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

• Mar 29th 2011, 06:08 PM
hurz
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
$\displaystyle \sum_{n=0}^{N-1} (exp(ix))^n = \frac {1 - exp(iNx)} {1 - exp(ix)}$
I confess that i don't understand this equality. Any explanation would be great for me. Thanks!
• Mar 29th 2011, 06:10 PM
Prove It
It's a finite geometric series of $\displaystyle \displaystyle N$ terms, with $\displaystyle \displaystyle a = 1, r = e^{ix}$.
• Mar 29th 2011, 06:20 PM
hurz
Quote:

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

Thank's! :)
• Mar 29th 2011, 06:20 PM
hurz
Quote:

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

Indeed! Thank's :)
• Mar 30th 2011, 08:45 AM
bkarpuz
Quote:

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

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