# Homework Help: Laurent Series

1. Apr 18, 2007

### moo5003

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

f(z) = sin(z)/[cos(z^3)-1]

Show that z=0 is a non-essential singularity with an ORD(f:0)=-5 and determine the co-eff. a_-1

2. Relevant equations

3. The attempt at a solution

So, after expanding the power series z^5f(z) I showed that h(0) = 2, ie removable and has order -5. My question comes when I'm finding the co-eff. I cant seem to split the function into explicit h(1/z) + g(z) form so its somewhat hard to find the a_-1 coeff. Any tips would be appreciated.

My power expansion thus far: (Sigma is always from n=0 to infinity for this)

f(z) = Sigma[(-1)^n/(2n+1)! * z^(2n+1)] / Sigma[(-1)^(n+1)/(2n+2)! * z^(6n+6) ]

How do I get a_-1 from this ?!:surprised

Alright, I think I may have gotten it:

Just written out the series goes something like:

[z-z^3/3!+z^5/5!-z^7/7!.....] / [-z^6/2!+z^12/4!.......]

The pattern I saw for the co-eff's of 1/z is as follows, tell me if you see the same thing :P.

-Sigma[(3n+5)!/(2n+2)!]

Last edited: Apr 18, 2007
2. Apr 18, 2007

### Dick

I have no idea what you are trying to do with the 'Sigmas' - but I don't think you have to do it. Just work with the explicit series you've written out. Convert the factorials to numbers. Now try to write out the first few terms of the quotient series. Hint: start by multiplying top and bottom by -2/z^6. If you keep enough terms you'll find one of the form c/z.

3. Apr 18, 2007

### Dick

BTW, it's not a removable singularity. It's a pole of order 5.