# Find the residues of the function

## Homework Statement

Find the residues of the function

## Homework Equations

f(z)=$\frac{1}{sin(z)}$ at z=0, $\frac{∏}{2}$, ∏

## The Attempt at a Solution

Since the function has a simple pole at z=0

I used: Res(f,0)=lim$_{z->0}$(z-0)$\cdot$$\frac{1}{sin(z)}$=1. This means the residue of the function at z=0 is 1.

And Res(f,$\frac{∏}{2}$)=lim$_{z->∏/2}$(z-$\frac{∏}{2}$)$\cdot$$\frac{1}{sin(z)}$=0. This means the residue of the function at z=$\frac{∏}{2}$ is 0.

However I think this method can not be applied on solving z=∏. How can I work it out?

Last edited:

vela
Staff Emeritus
Homework Helper
Try expanding sin z in a series about ##z=\pi##. Or, better yet, use ##\sin z = \sin[(z-\pi)+\pi]##. The basic idea is to get everything in terms of ##z-\pi##.

Try expanding sin z in a series about ##z=\pi##. Or, better yet, use ##\sin z = \sin[(z-\pi)+\pi]##. The basic idea is to get everything in terms of ##z-\pi##.

If I turn everything into z-∏, do we need to use the same method as I used to work out z=0 and z=$\frac{∏}{2}$?
But how? Turn the limit to z-∏→0?
Because I used computer to work out the answer of z=∏ which is -1. But I still cannot get this answer by myself.

vela
Staff Emeritus
$$\lim_{z \to \pi}\ (z-\pi)\frac{1}{\sin z}.$$ Rewriting the limit in terms of ##z-\pi## is simply to make the evaluation easier. I just realized you could simply apply L'Hopital's rule to do that and avoid unnecessary complications. If you had a more complicated function, however, writing things in terms of ##z-\pi## is often less work than using L'Hopital's rule.
$$\lim_{z\to\pi} \frac{z-\pi}{-\sin(z-\pi)}.$$ You might already recognize that limit, but it not, use the substitution ##w=z-\pi## to turn it into one you should recognize.