Let $\omega = exp(\frac{2 \pi i}{N})$ be an $Nth$ root of unity where $N$ is even, and further define $\omega_m = \omega^m$. Then I want to compute the value of the sum $$ \sum_{k=0}^{N-1} (-1)^k \omega_m^k $$
So far I have done, \begin{align*} \sum_{k=0}^{N-1} (-1)^k \omega_m^k & = 2 \sum_{k=0}^{\frac{N}{2}} \omega_m^{2k} - \sum_{k=0}^{N-1} \omega_m^k \\ &= 2 \sum_{k=0}^{\frac{N}{2}} \omega_m^{2k} - \begin{cases} 0 & m \equiv 0 (\mod N) \\ N & m \not\equiv 0 (\mod N) \end{cases} \\ \end{align*} So it suffices to find the first sum, which is where I am stuck. I can't think of a way to get a nice closed formula for the sum. Does anyone have any ideas on how to proceed, thanks in advance.
Just use$$\sum_k(-\omega_m)^k=\frac{1-(-\omega_m)^N}{1+\omega_m}=\frac{1+(-1)^{N+1}}{1+\exp\frac{2m\pi i}{N}}.$$