# Residues of reciprocal polynomials and functions involving reciprocal polynomials

1. Jan 10, 2010

### NewGuy

I have need to calculate the residues of some functions of the form $\frac{f(x)}{p(x)}$ where p(x) is a polynomial. To be more specific I have already calculated the 2 residues of $\frac{1}{x^2+a^2}$. That one was quite easy. Now I'm asked to calculate the residues of
$\left(\frac{1}{x^2+a^2}\right)^2$ and $\frac{z^2}{x^2+a^2}$
How would I do that? I have trouble splitting the fraction up into a series. Is there any general tips for calculating residues of functions with polynomials in the denominator?

2. Jan 11, 2010

### brahman

You just try to apply the Cauchy theorem. It's not so hard. For example, the function
$f(z) = \left( \frac{1}{z^2 + a^2}\right) ^2$​
has two singular point $z_{1,2} = \pm \, a i$, where $i^2 = -1$. As the definition of residue, we have
$Res \left[ f , z = ai \right] = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1}{(z^2 + a^2)^2} dz$​
with $\epsilon$ small enough. We write
$\frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1}{(z^2 + a^2)^2} dz = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1/(z+ai)^2}{(z - ai)^2} dz$​
and , by applying the Cauchy theorem, we have
$Res \left[ f , z = ai \right] = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1/(z+ai)^2}{(z - ai)^2} dz = \left. \frac{1}{1!} \frac{d}{dz} \left( \frac{1}{(z+ai)^2}\right) \right|_{z=ai} = \frac{1}{4a^3 i}$​
the first one
$Res [ f , z = - ai] = \frac{1}{ 2 \pi i } \oint _{|z + ai| = \epsilon} \frac{z^2/(z-ai)}{(z+ai)} dz = \left. \left( \frac{z^2}{(z-ai)^2}\right) \right|_{z=-ai} = \frac{a}{2i}$​
In fact, there're so many way to defeat these problems in textbook of complex analysis.