# complex function

• Apr 18th 2010, 09:19 AM
bhitroofen01
complex function
Hi everybody,

F(z)=$\displaystyle e^x$(cosy+isiny), i must show that F($\displaystyle z_1+z_2$)=F($\displaystyle z_1$).F($\displaystyle z_2$)
• Apr 18th 2010, 06:24 PM
hollywood
You'll find it a lot easier if you recognize that $\displaystyle \cos{y}+i\sin{y}=e^{iy}$.

Post again if you can't finish it from there.

- Hollywood
• Apr 19th 2010, 04:02 AM
HallsofIvy
The hard way to do it, if you don't have that formula yet, is to use trig identities.

$\displaystyle F(z_1+ z_2)= e^{x_1+ x_2}(cos(y_1+ y2)+ i sin(y_1+ y_2))$

Of course, $\displaystyle e^{x_1+ x_2}= e^{x_1}e^{x_2}$.

Also, $\displaystyle cos(y_1+ y_2)= cos(y_1)cos(y_2)- sin(y_1)sin(y_2)$ and $\displaystyle sin(y_1+ y_2)= sin(y_1)cos(y_2)+ cos(y_1)sin(y_2)$ so that $\displaystyle cos(y_1+ y_2)+ i sin(y_1+ y_2)$$\displaystyle = cos(y_1)cos(y_2)- sin(y_1)sin(y_2)+ i sin(y_1)cos(y_2)+ i cos(y_1)sin(y_2)$$\displaystyle = (cos(y_1)+ i sin(y_1))(cos(y_2)+ i sin(y_2))$
• Apr 19th 2010, 05:45 AM
tonio
Quote:

Originally Posted by bhitroofen01
Hi everybody,

F(z)=$\displaystyle e^x$(cosy+isiny), i must show that F($\displaystyle z_1+z_2$)=F($\displaystyle z_1$).F($\displaystyle z_2$)
Well, write $\displaystyle z_1=x_1+iy_1\,,\,z_2=x_2+iy_2\,,\,\,x_1,x_2,y_1,y_ 2\in\mathbb{R}$ , and now just evaluate $\displaystyle F(z_1+z_2)=F((x_1+x_2)+i(y_1+y_2))=e^{x_1+x_2}\lef t[\cos(y_1+y_2)+i\sin(y_1+y_2)\right]$ .
Now just a little algebra + some powers arithmetic + basic trigonometry to get that the above certainly equals $\displaystyle \left[e^{x_1}(\cos y_1 +i\sin y_1)\right]\cdot \left[e^{x_2}(\cos y_2+i\sin y_2)\right]$