# Inverse Binomial Expansion within Laurent Series?

Tags:
1. Jan 12, 2016

### sinkersub

1. The problem statement, all variables and given/known data
Find the Laurent Series of $f(z) = \frac{1}{z(z-2)^3}$ about the singularities z=0 and z=2 (separately).

Verify z=0 is a pole of order 1, and z=2 is a pole of order 3.

Find residue of f(z) at each pole.
2. Relevant equations

The solution starts by parentheses in the form (1 - az), where a is some constant.

$f(z) = \frac{-1}{8z(1-\frac{z}{2})^3}$

3. The attempt at a solution

The solution given expands $\frac{1}{(1-\frac{z}{2})^3}$ as follows, and I just don't understand how it does it! I'm simply trying to recognise what expansion this is, as I can see it's used later too! I understand almost all of the rest of the problem. A general series/hint to what series is used would be amazing!

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

I think it's a binomial expansion, but how does a binomial expansion work for negative powers?

Is it a binomial expansion?

sinkersub

2. Jan 12, 2016

### Staff: Mentor

Yes. See https://en.wikipedia.org/wiki/Binomial_series. The exponent doesn't need to be a positive integer. It can be negative or real, or even complex.

3. Jan 12, 2016

### sinkersub

Thanks for the clarification!