The function $$f(z)={1\over{(z-1)(z-2)}}$$ is to be expanded in Laurent Series centered at $0$ and converging in the annular region $\{1\lt |z| \lt 2\}$ .
How the book does it :
$$f(z)={1\over{(z-1)(z-2)}}\\={1\over {z-2}}-{1\over{z-1}}$$ Next,$${1\over{z-2}}=-{1\over 2}{1\over {1-{z/2}}}\\=-{1\over 2}\left(1+{z\over 2}+{z^2\over 4}+...\right)$$ and $${1\over {z-1}}={1\over z}\cdot{1\over{1-1/z}}\\={1\over z}+{1\over z^2}+{1\over z^3}+...$$
Thus we have our expansion . Done.
The alternative I was doing :
$$f(z)={1\over{(z-1)(z-2)}}\\={1\over {z-2}}-{1\over{z-1}}$$ Next, $${1\over{z-2}}={1\over z}\cdot {1\over {1-2/z}}\\={1\over z}\left( 1+{2\over z}+{4\over z^2}+{8\over z^3}+....\right)$$
and $${1\over {z-1}}=-{1\over {1-z}}\\=- \left(1+z+z^2+z^3+z^4+.... \right)$$
In both cases , we expand the series in sum of two series where one of them is in positive powers of $z$ and the other is in negative powers of $z$ . Are both of them correct or is there a particular reason that the book expands it the way it did $?$ Does the region of expansion gets to decide how the expansion should be done $?$
Thank you .
The Laurent expansion of a holomorphic function in an annulus is unique. So one of the answers must be wrong. In fact your expansion is wrong.
You use the fact that $$\frac1{1-r}=\sum_{n=1}^\infty r^n.$$ But that holds only for $|r|<1$. In both your expansions you use this formula for an $r$ with $|r|>1$. ($|2/z|>1$ and $|z|>1$ in the annulus in question.)