Since the degree of the numerator is equal to the degree of the denominator we divide the numerator by the denominator

$\therefore f(z) = \frac{z^2 - 1}{z^2 + 5z + 6} = 1 - \frac{5z + 7}{z^2 + 5z + 6}................(1)$

Let $\frac{-5z - 7}{z^2 + 5z + 6} = \frac{-5z - 7}{(z + 3)(z + 2)} = \frac{a}{z + 3} + \frac{b}{z + 2} \\ \therefore -5z - 7 = a(z + 2) + b(z + 3)$

put z = -3, a = -8

put z = -2, b = 3

$$\frac{-5z - 7}{(z + 3)(z + 2)} = \frac{-8}{z + 3} + \frac{3}{z + 2} \\ \therefore f(z) = \frac{z^2 - 1}{z^2 + 5z + 6} = 1 - \frac{5z + 7}{z^2 + 5z + 6} = 1 - \frac{8}{z + 3} + \frac{3}{z + 2} \\ f(z) = 1 - \frac{8}{z + 3} + \frac{3}{z + 2}$$

$\therefore$ f(z) is analytic in

• |z| < 2
• 2 < |z| < 3
• |z| > 3

i. |z| > 2

$\because$ |z| < 2, $\therefore$ |z| < 3

$f(z) = 1 - \frac{8}{z + 3} + \frac{3}{z + 2} \\ = 1 - \frac{8}{3} \cdot \frac{1}{1 + \frac{z}{3}} + \frac{3}{2} \cdot \frac{1}{1 + \frac{z}{2}} \\ = 1 - \frac{8}{3} \bigg(1 + \frac{z}{3}\bigg)^{-1} + \frac{3}{2} \bigg(1 + \frac{z}{2}\bigg)^{-1} \\ f(z) = 1 - \frac{8}{3} \bigg(1 - \frac{z}{3} + \bigg(\frac{z}{3}\bigg)^2 - \bigg(\frac{z}{3}\bigg)^3 + .............\bigg) + \frac{3}{2} \bigg(1 - \frac{z}{2} + \bigg(\frac{z}{2}\bigg)^@ - \bigg(\frac{z}{2}\bigg)^3 + ..........\bigg)$

This is required Taylor's series

ii. 2 < |z| < 3

We get $\bigg| \frac{2}{z} \bigg| \lt 1$ and $\bigg| \frac{z}{3} \bigg| \lt 1$

$f(z) = 1 - \frac{8}{z + 3} + \frac{3}{z + 2} \\ = 1 - \frac{8}{3} \cdot \frac{1}{1 + \frac{z}{3}} + \frac{3}{z} \cdot \frac{1}{1 + \frac{2}{z}} \\ = 1 - \frac{8}{3} \bigg( 1 + \frac{z}{3} \bigg)^{-1} + \frac{3}{z} \bigg(1 + \frac{2}{z} \bigg)^{-1} \\ f(z) = 1 - \frac{8}{3} \bigg( 1 - \frac{z}{3} + \bigg(\frac{z}{3} \bigg)^2 - \bigg(\frac{z}{3}\bigg)^3 + ....... \bigg) + \frac{3}{z} \bigg( 1 - \bigg(\frac{2}{z} \bigg) + \bigg(\frac{2}{z}\bigg)^2 - \bigg(\frac{2}{z}\bigg)^3 + .....\bigg)$

This is required Laurent Series

iii. |z| > 3

$\because$ |z| < 3, clearly, |z| > 2

$\therefore \bigg| \frac{z}{3} \bigg| \gt 1$ and $\bigg| \frac{z}{2} \bigg | \gt 1 \\ i.e. \bigg|\frac{3}{z} \bigg| \lt 1 $ and $\bigg|\frac{2}{z} \bigg| \lt 1$

$f(z) = 1 - \frac{8}{z + 3} + \frac{3}{z + 2} \\ = 1 - \frac{8}{z} \cdot \frac{1}{1 + \frac{3}{z}} + \frac{3}{z} \cdot \frac{1}{1 + \frac{2}{z}} \\ = 1 - \frac{8}{z} \bigg(1 + \frac{3}{z} \bigg)^{-1} + \frac{3}{z} \bigg(1 + \frac{2}{z} \bigg)^{-1} \\ f(z) = 1 - \frac{8}{z} \bigg(1 - \frac{3}{z} + \bigg(\frac{3}{z} \bigg)^2 - \bigg(\frac{3}{z} \bigg)^3 + .......................\bigg) + \frac{3}{z} \bigg(1 - \bigg(\frac{2}{z}\bigg) +\bigg( \frac{2}{z}\bigg)^2 - \bigg(\frac{2}{z} \bigg)^3 + ........\bigg)$

This is required Laurent Series