Solution
= Solution
For $|z|>2$, expand directly in inverse powers:
$$
f(z)=\frac1{z^2}\frac1{1-2/z}.
$$
Therefore the outer <Laurent series> is
$$
\boxed{f(z)=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+2}}},
\qquad |z|>2.
$$
= Solution
For $|z|>2$, expand directly in inverse powers:
$$
f(z)=\frac1{z^2}\frac1{1-2/z}.
$$
Therefore the outer <Laurent series> is
$$
\boxed{f(z)=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+2}}},
\qquad |z|>2.
$$