Find all possible Laurent series of $ f(z) = \cfrac{z^{2}-1}{z^{2} + 5 z + 6 } $.

Solution:

$f(z)=\cfrac{z^{2}-1}{z^{2} + 5 z + 6 }$

$f(z)=1-\left( \cfrac{5z+7}{z^{2} + 5 z + 6 } \right)$

Let, $\cfrac{-5z-7}{z^{2} + 5 z + 6 } = \cfrac{a}{z+3}+\cfrac{b}{z+2}$ ----------(1)

$-5z-7=a(z+2)+b(z+3)$

$-5z-7=az+2a+bz+3b$

$a+b=-5$ -----------(2)

$2a+3b=-7$ -----------(3)

Solve (2) and (3) simultaneously we get, a=-8, b=3,

$\cfrac{-5z-7}{z^{2} + 5 z + 6 } = \cfrac{-8}{z+3}+\cfrac{3}{z+2}$ -------------(4)

Put equation (4) in (1),

$f(z)=1-\cfrac{8}{z+3}+\cfrac{3}{z+2}$ -------------(5)

Case (i): When |z|<2,

$f(z)=1-\cfrac{8}{3 \left[\cfrac{z}{3}+1 \right] }+\cfrac{3}{2\left[ \cfrac{z}{2}+1 \right] }$

$f(z)= 1- \cfrac{8}{3} \left[ 1+ \cfrac{z}{3} \right] ^{-1} + \cfrac{3}{2} \left[ 1+ \cfrac{z}{2} \right] ^{-1} $

$f(z)= 1- \cfrac{8}{3} \left[ 1- \left( \cfrac{z}{3} \right) + \left( \cfrac{z}{3} \right)^{2} - ...... \right] + \cfrac{3}{2} \left[ 1- \left( \cfrac{z}{2} \right) + \left( \cfrac{z}{2} \right)^{2} - ...... \right]$

Case (ii): When 2<|z|<3

$f(z)=1-\cfrac{8}{3 \left[\cfrac{z}{3}+1 \right] }+\cfrac{3}{z\left[ 1+ \cfrac{2}{z} \right] }$

$f(z)= 1- \cfrac{8}{3} \left[ 1+ \cfrac{z}{3} \right] ^{-1} + \cfrac{3}{z} \left[ 1+ \cfrac{2}{z} \right] ^{-1} $

$f(z)= 1- \cfrac{8}{3} \left[ 1- \left( \cfrac{z}{3} \right) + \left( \cfrac{z}{3} \right)^{2} - ...... \right] + \cfrac{3}{z} \left[ 1- \left( \cfrac{2}{z} \right) + \left( \cfrac{2}{z} \right)^{2} - ...... \right]$

Case (iii): When |z|<3

$f(z)=1-\cfrac{8}{1+ z \left[\cfrac{3}{z} \right] }+\cfrac{3}{z\left[ 1+ \cfrac{2}{z} \right] }$

$f(z)= 1- \cfrac{8}{z} \left[ 1+ \cfrac{3}{z} \right] ^{-1} + \cfrac{3}{z} \left[ 1+ \cfrac{2}{z} \right] ^{-1} $

$f(z)= 1- \cfrac{8}{3} \left[ 1- \left( \cfrac{3}{z} \right) + \left( \cfrac{3}{z} \right)^{2} - ...... \right] + \cfrac{3}{z} \left[ 1- \left( \cfrac{2}{z} \right) + \left( \cfrac{2}{z} \right)^{2} - ...... \right]$