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

  1. #1
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    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)}<br />
    I confess that i don't understand this equality. Any explanation would be great for me. Thanks!
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  2. #2
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    It's a finite geometric series of \displaystyle N terms, with \displaystyle a = 1, r = e^{ix}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    It's a finite geometric series of \displaystyle N terms, with \displaystyle a = 1, r = e^{ix}.
    Thank's!
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  4. #4
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    Quote Originally Posted by Prove It View Post
    It's a finite geometric series of \displaystyle N terms, with \displaystyle a = 1, r = e^{ix}.
    Indeed! Thank's
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  5. #5
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    Quote Originally Posted by hurz View Post
    \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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