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

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

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)}
$

I confess that i don't understand this equality. Any explanation would be great for me. Thanks!

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

3. Originally Posted by Prove It
It's a finite geometric series of $\displaystyle N$ terms, with $\displaystyle a = 1, r = e^{ix}$.
Thank's!

4. 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

5. Originally Posted by hurz
$\sum_{n=0}^{N-1} (exp(ix))^n = \frac {1 - exp(iNx)} {1 - exp(ix)}
$

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