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I'm looking for a way to write the following sum as a closed form expression:

$$ \sum_{n=-N}^{N} e^{cn*i} $$

where $c$ is a constant and $i$ is the imaginary unit.

I found a closed form solution for $n= [0, N-1]$ at http://mathworld.wolfram.com/ExponentialSumFormulas.html , but I'm not sure how I can manipulate this to change the range from -N to N. Any help would be appreciated.

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This is just a geometric sequence. The first term is $e^{-cNi}$ and you keep multiplying by $e^{ci}$. You have $2N+1$ terms. the sum is then $$e^{-cNi}\frac{1-e^{ci(2N+1)}}{1-e^{ci}}$$

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    $\begingroup$ Sorry, can you explain why I would be multiplying by $e^{-ci}$ with each iteration? $\endgroup$ – Skipher Apr 9 at 15:04
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    $\begingroup$ Sorry, its $e^{ci}$. I'll fix the answer $\endgroup$ – Andrei Apr 9 at 15:07
  • $\begingroup$ My fundamental knowledge is lacking in understanding why I would multiply for each iteration of a summation. Could you explain why I multiply and where $e^{ci}$ comes from? Thank you so much your answer. $\endgroup$ – Skipher Apr 9 at 15:10
  • $\begingroup$ The sum looks like $a_0+a_0r+a_0r^2+...$ Look at the ratio of two consecutive terms: $\frac{e^{c(n+1)i}}{e^{cni}}=e^{ci}$ $\endgroup$ – Andrei Apr 9 at 15:13
  • $\begingroup$ Oh, I totally missed the connection to a geometric series. Thank you for explaining. $\endgroup$ – Skipher Apr 9 at 15:17
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$$\sum_{n=-N}^N e^{cni} = \sum_{n=-N}^0 e^{cni} + \sum_{n=1}^N e^{cni} = \sum_{n=0}^N e^{-cni} + \sum_{n=1}^N e^{cni} = \frac{1-e^{-ci(N+1)}}{1-e^{-ci}}+ e^{ci}\frac{1-e^{cNi}}{1-e^{ci}}$$

$$=\frac{1-e^{-ci(N+1)}}{1-e^{-ci}}+ \frac{e^{cNi}-1}{1-e^{-ci}} = \frac{e^{cNi}-e^{c(N+1)i}}{1-e^{-ci}} = -e^{c(N+1)i}$$

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