Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$ Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$.
Is the first step here to notice that
$$ \frac{1}{z-2} + \frac{1}{z-3} = \frac{2z-5}{(z-2)(z-3)}$$
and compute the Laurent series as a single term rather than in two parts?

Case I. $2<\lvert z\rvert<3$.
$$
\frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}=\sum_{n=1}^\infty\frac{2^{n-1}}{z^n}-\sum_{n=0}^\infty \frac{z^n}{3^{n+1}}.
$$
Case II.  $\lvert z\rvert>3$.
$$
\frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}+
\frac{1}{z}\cdot\frac{1}{1-\frac{3}{z}}=\sum_{n=1}^\infty\frac{2^{n-1}+3^{n-1}}{z^n}.
$$
