I am trying to find the Laurent series of the function $$f(z)=\frac{1}{z(z-1)(z-2)}$$in the rings: 1) $0<|z-1|<1$, 2) $1<|z-1|$, 3) $1<|z-2|<2 $
First I expressed $f$ as $$f(z)=\dfrac{1}{2z}-\dfrac{1}{z-1}+\dfrac{1}{2}\dfrac{1}{z-2}$$
In 1), we have $$-\dfrac{1}{z-1}=-1\dfrac{1}{z-1},$$ $$\dfrac{1}{2}\dfrac{1}{z-2}=\dfrac{-1}{2}(\dfrac{1}{1-(z-1)})$$$$=\sum_{k=0}^{\infty} \dfrac{-1}{2}(z-1)^k$$
I don't know how to conveniently express $\dfrac{1}{2z}$
I had the same problem in 2) for $\dfrac{1}{2z}$ and in 3) for $\dfrac{1}{2z}$ and $\dfrac{-1}{z-1}$
I would appreciate suggestions to rearrange these terms so as to express them as geometric series. Thanks in advance.