The statement is false as written; what about a constant map? I suppose you are assuming the map to be conformal; if so, then it's true and may be proven like this :

By Riemann's mapping theorem, we have a conformal map

where

is the open unit disc. Consider

given by

. Now

is a conformal map from the unit disc to itself, and therefore has the form

for some

. If

, then the fixed points of this transformation are the two solutions of

, which lie on the boundary of

(and hence they are

*not* fixed points of

, whose domain does not include the boundary). So the only possibility is

, i.e.

and

.