Thread: Complex exponentials

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

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$\displaystyle e^{z_1 + z_2} = e^{z_1} e^{z_2}$ 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.

Originally Posted by Deadstar

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.

Originally Posted by HallsofIvy

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.

Cheers for that but unfortunately the next question ask you to show that very relationship. While i can do it using series expansions etc, it means that I should be able to do the original question without it.

I'll leave the thread unsolved for anyone who wants to show me how the binomial theorem method does it. While I'm sure if I sat down for long enough I'd get it done I've been staring at it for so long now I just rage at it and need to work on other things...