Question about z transform
The variable $z$ is complex, and it can be represented either in polar form $z=re^{j\phi}$ or in terms of its real and imaginary parts. The polar form is preferred because the region of convergence (ROC) is determined by $r$, the magnitude of $z$. This becomes clear if you write the term inside the sum of the definition of the $\mathcal{Z}$-transform as

$$x[n]z^{-n}=x[n]r^{-n}e^{-jn\phi}\tag{1}$$

From $(1)$ it is clear that the value of $r$ determines the convergence of the series. This is why the ROC of the $\mathcal{Z}$-transform is specified in terms of $r=|z|$. For the bilateral $\mathcal{Z}$-transform the ROC is a ring ($a&lt;|z|&lt;b$, with $b&gt;a\ge 0$), whereas for the unilateral $\mathcal{Z}$-transform, the ROC is outside a circle centered in the origin of the complex plane ($|z|&gt;a$, $a&gt;0$).

Note that for the Laplace transform the complex variable $s$ occurs in the exponent, so it's the real part of $s$ that determines the ROC of the Laplace transform. These are the correspondences between the Laplace transform and the $\mathcal{Z}$-transform in terms of ROCs:

$$\begin{align}&amp;s\text{-plane }&amp;&amp;z\text{-plane}\\\\
&amp;\text{right half-plane} &amp;&amp; \text{region outside a circle (centered at the origin)}\\
&amp;\text{vertical strip} &amp;&amp; \text{ring (centered at the origin)}\\
&amp;\text{left half-plane} &amp;&amp; \text{region inside a circle (centered at the origin)}
\end{align}$$

The $j\omega$-axis in the $s$-plane corresponds to the unit circle in the $z$-plane, i.e. if the unit circle is part of the ROC, then the DTFT exists, and if the corresponding sequence is the impulse response of an LTI system, then the system is stable.