Computing the Laurent series of $\frac{1}{z^2 + z - 6}$ in the region $2 < |z| < 3$

Question

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?

Answer 1

$\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 )

Answer 2

The formula for a geometric series

$$\frac{1}{1 - r} = 1 + r + r^2 + \cdots$$

is valid only when $|r| < 1$. Since $2 < |z|$, we have $|2/z| < 1$. Hence, pulling out a factor of $z$ in the denominator,

$$\frac{1}{5}\frac{1}{z - 2} = \frac{1}{5z}\frac{1}{1 - 2/z} = \frac{1}{5z}\left( 1 + \frac{2}{z} + \frac{4}{z^2} + \cdots \right).$$

For the other term, as $|z| < 3$, we have $|z/3| < 1$. So similar to above, taking out a factor of $3$ in the denominator and writing it in the appropriate form,

$$\frac{1}{5}\frac{1}{z + 3} = \frac{1}{5\cdot3} \frac{1}{1 - (-z/3)} = \frac{1}{15}\left( 1 - \frac{z}{3} + \frac{z^2}{9} - \cdots \right).$$

In each case, we have taken out an appropriate factor in the denominator to ensure the common ratio $r$ satisfies $|r| < 1$. As a general rule, the number largest in absolute value in the denominator is factored out.