CVPL1

Find the partial fraction decomposition of $\frac{4z+4}{z(z-1)(z-2)^2}\,$ .

$\frac{4z+4}{z(z-1)(z-2)^2} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2} + \frac{D}{(z-2)^2}\,$

$4z+4=A(z-1)(z-2)^2 + Bz(z-2)^2 + Cz(z-1)(z-2) + D z(z-1)\,$

Plug in $z=0\,$ to get $A=-1\,$

Plug in $z=1\,$ to get $B=8\,$

Plug in $z=2\,$ to get $D=6\,$

Differentiate both sides of the equation once with respect to $z\,$ and plug in $z=2\,$ to get $C=-7\,$ .

Finally

$\frac{4z+4}{z(z-1)(z-2)^2} = \frac{-1}{z} + \frac{8}{z-1} - \frac{7}{z-2} + \frac{6}{(z-2)^2}\,$