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

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

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

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

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

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