Conjugate of the derivative...

• Apr 12th 2010, 11:51 AM
DJDorianGray
Conjugate of the derivative...
Hi everyone,
I need help with the following:

If $\displaystyle f$ is $\displaystyle C^1$ show that

$\displaystyle \left( \frac{\partial}{\partial z} f \right)^* = \frac{\partial}{\partial z^*} f^*$

where ^$\displaystyle *$ is the complex conjugate. I am not sure what $\displaystyle f^*$ is so I have no place to start. Any help is greatly appreciated. Thanks
• Apr 12th 2010, 11:56 AM
If ^* is the complex conjugate, then $\displaystyle f^*$ is the complex conjugate of f..... It may help you to write f(x,y) = u(x,y) + i v(x,y) and then use the facts that $\displaystyle x=\frac{z+z^*}2$ and $\displaystyle y=\frac{z-z^*}2$.
• Apr 12th 2010, 12:15 PM
DJDorianGray
Quote:

Originally Posted by maddas
If ^* is the complex conjugate, then $\displaystyle f^*$ is the complex conjugate of f..... It may help you to write f(x,y) = u(x,y) + i v(x,y) and then use the facts that $\displaystyle x=\frac{z+z^*}2$ and $\displaystyle y=\frac{z-z^*}2$.

Can I always write $\displaystyle f$ in such a form? That was my problem. I will try what you said in the meantime, thanks.
• Apr 12th 2010, 12:21 PM