Is there a more profound insight that can be seen going from the Z-transform:
$X(z) = \sum_{n=0}^{\infty}x[n]z^{-n}$
To the Laplace transform:
$ F(s) = \int_0^\infty f(x)e^{-sx}~dx$
That allows the region of convergence to be a plane or a circle?
Is it that we have changed a continuous variable to a discrete integer?
What is the major difference between the two equation (except for their form) that allows us to say that Z-transform has a circular region of convergence where Laplace transform has a planar region of convergence?
First, we need to understand the relationship between the Z-transform and the Laplace transform. The Z-transform is the discrete counterpart to the Laplace transform. I am probably not telling you anything too insightful yet. Now, the Laplace transform of a continuous time signal is $$ X(s) = \int_0^{\infty}x(t)e^{-st}dt\tag{1} $$ where $s = \sigma + j\omega$. We can derive the Z-transform by sampling the continuous signal. The sample signal is usually denoted $x(m)$ instead of $x(mT_s)$ where the period is $T_s = 1$. The the Z-transform is $$ X(e^s) = \sum_{m = 0}^{\infty}x(m)e^{-sm}.\tag{2} $$ Let $e^s = z$. Then $$ X(z) = \sum_{m = 0}^{\infty}x(m)z^{-m}.\tag{3} $$ Now, the frequency variables of Laplace and Z transforms are $s=\sigma +j\omega$ and $z=re^{j\omega}$, respectively. Both $s$ and $z$ are complex variables with real and imaginary parts. The vertical axis in the $s$ plane is $j\omega$ which is the frequency and the $\sigma$ axis corresponds to exponential growth or decay. Additionally, $j\omega$ axis in the $s$ plane is the location of the Fourier basis functions where the Fourier basis functions are $e^{-2\pi jfm}$. What does the plot in the $s$ plane look like for different frequencies? These are periodic horizontal lines. Now this periodic process can be applied to the $z$ plane with $z=re^{j\omega}$. The area of LHS of $s$ plane for $\sigma > 0$ or $r=e^{\sigma}<1$ is mapped into the area inside the unit circle which is the region of stable casual systems. The RHS of the $s$ plane $\sigma >0$ or $r=e^{\sigma}>1$ is mapped onto the outside of the unit circle and this region is the unstable system. Lastly, $\sigma = 0$ or $r=e^{\sigma}=1$ is mapped into the unit circle.