Calculating Residues of Reciprocal Polynomials

[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?

 

[brahman]

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?

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.