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## 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

## Homework Equations

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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