Solution
= Solution
In the middle annulus, expand the <pole> at one in negative powers and the <pole> at two in positive powers:
$$
-\frac1{z-1}=-\sum_{n=0}^\infty z^{-n-1},\qquad
\frac1{z-2}=-\sum_{n=0}^\infty2^{-n-1}z^n.
$$
Their <geometric series> convergence conditions are $|z|>1$ and $|z|<2$, respectively. The required <Laurent series> is
$$
\boxed{f(z)=-\sum_{n=0}^\infty z^{-n-1}-\sum_{n=0}^\infty2^{-n-1}z^n},\qquad1<|z|<2.
$$