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Is there an easy way to calculate the residue at one of the poles of a rational expression of the form $\frac{1}{1+z^n}$? I end up having to add a bunch of polar-form complex numbers in the denominator which I have no idea how to do except to convert them to rectangular coordinates, which becomes really messy. I feel like there ought to be simple pattern/rule for such a canonical expression.

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Rational expression of the form $\frac{1}{1+z^n}$ or $\frac{1}{1-z^n}$? –  Américo Tavares Mar 16 '12 at 19:00
    
oh shoot ya, not nth roots of unity, the former, I'll edit. –  Ron Jeremy Mar 16 '12 at 19:06

2 Answers 2

up vote 5 down vote accepted

The poles are at $\mathrm e^{(1+2k)\pi\mathrm i/n}$. The residue of $f(z)$ at a simple pole $z_0$ is the limit of $(z-z_0)f(z)$ as $z\to z_0$; you can find this limit using l'Hôpital's rule:

$$\lim_{z\to z_0}\frac{z-z_0}{1+z^n}=\lim_{z\to z_0}\frac1{nz^{n-1}}=\frac1{nz_0^{n-1}}=\frac1n\mathrm e^{-(n-1)(1+2k)\pi\mathrm i/n}=-\frac1n\mathrm e^{2k\pi\mathrm i/n}\;.$$

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oh nice, thanks. –  Ron Jeremy Mar 16 '12 at 19:27
    
You're welcome. –  joriki Mar 16 '12 at 19:35

Alternatively, with $\rm w^n=-1$:

$$\rm \frac{z-w}{z^n+1}=-w\frac{z/w-1}{(z/w)^n-1}=\frac{-w}{1+(z/w)+\cdots+(z/w)^{n-1}} \xrightarrow{z\to w}-\frac{w}{n}.$$

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