How to take the inverse of this signal?

Question

I have a little problem computing the inverse of this signal:

$$X(z) = \frac{(z-1)(z+\frac{3}{2})}{(z+\frac{j}{2})(z-\frac{j}{2})(z-\frac{1}{2})}$$ $$X(z^{-1}) = ?$$

I know how to take the inverse, $z=z^{-1}$ and then multiply the brackets and so on... But my problem is the numbers in the brackets, $1, -\frac{3}{2}, \frac{j}{2}$, are poles and zeros of a Pole-zero plot of the sequence $X(z)$.
And I think I loose that information, don't I?

Answer 1

if $X(z)$ has a zero/pole at $z=w$ then $X(z^{-1}$ will have a zero at $z=w^{-1}$, so you don't lose any information about the poles and zeros. You end up with:

$$\begin{align} X(z^{-1}) & = \frac{(z^{-1}-1)(z^{-1}+\tfrac{3}{2})}{(z^{-1}+\tfrac{j}{2})(z^{-1}-\tfrac{j}{2})(z^{-1}-\tfrac{1}{2})} \\ & = \frac{z(1-z)(1+\tfrac{3}{2}z)}{(1+\tfrac{j}{2}z)(1-\tfrac{j}{2}z)(1-\tfrac{1}{2}z)} \\ & = \frac{-\tfrac{3}{2}z(z-1)(z+\tfrac{2}{3})}{\tfrac{j^2}{8}(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\ & = -\frac{12}{j^2} \frac{z(z-1)(z+\tfrac{2}{3})}{(z + \tfrac{2}{j})(z - \tfrac{2}{j})(z-2)} \\ \end{align}$$

Note that $X(z)$ had a zero at infinity, and hence $X(z^{-1})$ has a zero at zero.