Obtain Laurent's series for $f(z)=\cfrac{1}{z(z+2)(z+1) }$ about $z=-2$.

Solution:

$f(z)=\cfrac{1}{z(z+2)(z+1)}$

$\therefore \cfrac{1}{z(z+2)(z+1)}=\cfrac{A}{z} + \cfrac{B}{z+1}+ \cfrac{C}{z+2}$

$1=A(z+1)(z+2)+Bz(z+2)+Cz(z+1)$

Put z=-1,

$ \Longrightarrow 1=B(-1)(-1+2) \Longrightarrow -1=B \Longrightarrow B=-1$

Put z=-2,

$ \Longrightarrow 1=C(-2)(-2+1) \Longrightarrow C=\cfrac{1}{2}$

Put z=0,

$ \Longrightarrow 1=A(0+1)(0+2) \Longrightarrow A=\cfrac{1}{2}$

$\therefore f(z) = \cfrac{1}{z(z+2)(z+1)}=\cfrac{1/2}{z} - \cfrac{1}{z+1}+ \cfrac{1/2}{z+2}$

Put $u=z+3 \Longrightarrow = u-2$

$\therefore f(z) = \cfrac{1/2}{u-2} - \cfrac{1}{u-1}+ \cfrac{1}{2u}$ ------------(1)

$(I)$ For $|u| \lt 1$,

Also, $|u| \lt 2$

$\therefore \left | \cfrac{u}{2} \right |\lt1$

Equation (1) becomes,

$\therefore f(z)= \cfrac{1}{2} \times \left [ \cfrac{1}{u-2} \right ] - \cfrac{1}{u-1} + \cfrac{1}{2} \left[ \cfrac{1}{u} \right]= \cfrac{1}{2} \times \cfrac{1}{2} \times \left [ \cfrac{1}{\cfrac{u}{2}-1} \right ] - \left[ \cfrac{1}{-(1-u)} \right] + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ \cfrac{1}{1-\cfrac{u}{2}} \right ] + \left[ \cfrac{1}{1-u} \right] + \cfrac{1}{2u} = -\cfrac{1}{4} \left [ 1-\cfrac{u}{2} \right ]^{-1} + \left[ 1-u \right]^{-1} + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ 1+\cfrac{u}{2} + \cfrac{u^{2}}{2^{2}} + ----- \right ] + \left[ 1+u+u^{2}+------ \right] + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ 1+\cfrac{(z+2)}{2} + \cfrac{(z+2)^{2}}{2^{2}} + ---- \right ] + \left[ 1+(z+2)+(z+2)^{2}+---- \right] + \cfrac{1}{2(z+2)}$

$(II)$ For $1\lt |u| \lt 2$,

$1 \lt |u| \ \& \ |u| \lt 2$

$\therefore \left | \cfrac{1}{u} \right |\lt1 \ \& \ \left | \cfrac{u}{2} \right |\lt1$

Equation (1) becomes,

$\therefore f(z)= \cfrac{1}{2} \times \cfrac{1}{2} \times \left [ \cfrac{1}{\cfrac{u}{2}-1} \right ] - \cfrac{1}{u} \left[ \cfrac{1}{1-\cfrac{1}{u}} \right] + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ \cfrac{1}{1-\cfrac{u}{2}} \right ] - \cfrac{1}{u} \left[ \cfrac{1}{1-\cfrac{1}{u}} \right] + \cfrac{1}{2u} = -\cfrac{1}{4} \left [ 1-\cfrac{u}{2} \right ]^{-1} - \cfrac{1}{u} \left[ 1- \cfrac{1}{u} \right]^{-1} + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ 1+\cfrac{u}{2} + \cfrac{u^{2}}{2^{2}} + ----- \right ] - \cfrac{1}{u} \left[ 1+ \cfrac{1}{u} + \cfrac{1}{u^{2}}+------ \right] + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ 1+\cfrac{u}{2} + \cfrac{u^{2}}{2^{2}} + ----- \right ] - \left[ \cfrac{1}{u} + \cfrac{1}{u^{2}} + \cfrac{1}{u^{3}}+------ \right] + \cfrac{1}{2u}$

$= -\cfrac{1}{4} \left [ 1+\cfrac{(z+2)}{2} + \cfrac{(z+2)^{2}}{2^{2}} + --- \right ] + \left[ \cfrac{1}{z+2} + \cfrac{1}{(z+2)^{2}} + \cfrac{1}{(z+2)^{3}}+--- \right] + \cfrac{1}{2(z+2)}$

$(III)$ For $|u| \gt 2$,

Also, $ |u| \gt 1$

$\therefore 1 \lt |u| \ \& \ 2 \lt |u|$

$\therefore \left | \cfrac{1}{u} \right |\lt 1 \ \& \ \left | \cfrac{2}{u} \right |\lt1$

Equation (1) becomes,

$\therefore f(z)= \cfrac{1}{2} \times \cfrac{1}{u} \times \left [ \cfrac{1}{1- \cfrac{2}{u}} \right ] - \cfrac{1}{u} \left[ \cfrac{1}{1-\cfrac{1}{u}} \right] + \cfrac{1}{2u}$

$=\cfrac{1}{2u} \left [ 1- \cfrac{2}{u} \right ]^{-1} - \cfrac{1}{u} \left[ 1- \cfrac{1}{u} \right]^{-1} + \cfrac{1}{2u}$

$= \cfrac{1}{2u} \left [ 1+\cfrac{2}{u} + \cfrac{2^{2}}{u^{2}} + ----- \right ] - \cfrac{1}{u} \left[ 1+ \cfrac{1}{u} + \cfrac{1}{u^{2}}+------ \right] + \cfrac{1}{2u}$

$= \left [ \cfrac{1}{2u}+\cfrac{1}{u^{2}} + \cfrac{2}{u^{3}} + ----- \right ] - \left[ \cfrac{1}{u} + \cfrac{1}{u^{2}} + \cfrac{1}{u^{3}}+------ \right] + \cfrac{1}{2u}$

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