# Homework Help: Calculating the residue of complicated expression

1. Jun 17, 2013

### VVS

1. The problem statement, all variables and given/known data

Hi, I want to calculate the residue of this expression:

2. Relevant equations

I know that the residue of a function with a pole of k-th order is given by this:

3. The attempt at a solution

I know that the function has infinite number of poles at k*∏, for k=-∞ to k=+∞
I think one has to expand the sine function or something like that but I don't know how it simplifies things.

2. Jun 17, 2013

### Office_Shredder

Staff Emeritus
Well the obvious first question is what is the order of the poles so you can use your formula?

3. Jun 17, 2013

### Ray Vickson

This function does not have just ONE residue, so it is wrong to talk about THE residue. It has residues at lots of different points. What, exactly, do you want?

4. Jun 17, 2013

### VVS

I'd say it is of third order. Do I need to expand sin(z) around pi?

5. Jun 17, 2013

### VVS

I know the answer, but I can't work it out myself. We want an expression for the residues as a function of k, since we have poles at multiples k*pi.

6. Jun 17, 2013

### Ray Vickson

Look first at the case k = 0. What do you think you should do? (You really should work this out for yourself; that is the only way to learn!)

Then look at k = ±1, ±2, etc.; these give different results from the k = 0 case. Do you see why?

7. Jun 17, 2013

### jackmell

You got them k's in there representing two different things, one of the worst things you can do in math. The formula for the residue is in terms of the order of the pole given as k. Me, I'd change the pi thing to $n\pi$. Ok, got that straight. Now what is the order of the pole? How you know it's third order or k=3? Do that later. Let's assume it is for $n\neq 0$, then by your formula, we could write:

$$\text{Res}\left(f(z),n\pi\right)= \lim_{z\to n\pi} \frac{1}{2}\frac{d^2}{dz^2}\left\{(z-n\pi)^2 f(z)\right\},\quad n=\pm 1, \pm 2,\cdots$$

We can't take that limit?

Last edited: Jun 17, 2013