic to , where is the upper half plane and . Here . acts on the freely discontinuously.

Any automorphism of can be lift to an automphism of which commutes with the action . In the matrix language, these automphisms are the matrix in commutes with . Some elementary calculation shows that these matrix have the form where . Any one of these is corresponding to a rotation of the annulus.