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

• February 20th 2011, 06:26 AM
hatsoff
prove the the conjugate of a holomorphic function f is differentiable iff f'=0
This probably has a simple solution, but I cannot find it so far...

Quote:

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!
• February 20th 2011, 07:52 AM
FernandoRevilla
If $f=u+iv$ and $z_0=x_0+iy_$ use:

$f'(z_0)=\dfrac{\partial u}{\partial x} (x_0,y_0) +i\dfrac{\partial v}{\partial x}(x_0,y_0)$

Fernando Revilla
• February 22nd 2011, 10:28 AM
hatsoff
Hmm. I guess I don't see how that helps us here.