# Calculating Residues

1. Mar 8, 2006

### Lyuokdea

I am having some problems with residue problems in complex analysis, they seem to be fairly simple problems, so my understanding of the method to solve them must be wrong.

1. Calculate the residue at z=0 of the function:
$$f(z) = \frac{sinh(z)}{z^4(1-z^2)}$$

I first Taylor expand sinh(z) to an adequate number of terms:

$$sinh(z) = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \frac{z^7}{7!} +...$$

now I divide the function through by z^4, I ignore the (1-z^2 component because at z=0 this is equal to 1. (This may be my bad assumption, but I don't see an alternative way to do the problem)

$$f(z) = z^{-3} + \frac{1}{6z} + \frac{z}{5!} + \frac{z^3}{7!} +...$$

I pull out the 1/z component only and get an answer that the residue is 1/6, but the book gives me an answer 7/6, what am I forgetting?

~Lyuokdea

2. Mar 9, 2006

### cepheid

Staff Emeritus
I'm pretty sure your assumption is wrong. You have to take a Laurent Series expansion first (about zero), you don't evaluate the function at zero, and you certainly can't pick and choose which parts of it to evaluate at zero.

3. Mar 9, 2006

### cepheid

Staff Emeritus
If you factor out the 1/(1-z^2), isn't that the sum of an infinite geometric series or something? Then you can just multiply terms...except that gets ugly. I dunno...

4. Mar 9, 2006

### cepheid

Staff Emeritus
Ignore this post

5. Mar 9, 2006

### TD

I don't think your assumption about leaving out the 1-z² factor is correct.
I believe taking the series of sinh(z) up to order 3 would do, then do a partial fraction decomposition. The terms with even powers of x can be left out since there are only odd powers in the nominator.

$$\frac{{z + \frac{{z^3 }}{6}}}{{z^4 \left( {z + 1} \right)\left( {z - 1} \right)}} = \frac{A}{{z^4 }} + \frac{B}{{z^3 }} + \frac{C}{{z^2 }} + \frac{D}{z} + \frac{E}{{z + 1}} + \frac{F}{{z - 1}} \to \frac{B}{{z^3 }} + \frac{D}{z} + \frac{E}{{z + 1}} + \frac{F}{{z - 1}}$$

Now you're only interested in D, which should give 7/6.