# Find the residues of the function

• DanniHuang
In summary, the student attempted to solve the homework problem by using the trig identity and L'Hopital's rule, but they realized that this might not be the most efficient method.

## 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:
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##.

vela said:
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.

You find the residue by evaluating the limit
$$\lim_{z \to \pi}\ (z-\pi)\frac{1}{\sin z}.$$ 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.

But say you did it with using the trig identity anyway. You should end up with
$$\lim_{z\to\pi} \frac{z-\pi}{-\sin(z-\pi)}.$$ You might already recognize that limit, but it not, use the substitution ##w=z-\pi## to turn it into one you should recognize.