Show that $\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} = \frac{16}{15}\frac{1}{(1-\frac{1}{4}z^{-1})}-\frac{16}{15}\frac{1}{(1-4z^{-1})}$

Question

Can anyone help me clarify how this rewriting is done?

$$\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} = \frac{16}{15}\frac{1}{(1-\frac{1}{4}z^{-1})}-\frac{16}{15}\frac{1}{(1-4z^{-1})}$$

Answer 1

Let $x=z^{-1}$

Then

\begin{align} \frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} =\frac{-4x}{(1-\frac{1}{4}x)(1-4x)} \end{align}

Suppose

\begin{align} &\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A}{(1-\frac{1}{4}x)}+\frac{B}{(1-4x)}\tag{1}\\&\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A(1-4x)+B(1-\frac{1}{4}x)}{(1-\frac{1}{4}x)(1-4x)}\\&-4x=A(1-4x)+B(1-\frac{1}{4}x)&\\&-4x=x\left(-4A-\frac{B}{4}\right)+A+B\end{align}

Comparing coefficients, you'll have $2$ equations involving $A$ and $B$.

$$-4A-\frac{B}{4}=-4 \tag{2}$$ $$A+B=0 \tag{3}$$

Solve $(2)$ and $(3)$ to get $A=\frac{16}{15}$ and $B=-\frac{16}{15}$ then replace them in $(1)$ and also revert $x$ back into $z^{-1}$ to get your final result.