Answer
$$\frac{z+1}{z^2(z-1)}=-\frac{2}{z}-\frac{1}{z^2}+\frac{2}{z-1}$$
Work Step by Step
$$\frac{z+1}{z^2(z-1)}$$
The partial fraction decomposition has the following form: $$\frac{z+1}{z^2(z-1)}=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z-1}$$
To find $A, B, C$, we clear fractions and get
$$z+1=Az(z-1)+B(z-1)+Cz^2$$ $$z+1=Az^2-Az+Bz-B+Cz^2$$ $$z+1=(A+C)z^2+(B-A)z-B$$
Equating coefficients of corresponding powers of $x$, we get
- $A+C=0$
- $B-A=1$
- $B=-1$
From $B=-1$, we get $A=B-1=-1-1=-2$
From $A=-2$, we get $C=-A=2$
Therefore, $$\frac{z+1}{z^2(z-1)}=-\frac{2}{z}-\frac{1}{z^2}+\frac{2}{z-1}$$
