What is
$\Large{\mathrm{res}_{e^{\frac{i\pi}n}}\left(\frac1{z^n+1}\right)?}$
My result doesn't agree with what WolframAlpha says.
I calculate it this way. We have
$\Large{z^n+1=(z-e^{\frac{i\pi}n})\big(z-e^{\frac{3i\pi}n})\cdots(z-e^{\frac{(2n-1)i\pi}n})}$
so, for $\Large y=z-e^{\frac{i\pi}n},$ we have
$\Large{(z^n+1)^{-1}=y^{-1}\ (y-e^{\frac{i\pi}n}(1-e^{\frac{2i\pi}n}))^{-1}\ \cdots\ (y-e^{\frac{i\pi}n}(1-e^{\frac{(2n-2)i\pi}n}))^{-1}}.$
I expand each of the factors around $y=0$, and obtain that each of them except the first has $\Large e^{\frac{i\pi}n}(1-e^{\frac{2ki\pi}n})$ as the constant coefficient of the series expansion. So the coefficient at $y^{-1}$ in the expansion of the whole product is
$\Large 1\cdot e^{\frac{i\pi}n}(1-e^{\frac{2i\pi}n})\cdot e^{\frac{i\pi}n}(1-e^{\frac{4i\pi}n}) \cdots e^{\frac{i\pi}n}(1-e^{\frac{(2n-2)i\pi}n}).$
This doesn't seem to be equal to $\Large-\frac1ne^{\frac{i\pi}n},$ which is what WolframAlpha gives.