Simplifying complex numbers in digital signal processing An exercise in my course in digital signal processing has a problem which leads to this expression. I've been calculating it for a while but I can't quite make out how they get to this answer. I have the transfer function $H(z)$
\begin{equation}
H(z)=\frac{\left(1-3z^{-1}\right)\left(1-4z^{-1}\right)}{\left(1-\frac{1}{3}z^{-1}\right)\left(1-\frac{1}{4}z^{-1}\right)}\cdot \left(1-\frac{1}{4}z^{-1}\right)=3^2\cdot4^2 \cdot \left(1-\frac{1}{4}z^{-1}\right)
\end{equation}
In order to figure out how they calculated it, I've removed $\left(1-\frac{1}{4}z^{-1}\right)$ from both sides and then calculating it backwards like this
\begin{equation}
\left(1-\frac{1}{3}z^{-1}\right)\left(1-\frac{1}{4}z^{-1}\right)\cdot3^24^2=3\left(3-z^{-1}\right)\cdot4\left(4-z^{-1}\right)=12\left(12-7z^{-1}+z^{-2}\right)\neq 1-7z^{-1}+12z^{-2}
\end{equation}
I'm suspecting this is incorrect, as the previous 3 out of 4 exercises has had the wrong answer, but as I don't feel confident with complex numbers I thought maybe I'm doing something wrong. Is there something in my way to calculate this backwards that should be done different due to complex numbers?
Thanks in advance!
 A: I don't know but the equality is just plain wrong.
For instance plug values in it $LHS(1)=9$ while $RHS(1)=108$.
Even simpler $LHS(3)=0$ while $RHS(3)$ obviously not zero.
Note also that $LHS(1/3)$ is not defined while $RHS(1/3)$ is defined.
The simplified expression I get is $$\frac {3(z-3)(z-4)}{z(3z-1)}$$
I'm surprised however that in your $H(\omega)$ the term $(1-\frac 14z^{-1})$ cancels between the numerator and denominator, are you sure about your writing ?
Secondly $H(\omega)$ should depend on $\omega$ but you have written it all in $z$, would it be that $\omega=\frac 1z$ ?
A: Why not just plain compute it? It's only a second order polynomial.
$$H(\omega)=\frac{\left(1-3z^{-1}\right)\left(1-4z^{-1}\right)}{\left(1-\frac{1}{3}z^{-1}\right)\left(1-\frac{1}{4}z^{-1}\right)}\cdot \left(1-\frac{1}{4}z^{-1}\right)$$
$$=\frac{1-7z^{-1}+12z^{-2}}{1-\frac{7}{12}z^{-1}+\frac{1}{12}z^{-2}  }\cdot \left(\frac{1-\frac{1}{4}z^{-1}}{1}\right)$$
And it should be clear what that is already. But if you want to simplify it further multiply both sides of the RHS with the denumerator of the LHS.
