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Is there any simple expression for the sum: $$ S = \sum_{n = 0}^{N-1} \frac{1}{a + e^{2 \pi i n / N}} $$ where $ N $ is a positive integer and $ a $ is some real number. It feels to me like there should be, but I am unable to find it. Similarly, any formula for $$ P = \prod_{n=0}^{N-1} \left( a + e^{2 \pi i n / N} \right) $$ would be useful since $ S = \frac{d}{da} \log P $.

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Yes. The roots of $x^N - 1$ are the $N$-th roots of unity, so $$x^N-1 = \prod_{n=0}^{N-1} (x - e^{2 \pi i n / N}).$$ To get the plus sign, replace $x$ with $-x$: $$(-x)^N -1 = \prod_{n=0}^{N-1} (-x - e^{2 \pi i n / N}) = (-1)^n\prod_{n=0}^{N-1} (x + e^{2 \pi i n / N}) .$$ This means $P = a^n -1$ if $n$ is even, or $a^n +1$ if $n$ is odd.

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  • $\begingroup$ Thank you for your answer. Unfortunately, I posted the wrong question and the formula I really want is slightly different. Thank you anyway. $\endgroup$ – user122520 Jan 22 '14 at 3:40

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