# $\mathcal Z$-transform, ROC of a system from dividing to others

TRUE / FALSE

Given three systems with rational $\mathcal Z$-transform. Systems A and B are not stable with $\mathcal Z$-transform $H(z)$, $G(z)$ respectively. A and B have no common poles. System's C $\mathcal Z$-transform is $F(z)=H(z)/G(z)$, then C is also unstable?

False:

\begin{align} H(z)&= \frac{1}{1-2z^{-1}}: \textrm{ROC} \quad\lvert z\rvert>2\\ G(z)&= \frac{1}{1-3z^{-1}}: \textrm{ROC} \quad \lvert z\rvert>3\\ F(z)&= \frac{1-3z^{-1}}{1-2z^{-1}}: \textrm{ROC} \quad \lvert z\rvert<2 \end{align}

It is the answer, but doesn't $F(z)$ ROC must be like that of $H(z$), or the division allows to choose the ROC I want?

I agree with your doubts. The $\mathcal{Z}$-transform
$$\hat{G}(z)=1/G(z)=1-3z^{-1}$$
has the ROC $|z|>0$. The ROC of the multiplication of $H(z)$ and $\hat{G}(z)$ must be the intersection of the ROCs of $H(z)$ and $\hat{G}(z)$ (unless there occur pole-zero cancellations, which is not the case here). Since the ROC of $H(z)$ is $|z|>2$, the ROC of $F(z)=H(z)\hat{G}(z)$ must also be $|z|>2$.
Another way to think about it is that since $H(z)$ and $\hat{G}(z)$ are both causal, also $F(z)$ must be causal, i.e., its ROC must be outside a circle in the complex plane.