= Solution
{c}
Start with the <partial fraction decomposition>
$$
\frac1{(z+1)(z-2)}=-\frac1{3(z+1)}+\frac1{3(z-2)}.
$$
For $1<|z|<2$, expand the first fraction in negative powers and the second in nonnegative powers using the <geometric series>:
$$
\frac1{z+1}=\sum_{n=0}^\infty(-1)^nz^{-n-1},\qquad
\frac1{z-2}=-\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}.
$$
Thus the <Laurent expansion> on this <annulus> is
$$
\boxed{\frac1{(z+1)(z-2)}=-\frac13\sum_{n=0}^\infty(-1)^nz^{-n-1}-\frac13\sum_{n=0}^\infty\frac{z^n}{2^{n+1}},\qquad1<|z|<2.}
$$
For $|z|>2$, both fractions expand in negative powers:
$$
\frac1{z-2}=\sum_{n=0}^\infty2^nz^{-n-1}.
$$
Combining the two <geometric series> gives
$$
\boxed{\frac1{(z+1)(z-2)}=\frac13\sum_{n=0}^\infty\bigl(2^n-(-1)^n\bigr)z^{-n-1},\qquad|z|>2.}
$$
The $z^{-1}$ coefficient is zero, consistently with the original <rational function> being $O(z^{-2})$ at infinity. Each displayed <series> converges absolutely on its stated <annulus> and uniformly on every <compact subset> of that <annulus>.
Back to article page
