# Classifying Singularities and the Laurent Series

## Homework Statement

Classify the singularities of $\frac{1}{z^2sinh(z)}$ and describe the behaviour as z goes to infinity

Find the Laurent series of the above and find the region of convergence

N/A

## The Attempt at a Solution

I thought these two were essentially the same thing, but I'm being given a fair few marks for each question, so I'm convinced I must be over-simplifying it

For the first part, I used to Laurent expansion to get:

$\frac{1}{z^2sinh(z)} = \frac{1}{z^2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}$
$\frac{1}{z^2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}= \frac{1}{z^3(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)}$
$\frac{1}{z^3(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)}=\frac{1}{z^3}(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)^{-1}$
$\frac{1}{z^3}(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)^{-1}=\frac{1}{z^3}(1-\frac{z^2}{3!}-\frac{z^4}{5!}+(\frac{z^2}{3!}+\frac{z^4}{5!})^2+...)$
$= \frac{1}{z^3} -\frac{1}{6z}+\frac{3z}{40}+....$

Hence, I got there to be an isolated singularity of order 3 at z=0, a simple pole of at z=0 and an essential singularity at $z=\infinity$

So the behaviour as z goes to infinity is that the function is undefined? (I showed this by redefining a new variable $\frac{1}{z}$

I just can't see what else to do, but for the number of marks it is worth, I don't feel I have done enough work

Also, Im not sure how to find the region of convergence

Many thanks

EDIT: Just had a thought- For the first part I could have written $\frac{1}{z(i)(iz-\frac{1}{zi})}$ ?

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stevendaryl
Staff Emeritus
Just a question for you: Is there only one value of $z$ where $sinh(z) = 0$?

Just a question for you: Is there only one value of $z$ where $sinh(z) = 0$?

Oh no of course! There are also poles every $n \pi i$ With order $\frac{1}{n^2 \pi^2}$ ?

stevendaryl
Staff Emeritus
Oh no of course! There are also poles every $n \pi i$ With order $\frac{1}{n^2 \pi^2}$ ?
Okay, that needs to be part of the answer to your homework problem. What is the expansion about $z=n\pi\ i$ for $n \neq 0$?

Okay, that needs to be part of the answer to your homework problem. What is the expansion about $z=n\pi\ i$ for $n \neq 0$?
Can we not just evaluate it at these points? Since the rest of the function will be analytic?

stevendaryl
Staff Emeritus
Can we not just evaluate it at these points? Since the rest of the function will be analytic?
You wrote down an expansion in powers of $z$. You need an expansion in powers of $z-n\pi i$ for the other poles.

If you have a pole at $z=p$, then the Laurent series is in powers of $(z-p)$, not powers of $z$

You wrote down an expansion in powers of $z$. You need an expansion in powers of $z-n\pi i$ for the other poles.

If you have a pole at $z=p$, then the Laurent series is in powers of $(z-p)$, not powers of $z$

Okay, so we can expand $\frac {1}{z}= \frac{1}{n \pi i +z- n \pi i} = \frac{1}{n \pi i} (1+\frac{z-n \pi i}{n\pi i})^{-1}$
$= \frac{1}{n \pi i}(1-\frac{z-n \pi i}{n\pi i}+\frac{(z-n \pi i)^2}{(n\pi i)^2}+...)$

How would we expand sinhx such that the brackets of $z-n \pi i$ remain in the denominator

Many thanks

stevendaryl
Staff Emeritus
Okay, so we can expand $\frac {1}{z}= \frac{1}{n \pi i +z- n \pi i} = \frac{1}{n \pi i} (1+\frac{z-n \pi i}{n\pi i})^{-1}$
$= \frac{1}{n \pi i}(1-\frac{z-n \pi i}{n\pi i}+\frac{(z-n \pi i)^2}{(n\pi i)^2}+...)$

How would we expand sinhx such that the brackets of $z-n \pi i$ remain in the denominator

Many thanks
Well, since $sinh$ is periodic, the expansion in powers of $z$ is the same as the expansion in terms of $z - n \pi i$.

Well, since $sinh$ is periodic, the expansion in powers of $z$ is the same as the expansion in terms of $z - n \pi i$.
So $sinhz= (z-n\pi i)+ \frac{(z-n \pi i)^3}{3!} +....$ ?

stevendaryl
Staff Emeritus
So $sinhz= (z-n\pi i)+ \frac{(z-n \pi i)^3}{3!} +....$ ?
Yes. You can see that by letting $q = z - n \pi i$. Then $sinh(z) = sinh(q + n \pi i) = sinh(q) = q + \frac{q^3}{3!} + ...$
Yes. You can see that by letting $q = z - n \pi i$. Then $sinh(z) = sinh(q + n \pi i) = sinh(q) = q + \frac{q^3}{3!} + ...$
Ah okay. Does this mean that strictly speaking this only has a taylor series about $n \pi i$ ? Also when calculating the residue (slightly off topic I'm afraid) about $n \pi i$ do you need to expand $sinhz$ or can we simply 'read off' the residue?