Solution
= Solution
Partial fractions give $f(z)=-1/(z-1)+1/(z-2)$. In the interior <disk>, expand both terms as <geometric series>:
$$
-\frac1{z-1}=\sum_{n=0}^\infty z^n,\qquad
\frac1{z-2}=-\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}.
$$
Thus the <Laurent series> is a <Taylor series> in this region:
$$
\boxed{f(z)=\sum_{n=0}^\infty(1-2^{-n-1})z^n},\qquad |z|<1.
$$
The nearer <pole> at $z=1$ determines the radius of convergence.