Does the ROC (Z-Transform) always start/end from/in a pole?

Question

Can I assume that the R.O.C. from a Z-Transform will always start from or end in a pole or 0/infinity?

Using an example:

$$H(z)=\frac{(z+1)(z-1)(z+j)(z-j)}{z^4}\space,\space\space j=\sqrt{(-1)}$$

So we have 4 zeros in the unit circle equally spaced by a $\frac{\pi}{2}$ angle and 4 poles in the origin ($z=0$). If not specified, can I assume the the ROC wil necessarily start from exclusive zero and go towards infinity? Or maybe possible to exist a ROC which is a ring in the Z-plane (which will not start/end in a pole)?

Answer 1

The region of convergence of a power series of a meromorphic function $f:\mathbb{C}\to\mathbb{C}$ will always be a open disk with an essential singularity on the boundary. The z-transform of an LTE system is essentially an expansion of $H(z^{-1})$ into a power series of $z^{-1}$. $H$ is meromorphic and therefore the region of convergence of the power series will always end at a pole of $H(z^{-1})$.

Answer 2

Since the value of Z-transform tends to infinity at poles, poles can not be a part of ROC. So ROC either starts from a pole or end in a pole. For example,

1. If $x[n]$ is a right-sided sequence, then the ROC extends outward from the outermost pole in $X(z)$
2. If $x[n]$ is a left-sided sequence, then the ROC extends inward from the innermost pole in $X(z)$
3. If $x[n]$ is a two-sided sequence, the ROC will be a ring in the z-plane that is bounded on the interior and exterior by a pole
4. If $x[n]$ is a finite-duration sequence, then the ROC is the entire z-plane, except possibly $z=0$ or $|z|=\infty$

From all these cases, one can say that poles decides the boundary of ROC.

The example you provided is a finite-duration sequence and its ROC is entire z-plane except at z =0.