Solution
= Solution
Outside both <poles>, the two <geometric series> use negative powers:
$$
-\frac1{z-1}=-\sum_{n=0}^\infty z^{-n-1},\qquad
\frac1{z-2}=\sum_{n=0}^\infty2^n z^{-n-1}.
$$
Hence
$$
\boxed{f(z)=\sum_{n=0}^\infty(2^n-1)z^{-n-1}},\qquad |z|>2.
$$
The coefficient of $z^{-1}$ vanishes, as it must since the <rational function> starts with $z^{-2}$ at infinity.