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Let $z=e^{i\theta}$ with $\theta \in [0,2\pi[$ . Consider the sum $$ \sum_{n=1}^{N} (e^{i\theta})^n. $$ How could this be equal to $$ \frac{1-e^{iN+T\theta}}{1-e^{i\theta}} \quad ? $$ I tried to apply the sum as if it were $z^n$ in place of $e^{i\theta}$, but I got a slightly different expression, that is
$$ \frac{1-e^{i\theta N}}{1-e^{i\theta}}. $$ Any idea?

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Are you sure the "$+T$" belongs there at all? I don't see any $T$ in the original expression. – Owen Biesel Oct 8 '12 at 20:55
No, unfortunately I have no informations about $T$. I was looking for a summation that can be put in that form, but I am beginning to suspect that the equality we talk about is a mistake/typo. – Flast9 Oct 8 '12 at 21:08


Use geometric progression formula $1+q+q^2+\ldots+q^n=\dfrac{1-q^{n+1}}{1-q}$ for $\theta \notin \{0,\,2\pi \} $ and separately calculate sum $\sum\limits_{n=1}^{N} (e^{i\theta})^n$ for the case $e^{i \theta}=1.$

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