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Suppose $F_n=1/(1+z^n)$ for $n$ even. I'm curious, why is the sum of residues of $F_n$ in the upper half plane a geometric series whose sum is $1/[in\sin(\pi/n)]$?

I know that if $f(z)=\frac{P(z)}{Q(z)}$ has a simple root $a$ of $Q(z)$, then $\text{Res}[f(z),a]=\frac{P(a)}{Q'(a)}$. Hence if $p$ is a pole of $F_n$, then $$ \text{Res}[F_n,p]=\frac{1}{np^{n-1}}=\frac{p}{np^n}=-\frac{p}{n}. $$

By Cauchy's Integral Formula, the sum of the residues in the upper half plane is $$ \sum_{y>0}\text{Res}[F_n,z]=\frac{1}{2\pi i}\int_{-\infty}^\infty F_n(x)dx=\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{dx}{1+x^n}. $$

I don't know how to proceed in showing this is a geometric series which sums to $1/[in\sin(\pi/n)]$. I'd appreciate suggestions on how to reach the conclusion. Thanks.

Yong Pan
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1 Answers1

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The poles of $(1+z^n)^{-1}$ occur at the $n^{\text{th}}$ roots of $-1$, namely

$$z_k = e^{i\pi(2k+1)/n}.$$

Those poles in the upper half-plane are given by $0 \leq k \leq \frac{n}{2}-1$. As you noticed, the residue at $z_k$ is

$$-\frac{1}{n}e^{i\pi(2k+1)/n},$$

so that the sum of the residues in the upper half-plane is

$$\begin{align} \sum_{k=0}^{n/2-1} - \frac{1}{n} e^{i\pi(2k+1)/n} &= - \frac{1}{n} e^{i\pi/n} \sum_{k=0}^{n/2-1} \left(e^{i 2 \pi/n}\right)^k \\ &= -\frac{1}{n} e^{i \pi/n} \frac{\left(e^{i 2 \pi/n}\right)^{n/2}-1}{e^{i 2 \pi/n}-1} \\ &= \frac{1}{n} e^{i \pi/n} \frac{2}{e^{i 2 \pi/n}-1} \\ &= \frac{1}{n} \cdot \frac{2}{e^{i\pi/n} - e^{-i\pi/n}} \\ &= \frac{1}{i n} \cdot \frac{2 i}{e^{i\pi/n} - e^{-i\pi/n}} \\ &= \frac{1}{i n \sin(\pi/n)}. \end{align}$$

Antonio Vargas
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  • Thanks for your calculation! – Yong Pan Apr 20 '12 at 21:59
  • @Antonio Vargas : I can't understand why k must be between 0 and n/2-1, please help me. – Amir Alizadeh Dec 29 '12 at 16:09
  • @AmirAlizadeh, The argument of $z_k$ increases as $k$ increases. We have $\arg z_0 = \pi/n > 0$ and $\arg z_{n/2-1} = \pi(n-1)/n < \pi$. This means that $0 < \arg z_k < \pi$ for all $0 \leq k \leq n/2-1$. But if $k$ is any larger than $n/2-1$ then the argument lies in $(\pi,2\pi)$. Indeed, $\arg z_{n/2} = \pi(n+1)/n > \pi$ and $\arg z_{n-1} = \pi(2n-1)/n < 2\pi$, so that $\pi < \arg z_k < 2\pi$ for all $n/2 \leq k \leq n-1$. We conclude that the only residues in the upper half-plane are the ones given by $0 \leq k \leq n/2-1$. – Antonio Vargas Dec 29 '12 at 19:57