Lets look into the relation between the Z-transform and the Fourier Transform:

\begin{nicebox}
\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty} x(n) e^{-i\omega n} \\
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} x(n) z^{-n} \\
\end{align*}
\end{nicebox}

There is an affiliated $RoC(x)$ with the Z-Transform. If the $RoC$ included the unit circle, then the following statement is true:

\begin{nicebox}
$$
X(\omega) = \left. \hat{X}(z) \right|_{z = e^{i\omega}}
$$
\end{nicebox}

\begin{example}

Lets look at the function $x(n) = \alpha^n u(n)$ for two cases, the first where $\abs{\alpha} \lt 1$, and the other where $\abs{\alpha} \gt 1$.

\begin{align*}
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) z^{-n} \\
\hat{X}(z) &= \sum \limits_{n = 0}^{\infty} \lr{\alpha z^{-1}}^n \\
\hat{X}(z) &= \frac{1}{1- \alpha z^{-1}} \quad \mbox{ if } \abs{\alpha z^{-1}} \lt 1 \\
\end{align*}

This means that $RoC(x): \abs{z} \gt \abs{\alpha}$.


I) $\abs{\alpha} \lt 1$ We have a fourier transform.

This works because the $RoC$ includes all values such that $\abs{z} \gt \abs{\alpha}$. Since $1 \gt \abs{\alpha}$, 1 is a value in the $RoC$.

\begin{align*}
X(\omega) &= \left. \hat{X}(z) \right|_{z = e^{i\omega}} \\
X(\omega) &= \frac{1}{1-\alpha e^{-i\omega }} \\
\end{align*}


II) $\abs{\alpha} \gt 1$ This function doesn't have a fourier transform.

Since $\abs{\alpha} \gt 1$, then $\abs{z} \gt \abs{\alpha} \gt 1$. This means that the unit circle is excluded from the region of convergence.

\end{example}

Relation to Fourier Transform