<s>[INST] Finding the Laurent series of a given function How do I find the Laurent series expansion for the function
$\frac{z}{(z+1)(z+2)}$ 
which converges for $1 < |z| < 2$ and diverges elsewhere.
I have done the partial fractions for $f(z) = \frac{2}{z+2} - \frac{1}{z+1}$.
 [/INST] As you found already, $f(z) = \frac{2}{2+z} - \frac{1}{1+z}$
Let's look at each fraction separately:  
$-\frac{1}{1+z}$ is analytic in $|z| \gt 1$ and since $|\frac{1}{z}| \lt 1$ we have:
$$
-\frac{1}{1+z} = -\frac{1}{z}\cdot\frac{1}{1-(-\frac{1}{z})} = -\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}-\dots) = -\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+\dots
$$
$\frac{2}{2+z}$ is analytic in $|z|\lt2$, we have $|\frac{z}{2}| \lt 1$ and we can represent it with it's Taylor series:
$$
\frac{2}{2+z} = \frac{1}{1-(-\frac{z}{2})} = 1 -\frac{z}{2}+\frac{z^2}{4}-\dots
$$
Therefore the Laurent series:
$$
\dots -\frac{1}{z^3}+\frac{1}{z^2}-\frac{1}{z}+1-\frac{z}{2}+\frac{z^2}{4}-\dots
$$
Converges in $1 \lt |z| \lt 2$
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