# Classifying Singularities and the Laurent Series

1. Jun 10, 2016

### Physgeek64

1. The problem statement, all variables and given/known data
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

2. Relevant equations

N/A
3. 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 $\frac{1}{z(i)(iz-\frac{1}{zi})}$ ?

Last edited: Jun 10, 2016
2. Jun 11, 2016

### stevendaryl

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

3. Jun 11, 2016

### Physgeek64

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

4. Jun 11, 2016

### stevendaryl

Staff Emeritus
Okay, that needs to be part of the answer to your homework problem. What is the expansion about $z=n\pi\ i$ for $n \neq 0$?

5. Jun 11, 2016

### Physgeek64

Can we not just evaluate it at these points? Since the rest of the function will be analytic?

6. Jun 11, 2016

### stevendaryl

Staff Emeritus
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 $z$

7. Jun 11, 2016

### Physgeek64

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

8. Jun 11, 2016

### stevendaryl

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

9. Jun 11, 2016

### Physgeek64

So $sinhz= (z-n\pi i)+ \frac{(z-n \pi i)^3}{3!} +....$ ?

10. Jun 11, 2016

### stevendaryl

Staff Emeritus
Yes. You can see that by letting $q = z - n \pi i$. Then $sinh(z) = sinh(q + n \pi i) = sinh(q) = q + \frac{q^3}{3!} + ...$

11. Jun 11, 2016

### Physgeek64

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?

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