Anyone familiar with residue theory

[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

 

[shmoe]

if we have a pole of order 1 then
Res[f(z),z0] = lim as z -> z0 of (z - z0)f(z)

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.

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

Where did z+2 come from?

 

[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

 

[shmoe]

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

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.

 

[tilika123]

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

 

[tilika123]

correct me if I am wrong but with residues at poles you don't 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.

 

[tilika123]

sorry thanks for the help

 

[shmoe]

correct me if I am wrong but with residues at poles you don't actually have to use laurent series you could just use residue at poles formulas.

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

happy to help.