Let
be a one to one analytic function whose domain is a region
Then
is analytic on
and
for![]()
Proof: We must show that
for
and
is continuous on![]()
To prove the first part note that if
for some
then by the local mapping theorem
is many to one near
contradicting that
is one to one on![]()
To prove the second part let
and put
We must show that for each there is a
such that![]()

We know that
is an open set by the Open Mapping Theorem so there exists
such that![]()
This implies that![]()