We can write $\hat{R}(z)$ as a rational function of a polynomial $P(z)$ and $Q(z)$:

$$
\hat{R}(z) = \frac{P(z)}{Q(z)} = \frac{b_M z^M + b_{M-1}z^{M-1} + \cdots + b_0}{a_N z^N + a_{N-1} z^{N-1} + \cdots + a_0}
$$

For rational Z-transforms, we say that

\begin{enumerate}
\item $\hat{R}$ is strictly proper is $M \lt N$ (the power in the numerator is less than the power in the denominator).
\item $\hat{R}$ is proper if $M \leq N$.
\item $\hat{R}$ is improper if $M \gt  N$.
\item The roots of $Q(z)$ are the poles.
\item The roots of $P(z)$ are the zeros.
\item If $M \lt N \Rightarrow M - N $ poles at $\infty$.
\item If $N \lt M \Rightarrow N - M $ zeros at $\infty$.
\item The number of poles is equal to the number of zeros if we account for what happens at infinity.
\end{enumerate}

In general, the set of signals that produce rational Z-transforms are obtained from, and only from, linear combinations of $\alpha^n u(n)$, $\alpha^n u(-n)$, $n^k \alpha^n u(n)$, and $n^k \alpha^n u(-n)$, where $k \in \N$, and shifted versions thereof.

\begin{example}

\begin{align*}
\hat{R}(z) &= \frac{(z-1)(z-2)}{(z-3)} \\
\end{align*}
Here $M \gt  N$, so we have a pole at infinity. Poles are $3,\infty$. Zeros are $1,2$.

\end{example}

\begin{example}

\begin{align*}
\hat{R}(z) = \frac{(z-3)}{(z-1)(z-2)} \\
\end{align*}

Here $N\gt M$, so we have a zero at infinity. Poles are $1,2$. Zeros are $3,\infty$.

\end{example}

\begin{example}

\begin{align*}
\hat{R}(z) &= \frac{z^2 + 1}{z} \\
\end{align*}

Can $r$ be causal?

\begin{align*}
\hat{R}(z) &= \frac{z^2 + 1}{z} \\
\hat{R}(z) &= \frac{(z-i)(z+i)}{z} \\
\end{align*}

We have zeros at $\pm i$ and poles at $0,\infty$. If there is a pole at infinity, then $r$ cannot be causal. If there is a pole at 0, then $r$ cannot be anti-causal. We can also see this fact using an inverse transform $\hat{R}(z) = z + \frac{1}{z} \iztp \delta(n-1) + \delta(n+1)$.

\end{example}

Rational Z-Transforms