complex function

**bhitroofen01** · April 18th 2010, 09:19 AM

Hi everybody,

F(z)= $e^x$ (cosy+isiny), i must show that F( $z_1+z_2$ )=F( $z_1$ ).F( $z_2$ )
can you help me please???

**hollywood** · April 18th 2010, 06:24 PM

You'll find it a lot easier if you recognize that $\cos{y}+i\sin{y}=e^{iy}$ .

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

- Hollywood

**HallsofIvy** · April 19th 2010, 04:02 AM

The hard way to do it, if you don't have that formula yet, is to use trig identities.

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

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

Also, $cos(y_1+ y_2)= cos(y_1)cos(y_2)- sin(y_1)sin(y_2)$ and $sin(y_1+ y_2)= sin(y_1)cos(y_2)+ cos(y_1)sin(y_2)$ so that $cos(y_1+ y_2)+ i sin(y_1+ y_2)$ $= cos(y_1)cos(y_2)- sin(y_1)sin(y_2)+ i sin(y_1)cos(y_2)+ i cos(y_1)sin(y_2)$ $= (cos(y_1)+ i sin(y_1))(cos(y_2)+ i sin(y_2))$

**tonio** · April 19th 2010, 05:45 AM

Originally Posted by bhitroofen01

Hi everybody,

F(z)= $e^x$ (cosy+isiny), i must show that F( $z_1+z_2$ )=F( $z_1$ ).F( $z_2$ )
can you help me please???

Well, write $z_1=x_1+iy_1\,,\,z_2=x_2+iy_2\,,\,\,x_1,x_2,y_1,y_ 2\in\mathbb{R}$ , and now just evaluate $F(z_1+z_2)=F((x_1+x_2)+i(y_1+y_2))=e^{x_1+x_2}\lef t[\cos(y_1+y_2)+i\sin(y_1+y_2)\right]$ .

Now just a little algebra + some powers arithmetic + basic trigonometry to get that the above certainly equals $\left[e^{x_1}(\cos y_1 +i\sin y_1)\right]\cdot \left[e^{x_2}(\cos y_2+i\sin y_2)\right]$

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