# Complex Analysis-Maximum Principle

• Mar 7th 2010, 11:06 PM
WannaBe
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 $z=e^{i\gamma}$ where $0<= \gamma <=\pi$
then f is a constant at the unit circle...

Thanks !
• Mar 8th 2010, 06:38 AM
Opalg
Quote:

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 $z=e^{i\gamma}$ where $0<= \gamma <=\pi$
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 $g(\overline{z}) = \overline{g(z)}$, to extend g to an analytic function on the cut plane $\mathbb{C}\setminus(-\infty,0]$. This extended function is constant on the positive real axis and hence constant everywhere.
• Mar 8th 2010, 07:48 AM
WannaBe
Wow thanks a lot!