# Complex Numbers

• March 23rd 2011, 01:41 AM
sebko
Complex Numbers
have tried to do this without much success.

Find imaginary party v=v(x,y) such that
f(z)=u+i*v
for a u(x,y)=e(x)cos(y)

I have tried something using the cartesian form of the cauchy riemann equations. and ended up getting . exp(x)*sin(y)+K, but i think im totally of.
• March 23rd 2011, 04:14 AM
FernandoRevilla
Quote:

Originally Posted by sebko
I have tried something using the cartesian form of the cauchy riemann equations. and ended up getting . exp(x)*sin(y)+K, but i think im totally of.

That is right, $u=e^x\cos y$ is harmonic on $\mathbb{R}^2$ so, there exists $v$ such that $f=u+iv$ is holomorphic in $\mathbb{C}$ . Using the Cauchy Riemann equations we obtain $v=e^x\sin y+k\;(k\in\mathbb{R})$ . Then, $f(z)=e^z+ki$ .
• March 23rd 2011, 06:58 AM
sebko
Muchas Gracias!
You have made it very clear