For each of the following functions find the first few terms of each of the Laurent series about the origin, that is, one series for each annular ring between singular points. Find the residue of each function at the origin.
The function is...
$$\frac{1}{z(z-1)(z-2)^2 }$$
I did partial fraction expansion and got $\frac{1}{z-1}- \frac{1}{4z} - \frac{3}{4(z-2)} + \frac{1}{2(z-2)^2}$ but am not sure where to go from here... Any help would be appreciated.
For the linear terms, $\frac1{z-a} =\frac{-1}{a}\frac1{1-z/a} =\frac{-1}{a}\sum_{n=0}^{\infty} (-1)^n\frac{z^n}{a^n} $.
For higher powers $\frac1{(z-a)^m} =\frac{(-1)^m}{a^m}\frac1{(1-z/a)^m} =\frac{(-1)^m}{a^m}\sum_{n=0}^{\infty} (-1)^n\binom{-m}{n}\frac{z^n}{a^n} $
Note that $(-1)^n\binom{-m}{n} =\frac{(-1)^n}{n!}\prod_{k=0}^{n-1} (-m-k) =\frac{1}{n!}\prod_{k=0}^{n-1} (m+k) =\frac{1}{n!}\frac{(m+n-1)!}{(m-1)!} =\binom{m+n-1}{n} $.