I'm trying to find Laurent series of the following function at $$\mid z\mid<1$$ My function is as follows: $$f(z)=\dfrac{1}{z(z-1)(z-2)}=\dfrac{1}{z}\left(\dfrac{1}{(z-2)}+\dfrac{1}{(1-z)}\right)$$
My attempt was like this : $$\dfrac{1}{z}\left(\dfrac{-1}{1+(1-z)}+\sum_{k=0}^{n}z^k\right)=\dfrac{1}{z}\left(\dfrac{-1}{1-\omega}+\sum_{k=0}^{n}z^k\right)$$ So by calling $$\omega=z-1$$ I made it Taylor series expandable. But I found the wrong answer. Is it because by changing variable I expended my function for point 1 ?
HINTS:
Note that for $|z|<1$,
$$\frac{1}{1-z}=\sum_{n=0}^\infty z^n$$
and for $|z|<2<1$
$$\frac{1}{2-z}=\frac12\sum_{n=0}^\infty \left(\frac{z}2\right)^n$$