# Hard Laurent series. A little lost.

1. May 9, 2010

### daoshay

1. The problem statement, all variables and given/known data
find the Laurent series for

$$\frac{z+2}{z^{5}-8z^{2}}$$ in 2<|z|<$$\infty$$

2. Relevant equations

3. The attempt at a solution
Well, I factored out z$$^{5}$$ in the denominator, which left me with a geometric sum (since |z|>2). I've come up with $$\frac{z+2}{z^{5}}$$$$\sum$$$$\frac{8}{z^{3}}$$$$^{n}$$

I'm not sure if I'm going in the right direction, but if I am, could someone please give me a hint at what to look at next.

2. May 9, 2010

### vela

Staff Emeritus
Looks good so far. Just multiply out and collect terms now.

3. May 9, 2010

### daoshay

When you say multiply out and collect terms, do you mean I need to write it out as:

8(z+2)/z^8 + 8^2(z+2)/z^11 + .... (tried to type it in latex, but it calculated it back to f(z)... impressive!)

and that is the Laurent series?

4. May 9, 2010

### vela

Staff Emeritus
You need to multiply it out completely so you're just left with a constant times a power of z for each term.

5. May 10, 2010

### daoshay

I think I've got it.

Sum from n=3 -> infinity of 2^n/z^(n+4) - Sum from n=3 -> infinity of 2^(3n+2)/z^(3n+6)

because when I multiply and combine terms I get 2^3/z^7+2^4/z^8+2^6/z^10+2^7/z^11+...

I'm sorry about my format, latex is too smart for me.

6. May 10, 2010

### daoshay

Now that I've got the infinite sum, should I write it out as the difference of two summations? Or is it typically acceptable to write it out as the sum of terms with a clear pattern? I will go and ask my professor about his preferred notation anyhow.

7. May 10, 2010

### vela

Staff Emeritus
It's a judgment call, so asking your professor is a good idea in case he or she wants it expressed in a certain way.