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?

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

http://math.stackexchange.com/questions/402861/frac-4z-11-frac14z-11-4z-1-frac1615-frac11-f — jimjim, May 26 '13 at 13:01

score 0 · Accepted Answer · answered May 26 '13 at 12:18

0

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

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

answered May 26 '13 at 12:18

Matt L.

10,636

Ah I see. I thought you where suppose to multiply both terms in the denominator. Thank you. – Attaque May 26 '13 at 12:25

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

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