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