Originally Posted by

**thespian** This is another past exam question, which we have not been given anwers to, so i was hoping someone could check my answer, so i know if i am on the right track.

Find the first three non zero terms of the laurent series on an annulus centred at zero.

f(z)=(1/(z²(1-z)²))

Now the sinularities are at 0, and 1, and they are both double poles.

I did it as follows:

0<|z|<1

(1/(z²(1+(-z)²))=(1/(z²))*∑{n=0}to{∞} zⁿ]

I don't understand what you did here: certainly $\displaystyle \frac{1}{z^2(1-z)^2}\neq \frac{1}{z^2(1+(-z)^2)}...!$

So you can do $\displaystyle \frac{\frac{1}{z^2}}{(1-z)^2}=\frac{1}{z^2}\left(1+z+z^2+...\right)^2 =\frac{1}{z^2}(1+2z+3z^2+4z^3+...)=\frac{1}{z^2}+\ frac{2}{z}+3+4z+...$

Tonio

and i used the binomial theorm to get the cooeficients of z.

This gave me:

(1/(z²))-(2/z)+3-4z

Does any one know a way that i can check my solution, to make sure i have not stuffed it up.