# Anyone familiar with residue theory

1. Dec 11, 2005

### tilika123

from my understanding we use residue theory when we have poles.
The question i have is
if f(z) = 1/(1-Z^2) has two poles at 1, -1 each of order 1
then does
Res[f(z),-1] = lim as z -> -1 of (z+1)(f(z)) = -1/2

if we have a pole of order 1 then

Res[f(z),z0] = lim as z -> z0 of (z - z0)f(z)

or does
Res[f(z),-1] = lim as z -> -1 of (z+2)(f(z)) = undefined

Last edited: Dec 11, 2005
2. Dec 11, 2005

### shmoe

This is correct. So for the residue of a pole of order one at z0=-1 you look at the limit of (z+1)f(z) as z->-1, which is what your professor appears to have done, though check the sign carefully.

Where did z+2 come from?

3. Dec 11, 2005

### tilika123

from the formula in the text it states (z-z0)(f(x)) so if zo = -1 and z = z+ 1 would that not = z+2

4. Dec 11, 2005

### shmoe

z=z+1? That ain't true. If z0=-1 then z-z0=z-(-1)=z+1.

remember the reside at z0 is the coefficient of 1/(z-z0) in the Laurent series at z0. A simple pole means

$$f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1 (z-z_0)^1+a_2 (z-z_0)^2+\ldots$$

It's $$a_{-1}$$ you're after. Multiply f(z) by (z-z0) and let z->z0, poof the other terms vanish.

5. Dec 11, 2005

### tilika123

yes i just caught my mistake z is actually just a variable where z0 is where the pole is at

6. Dec 11, 2005

### tilika123

correct me if im wrong but with residues at poles you dont actually have to use laurent series you could just use residue at poles formulas. I have not really gone over laurent series but it looks confusing.

7. Dec 11, 2005

### tilika123

sorry thanks for the help

8. Dec 11, 2005

### shmoe

That's correct, but understanding where these formulas come from is a good thing.

happy to help.