# 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.