How to show that $\frac{1}{(1-\frac{1}{4}z^{-1})(1-\frac{1}{4}z)} = \frac{-4z^{-1}}{(1-\frac{1}{4}z^-1)(1-4z^{-1})}$

Question

Can anyone help me clarify what rule is used in this rewriting of this fraction?

$$\frac{1}{\left(1-\dfrac{1}{4}z^{-1}\right)\left(1-\dfrac{1}{4}z\right)} = \frac{-4z^{-1}}{\left(1-\dfrac{1}{4}z^{-1}\right)\left(1-4z^{-1}\right)}$$

Answer 1

Multiply both numerator and denominator with $-4z^{-1}$. For the term $(1-z/4)$ in the denominator you get

$$-4z^{-1}(1-z/4)=-4z^{-1}+1=(1-4z^{-1})$$