Computing the Laurent series of $\frac{1}{z^2 + z - 6}$ in the region $2 < |z| < 3$ 
Make a series expansion of $f(z)=\dfrac{1}{z^2+z-6}$ valid in the region $2<|z|<3$.

By partial fractions,
$$f(z) = \frac{1}{(z-2)(z+3)} = \frac{1}{5(z-2)}-\frac{1}{5(z+3)}.$$
From here, how are these fractions expanded into a geometric series?


$\forall|x|\lt1$$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

$$\frac{1}{5(z-2)}=\frac {1}{5z}\cdot\frac{1}{1-\frac{2}{z}}=\frac {1}{5z}\left(1+\frac2z+\frac{4}{z^2}+\cdots\right)$$
$$\frac{1}{5(z+3)}=\frac{1}{3\cdot5}\cdot\frac{1}{\frac{z}{3}+1}=\frac{1}{15}\left(1-\frac z3+\frac{z^2}{9}-\cdots\right)$$
Why $\dfrac2z$ and $\dfrac z3$ See Following graph for ( Visualization )

