# What is the residue of $\frac 1 {z^n+1}$ at $e^{\frac{i\pi}n}$?

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.

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Is it $1/(z^3+1)$ or $1/(z^n+1)$? – Julián Aguirre Nov 25 '12 at 20:21
@JuliánAguirre It should be $n$, sorry. I was trying do it for $n=3$ on a piece of paper and got confused. – Bartek Nov 25 '12 at 20:22

Proposition. Let $f$ and $g$ be holomorphic functions on a neighborhood of $a\in\mathbb{C}$ such that $f(a)\ne0$ and $a$ is a simple zero of $g$. Then a is a simple pole of $f/g$ and $$\operatorname{Res}\Bigl(\frac{f}{g},a\Bigr)=\frac{f(a)}{g'(a)}.$$ Since $e^{\tfrac{i\pi}{n}}$ is a simple zero of $1/(z^n+1)$, we get $$\operatorname{Res}\Bigl(\frac{1}{z^n+1},e^{\tfrac{i\pi}{n}}\Bigr)=\frac{1}{n}\,e^{-\tfrac{i\pi(n-1)}{n}}.$$
I understand what you are doing, but I wasn't feeling like checking all the computations. I think there are some errors. Let $\omega=e^{i\pi/n}$. From $y=z-\omega$ it follows $z-e^{(2k+1)i/n}=y+\omega(1-\omega^{2k})$ and the constant term in the power series expansion of $(y+\omega(1-\omega^{2k}))^{-1}$ around $y=0$ is $\omega(1-\omega^{2k}))^{-1}$. – Julián Aguirre Nov 26 '12 at 13:05