written 5.0 years ago by | • modified 5.0 years ago |
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 $|z|\gt2$
=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[\frac{1}{1-\frac{1}{z}}]-\frac{1}{12z}[\frac{1}{1+\frac{2}{z}}]$
=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1-\frac{1}{z}]^{-1}-\frac{1}{12z}[1+\frac{2}{z}]^{-1}$
=$\frac{-1}{4}[\frac{1}{z}]-\frac{1}{2}[\frac{1}{z^2}]+\frac{1}{3z}[1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}----]- \frac{1}{12z}[{1-\frac{2}{z}+\frac{2^2}{z^2} -\frac{2^3}{z^3}+....]}$