I was calculating a complex integral, and got an answer which is $$2\pi i\left(\text{Res}_{z=e^{\pi i/6}}f(z)+\text{Res}_{z=e^{3\pi i/6}}f(z)+\text{Res}_{z=e^{5\pi i/6}}f(z)\right)$$ where $f(z)=\dfrac{1}{z^6+1}$.
It is easy to calculate each residue, e.g. $\text{Res}_{z=e^{\pi i/6}}f(z)=\lim_{z\rightarrow e^{\pi i/6}}\dfrac{z-e^{\pi i/6}}{z^6+1}$, and similarly for the other two.
But how can I simplify the whole expression?
