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**Opalg** That will give you the Laurent series that converges when |z|<2. In the region |z|>2 you need to write it as

$\displaystyle \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac1z\Bigl(1+\tfrac2z + \bigl(\tfrac2z)^2+\ldots\Bigr)$.

In general, the Laurent series for $\displaystyle \frac1{z-a}$ will consist of powers of $\displaystyle \tfrac za$ in the region |z|<a, and powers of $\displaystyle \tfrac az$ in the region |z|>a.

So for the function $\displaystyle \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)} + \frac {1}{3{(z+1)^2}} $ there are three separate Laurent expansions, obtained by adding the power series expansions of the partial fractions. In the region |z|<1 you need to use positive powers of z/2 for the first fraction, and positive powers of z for the other two fractions. In the region 1<|z|<2 you need to use positive powers of z/2 for the first fraction, and negative powers of z for the other two fractions. In the region |z|>2 you need to use negative powers of z/2 for the first fraction, and negative powers of z for the other two fractions.