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.