how to show that \sum_{n=0}^{N-1} (exp(ix))^n = \frac {1 - exp(iNx)} {1 - exp(ix)}

**hurz** · March 29th 2011, 06:08 PM

Hello guys.
I started to study complex numbers on this semester. I'm stuck on a problem. I have the solution of this problem, but i cannot understand one of the equalities. My dificulty is how to show that
$\sum_{n=0}^{N-1} (exp(ix))^n = \frac {1 - exp(iNx)} {1 - exp(ix)}<br />$
I confess that i don't understand this equality. Any explanation would be great for me. Thanks!

**Prove It** · March 29th 2011, 06:10 PM

It's a finite geometric series of $\displaystyle N$ terms, with $\displaystyle a = 1, r = e^{ix}$ .

**hurz** · March 29th 2011, 06:20 PM

Originally Posted by Prove It

It's a finite geometric series of $\displaystyle N$ terms, with $\displaystyle a = 1, r = e^{ix}$ .

Thank's!

**hurz** · March 29th 2011, 06:20 PM

Originally Posted by Prove It

It's a finite geometric series of $\displaystyle N$ terms, with $\displaystyle a = 1, r = e^{ix}$ .

Indeed! Thank's

**bkarpuz** · March 30th 2011, 08:45 AM

Originally Posted by hurz

$\sum_{n=0}^{N-1} (exp(ix))^n = \frac {1 - exp(iNx)} {1 - exp(ix)}<br />$

By the way, you can use a backslash (\) before $exp$ to get the following operator view $\exp$ .

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