# Evaluating Integrals using the Residue THM

• brianhawaiian
In summary, the function sinz/z2 has a simple pole at z = 0. The rules state that if a function has a simple pole at z0, the residue is equal to the limit as z approaches z0 of (z - z0) * f(z). Integrating first is not necessary. The residue can also be calculated using the other rules, but they will give the same result in this case.
brianhawaiian

## Homework Statement

integral |z|=1 of sinz/z2dz

## Homework Equations

Rule #1 if f(z) has a simple pole at z0, then
Res[f(z),z0] = lim(as z goes to z0) (z - z0)*f(z)

Rule #2 if f(z) has a double pole at z0, then
Res[f(z),z0] = lim(as z goes to z0)d/dz (z - z0)2*f(z)

Rule #3 If f(z) and g(z) are analytic at z0, and if g(z) has a simple zero at z0 then,
Res[f(z)/g(z), z0] = f(z0)/g'(z0)

Rule #4 If g(z) is analytic and has a simple zero at z0 then,
Res[1/g(z), z0] = 1/g'(z0)

## The Attempt at a Solution

Just confused on which one to use? Would I integrate first? And if so what would my z0 be in Rest[..., z0]

Thanks

What kind of of pole does the function sinz/z2 have and where is it? Although I think 1 and 2 will give the same result.

phsopher said:
What kind of of pole does the function sinz/z2 have and where is it? Although I think 1 and 2 will give the same result.

double pole at i/2 and -i/2?

Edit: Actually simple pole at z = 0?

Last edited:
Yep.