Homework Help: Find residue of sinz/z^4

1. May 23, 2012

zezima1

1. The problem statement, all variables and given/known data
Calculate the residue of sinz/z^4

2. Relevant equations
The residue for a pole of order m at 0 is given my:
lim z->0 [1/(m-1)!dm-1/dzm-1[(zmf(z)]]

3. The attempt at a solution
Clearly the pole has order 3:
So we get:
Re(0) = limz->0[1/2d2/dz2[sinz/z]]
I get that the only term in the limit that doesn't go to zero is:
Re(0) = limz->0[-1/2sinz/z] = -1/2
But I should get -1/3!
Apparantly something goes wrong somewhere as I should get a tree from differentiation. Can anyone see where that is? :S

2. May 23, 2012

Office_Shredder

Staff Emeritus
You didn't take the second derivative of sin(z)/z correctly

3. May 23, 2012

zezima1

hmm I thought I might have done something wrong there. Let's go over it:

(sinz/z)'' = (cosz/z-sinz/z2)' = -sinz/z - 2cosz/z2+3sinz/z3

Now as we let z->0 the only term that survives is sinz/z which ->-1. Where is my mistake?

4. May 23, 2012

scurty

Your last term should be $\frac{2 \sin{(z)}}{z^3}$. Go through that derivative again to find your mistake.

Also, the other terms do approach values, they don't "disappear." Use l'Hospital's Rule.

5. May 23, 2012

aaaa202

okay yes I see what I did wrong now. I accidently took the terms involving z of a higher order than one to go to zero when they really diverge.

6. May 23, 2012

scurty

Well, no, they don't diverge. They converge to specific points (like I said, use l'Hospital's Rule). Also, you didn't take the second derivative properly which would yield bad results as well.

7. May 23, 2012

HallsofIvy

Are you required to use that formula? Often the simplest way to find the residue of a function, at a given value of z, is to use the fact that the residue is the coefficient of $z^{-1}$ in the Laurent series for f(z) about the given value.

Here, we can start with the Taylor's series for sin(z): $z- (1/6)z^3+ \cdot\cdot\cdot$ so that
$$\frac{sin(z)}{z^4}= \frac{1}{z^3}- \frac{1}{6z}+ \cdot\cdot\cdot$$
so that the residue is, as you say, -1/6.