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using sum of geometric progression and inverse DFT we can show: ${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{\exp(2\pi i\frac{j}{k})-a}=k\frac{a^{u(mod\ k)}}{1-a^{k}}}$

My question is how to compute

${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{(\exp(2\pi i\frac{j}{k})-a)^m}}$ here $m,u,k$ are natural numbers

1 Answers 1

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This is

$\frac{1}{n!} \frac{\mathrm{d}^{n}}{\mathrm{d}a^{n}} S (a)\;,$

where $n=m-1$ and $S(a)$ is the sum you computed.