prove the the conjugate of a holomorphic function f is differentiable iff f'=0

**hatsoff** · February 20th 2011, 06:26 AM

This probably has a simple solution, but I cannot find it so far...

Originally Posted by problem statement

Suppose the complex-valued function $f$ is holomorphic about zero. Show that $g:=\overline{f}$ , i.e. the conjugate of $f$ , is differentiable at $z$ if and only if $f'(z)=0$ .

The first direction follows fairly directly from the Cauchy-Riemann equations. However, I still need to show that $f'(z)=0$ implies $g'(z)$ exists, and I can't seem to make it happen. Any help would be much appreciated.

Thanks!