Laurent series for $\frac{1}{z(z-1)}$ in the domain $1 < |z-2| < 2$ 
Expand $f(z) = \cfrac{1}{z(z-1)}$ in a Laurent series valid for $1 < |z-2| < 2$

First of all I wrote the fraction in partial fraction form:
$\cfrac{1}{z(z-1)} = \cfrac{-1}{z} + \cfrac{1}{z-1}$
Then, I tried manipulating the fraction to get (z-2) part:
$\cfrac{-1}{z} = \cfrac{-1}{(z-2)+2} = \cfrac{1}{2}\cfrac{-1}{1+(\cfrac{z-2}{2})}$
Now, $|\cfrac{z-2}{2}| < 1$
Thus, I can apply the series expansion of:
$\cfrac{1}{1+x} = 1 - x + x^2 - ...$, when $|x| < 1$
Thus:
$\cfrac{-1}{z} = \cfrac{-1}{2}[1 - (\cfrac{z-2}{2})^2 + ...]$
$\implies \cfrac{-1}{z} = \cfrac{-1}{2}\sum_{n = 0}^{\infty}(-1)^n(\cfrac{z-2}{2})^n$
$\cfrac{-1}{z} = \sum_{n=0}^{\infty}(-1)^{n+1}\cfrac{(z-2)^n}{2^{n+1}}\tag{1}$
But, how can I manipulate the second fraction?
Thanks

$$\frac{-1}{z} = \frac{1}{-2-(z-2)} = -\frac12\;\; \frac{1}{1 - (\frac{2-z}{2})}$$
$$= -\sum_{n=0}^{\infty}(-1)^n\frac{(z-2)^n}{2^{n+1}} $$
$$\frac{1}{z-1} = \frac{1}{1-(2-z)} =\frac{1}{z-2}\;\; \frac{1}{1 - ({2-z})^{-1}}$$
