# Residue of a ratio of polynomials

## Homework Statement

The problem is to find the inverse laplace of $$\frac{s^2-a^2}{(s^2+a^2)^2}$$

I am supposed to use the residue definition of inverse laplace (given below)

The poles of F(s) are at ai and at -ai and they are both double poles.

## Homework Equations

$$f(t) = \sum_{j=1}^{N}Res[F(s)*e^{st};s_{j}]$$

Where $$s_{j}$$ are the poles of F(s)

$$Residue(f(z)) = \frac{1}{(m-1)!}lim_{z->z0}\frac{d^{m-1}}{dz^{m-1}}((z-z0)^{m}f(z))$$

where z0 is a pole of order m of f(z)

## The Attempt at a Solution

The poles of F(s) are at ai and at -ai and they are both double poles.

If you can just help find the residue of ai I'm sure I can figure out -ai.

$$Residue[F(s);ai] = lim_{s->ai}$$$$\frac{d}{ds}\frac{(s-ai)^2(s^2-a^2)}{(s-ai)^2*(s+ai)^2}$$

$$= lim_{s->ai}$$$$\frac{d}{ds}\frac{s^2-a^2}{(s+ai)^2}$$

using quotient rule I get:

$$= lim_{s->ai}$$$$\frac{2s(s+ai)^2-2(s+ai)(s^2-a^2)}{(s+ai)^4}$$

substitute ai for s

$$=\frac{2ai(2ai)^2-2(2ai)(-2a^2)}{(2ai)^4}$$

$$=\frac{2ai(-4a^2)+8a^3i}{16a^4} = 0$$

But it can't equal zero right? I also get zero when I try to calculate the residue at -ai.

Related Calculus and Beyond Homework Help News on Phys.org
To compute residues the limit formula is (in general) not recommended. It is messy and you can easily make mistakes that you then can't correct easily by backtracking the calculations. Instead, you should compute the Laurent expansion around the poles using known series expansions.

E.g. to find the residue at s = i a, you put

s = i a + t

and compute the coefficient of 1/t:

[2iat + t^2 - 2a^2]/[t^2 + 2iat]^2 =

1/(2iat)^2 [2iat + t^2 - 2a^2] 1/[1+t/(2ia)]^2 =

1/(2iat)^2 [2iat + t^2 - 2a^2] [1-t/(ia) + O(t^2)]

So, we see that the residue is:

-i/(4a)

vela
Staff Emeritus