Thread: Complex Analysis-Maximum Principle

Let f be an analytic function at the open unit circle and continous at |z|=1.

Prove: If f(z)=1 on the upper half of the unit circle
( for where
then f is a constant at the unit circle...



I've no idea about this question.., I'll be delighted to get some guidance....


Thanks !

Originally Posted by WannaBe

Let f be an analytic function at the open unit circle and continous at |z|=1.

Prove: If f(z)=1 on the upper half of the unit circle
( for where
then f is a constant at the unit circle...

Using a conformal map from the unit disk to the upper half-plane that takes the upper part of the unit circle to the positive real axis, you can replace the problem by one in which g is an analytic function on the upper half-plane, continuous on the real axis and such that g(z)=1 on the positive real axis. Then apply the reflection principle, defining , to extend g to an analytic function on the cut plane . This extended function is constant on the positive real axis and hence constant everywhere.

Wow thanks a lot!