Solution:

Given $f(z) = \frac{1}{z^2(z-1)(z+2)}$

$f(z) = \frac{1}{z^2(z-1)(z+2)}$= $\frac {A}{z}+\frac {B}{z^2}+\frac {C}{z-1}+\frac {D}{z+2}$----(1)

$1=Az(z-1)(z+2)+B(z-1)(z+2)+C(z+2)z^2+D(z-1)z^2$-----(2)

when z=0 in equation (2),

$1=0+B(-1)(2)+0+0$

$B=\frac{-1}{2}$

when z=1 in eqn (2)

$1=0+0+0+c(3)(1)^2+0$

$C=\frac{1}{3}$

when z= -2 in eqn (2)

$1= 0+0+0+D(-3)(-2)^2$

$1= -12D $

$D=\frac{-1}{12}$

To find A ,further simplify eqn(2)

$1= Az(z^2+z-2)+B(z^2+z-2)+c(z^3+2z^2)+D(z^3-z^2)$

$1=Az^3+Az^2-2Az+Bz^2+Bz-2B+cz^3+2cz^2+Dz^3-Dz^2$

by equating L.H.S. and R.H.S.

We get

A+C+D=0-----(3)

Put C and D values in eqn(3)

$A+\frac{1}{3}-\frac{1}{12}=0$

$A=\frac{-3}{12}$

$A=\frac{-1}{4}$

Put A,B,C,D values in eqn(1)

f(z)=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3}[\frac{1}{z-1}]-\frac{1}{12}[\frac{1}{z+2}]$----(4)

i) When $0\lt|z|\lt1,$

from eqn(4)

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]-\frac{1}{3}[\frac{1}{1-z}]-\frac{1}{24}[\frac{1}{1+z/2}]$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]-\frac{1}{3}[1-z]^{-1}-\frac{1}{24}[1+\frac{z}{2}]^{-1}$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]-\frac{1}{3}[1+z+z^2+----]-\frac{1}{24}[{1-\frac{z}{2}+ \frac{z^2}{2^2} -\frac{z^3}{2^3}+....]}$

ii) When $1\lt|z|\lt2$

from eqn(4)

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[\frac{1}{1-\frac{1}{z}}]-\frac{1}{24}[\frac{1}{1+\frac{z}{2}}]$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1-\frac{1}{z}]^{-1}-\frac{1}{24}[1+\frac{z}{2}]^{-1}$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}----]- \frac{1}{24}[{1-\frac{z}{2}+\frac{z^2}{2^2} -\frac{z^3}{2^3}+....]}$

iii) when $|z|\gt2$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[\frac{1}{1-\frac{1}{z}}]-\frac{1}{12z}[\frac{1}{1+\frac{2}{z}}]$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1-\frac{1}{z}]^{-1}-\frac{1}{12z}[1+\frac{2}{z}]^{-1}$

=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}----]- \frac{1}{12z}[{1-\frac{2}{z}+\frac{2^2}{z^2} -\frac{2^3}{z^3}+....]}$