# Thread: Finding the residue of a given function

1. ## Finding the residue of a given function

Hi

I was wondering whether someone would be able to have a look at this question and my answer to see if I am correct. If I am not, or the solution requires more information, I would greatly appreciate it if you could show me how solve this problem.

QUESTION:

Find Res(g,0) for g(z) = z^-2 . cosh(z) (I mean z squared multiplied by cosh(z))

MY SOLUTION

= cosh(z) / z^2

cosh(z) = 1 + z^2/2! + z^4/4! + z^6/6!

cosh(z)/z^2 = 1/z^2 + 1/2! + z^2/4! + z^4/6! + .....

This has a pole of order 2 at z = 0

Res(g,0) = 1/2

Thank you
Doug

2. No.. Do you remember what residue is?

3. Originally Posted by dougwm
Hi

I was wondering whether someone would be able to have a look at this question and my answer to see if I am correct. If I am not, or the solution requires more information, I would greatly appreciate it if you could show me how solve this problem.

QUESTION:

Find Res(g,0) for g(z) = z^-2 . cosh(z) (I mean z squared multiplied by cosh(z))
$\cosh z = 1 + \frac{z^2}{2!}+...$ so $\frac{\cosh z}{z^2} = \frac{1}{z^2}+\frac{1}{2!}+...$.
The coefficient for $z^{-1}$ is $0$.
Thus, what does this mean ?

4. That it is a removable singularity at z = 0? Therefore the residue of this is 0?

5. Originally Posted by dougwm
That it is a removable singularity at z = 0? Therefore the residue of this is 0?
It is not a removable singularity. It is a singularity it just happens to have the residue 0.

6. How come the residue is 0? Is it directly from the definition?

I dont understand how you would work out the residue for this equation

7. See this post to see a derivation of Residue Theorem and why $a_{-1}$ is called residue.

It's called residue because when you integrate the Laurent series of f(z), only the $a_{-1}\frac{1}{z-z_0}$ term is left.

Definition: Residue is the negative first terms coefficient.

Can you see any term in the form $a_{-1}\frac{1}{z-z_0}$ in your expansion?

8. no, there is only: 1/z^2 and then 1/2! therefore thers is no 1/z with coefficient...

so the residue is zero, but it is not a removable singularity as there is 1/z^2 (therefore not all principle parts are 0) so it is just a singularity.

9. Exactly. Such singularities are called poles. See here.

10. How about this, since $\lim_{z\to 0}\left(z^2\frac{\cosh(z)}{z^2}\right)=1$, then the function has a pole of order 2 at the origin. Thus:

$\mathop\text{Res}_{z=0}\left\{\frac{\cosh(z)}{z^2} \right\}=\frac{d}{dz} cosh(z)\Bigg|_0=0$