Inverse Binomial Expansion within Laurent Series?

[sinkersub]

Homework Statement

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.

Homework 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}

The Attempt at a Solution

[/B]
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

 

[Mark44]

Homework Statement

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.

Homework 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}

The Attempt at a Solution

[/B]
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?

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.

 

[sinkersub]

Thanks for the clarification!