# Complex exponentials

• Mar 28th 2010, 01:29 PM
Complex exponentials
This is easily the simplest question in all tutorials I have yet it's the one that I can't do...

Show that

\$\displaystyle e^{z_1 + z_2} = e^{z_1} e^{z_2}\$ where \$\displaystyle z_1\$, \$\displaystyle z_2\$ are complex numbers.

I'm using the binomial theorem but it's going in a huge mess and I'm getting nowhere. Can someone show me how it's done or show me a simpler way, binomial theorems just making me rage.
• Mar 28th 2010, 01:32 PM
HallsofIvy
Quote:

This is easily the simplest question in all tutorials I have yet it's the one that I can't do...

Show that

\$\displaystyle e^{z_1 + z_2} = e^{z_1} e^{z_2}\$ where \$\displaystyle z_1\$, \$\displaystyle z_2\$ are complex numbers.

I'm using the binomial theorem but it's going in a huge mess and I'm getting nowhere. Can someone show me how it's done or show me a simpler way, binomial theorems just making me rage.

Perhaps it would be better to use the fact that if z= x+ iy, then [tex]e^z= e^{x+ iy}= e^x(cos(y)+ i sin(y)). Now use the fact that, for real numbers \$\displaystyle x_1\$ and \$\displaystyle x_2\$ \$\displaystyle e^{x_1+ x_2}= e^{x_1}e^{x_2}\$ and the sum formulas for the sine and cosine.
• Mar 28th 2010, 01:46 PM