Question

f(z)= z/(z+1)(z+1)^2

Answer 1

The form of the partial fraction decomposition is;

$\displaystyle \frac{z}{\left(z + 1\right)^{3}}=\frac{A}{z + 1}+\frac{B}{\left(z + 1\right)^{2}}+\frac{C}{\left(z + 1\right)^{3}}=\frac{z}{\left(z + 1\right)^{3}}=\frac{\left(z + 1\right)^{2} A + \left(z + 1\right) B + C}{\left(z + 1\right)^{3}}\\ \Rightarrow z=\left(z + 1\right)^{2} A + \left(z + 1\right) B + C\\$

Expanding the right hand side yields;

$\displaystyle z=z^{2} A + 2 z A + z B + A + B + C\\ \text{Collecting like terms yields;}\\ z=z^{2} A + z \left(2 A + B\right) + A + B + C\\ \text{Comparing yields the following system of linear equations:}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\begin{cases} A = 0\\2 A + B = 1\\A + B + C = 0 \end{cases}\\ \text{Solving the above equations yields: }A=0,\ B=1,\ C=-1\\ \therefore\frac{z}{\left(z + 1\right)^{3}}=\frac{0}{z + 1}+\frac{1}{\left(z + 1\right)^{2}}+\frac{-1}{\left(z + 1\right)^{3}}=\frac{z}{\left(z + 1\right)^{3}}=\frac{1}{\left(z + 1\right)^{2}}-\frac{1}{\left(z + 1\right)^{3}}$