Making $-{{\pi i}\over n} e^{\alpha i}({{1 - e^{2 n \alpha i}\over{1-e^{2 \alpha i}}}})={\pi \over {n sin(\alpha)}}$; $\alpha={{2m+1}\over{2n}} \pi$ As part of a (much) longer problem in complex analysis, I need to show that the equality mentioned in the title makes sense, but I can't seem to find the right algebra tricks to get from point A to point B. I've tried distributing $i e^{\alpha i}$ over the numerator, substituting in what $\alpha$ is equal to, and expanding every exponential into $cos(\theta) + sin(\theta)i$, none of which seems particularly helpful in arriving at the desired conclusion. Any help would be appreciated.
More information: I arrived at this equation by saying $Res_{z=c_k} {z^{2m} \over {z^{2n}+1}}=-{1 \over {2 n}} e^{i(2k + 1)\alpha}$ and using the given summation formula $\sum_{k=0}^{n-1} z^k={{1-z^n}\over{1-z}}$ to evaluate the sum $2 \pi i \sum_{k=0}^{n-1} Res_{z=c_k} {z^{2m} \over {z^{2n}+1}}$, which is equivalent by some factoring to $-{{\pi i}\over n} e^{\alpha i} \sum_{k=0}^{n-1} (e^{2i \alpha})^k$.

Just note that
$$
\frac{ie^{\alpha i}}{1-e^{2\alpha i}}=\frac{i}{e^{-\alpha i}-e^{\alpha i}}=\frac{-i/2i}{(e^{\alpha i}-e^{-\alpha i})/2i}=\frac{-\frac12}{\sin\alpha}
$$
and that
$$e^{2n\alpha}-1 =e^{(2m+1)\pi}-1=-1-1 =-2$$
and the rest follows immediately.
