Proving if $f_{\omega}:\mathbb{Z} \xrightarrow[ ]{} \mathbb{C}$ as $f_{\omega} (n) := e^{2\pi i \omega n}$ is periodic

Question

For $\omega \in \{0,1)$ we define the following function:

$f_{\omega}:\mathbb{Z} \xrightarrow[ ]{} \mathbb{C}$ as $f_{\omega} (n) := e^{2\pi i \omega n}$

Can this be proved $f_{\omega} \Leftrightarrow \omega \in \mathbb{Q} $ ?

Thanks.

Answer 1

\begin{align*}f_\omega\text{ is periodic} &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\forall n\in\mathbb{Z} :& f_\omega(n) = f_\omega(n + T) \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\forall n\in\mathbb{Z} :& e^{2\pi i \omega n} = e^{2\pi i \omega (n+ T)} \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\forall n\in\mathbb{Z} :& e^{2\pi i \omega n} = e^{2\pi i \omega n}e^{2\pi i \omega T} \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& && 1 = e^{2\pi i \omega T} \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\exists k\in\mathbb{Z}:& 2\pi i k = 2\pi i \omega T \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\exists k\in\mathbb{Z}:& k = \omega T \\ &\Leftrightarrow& &\exists T\in\mathbb{Z}\setminus\{0\}:& &\exists k\in\mathbb{Z}:& \frac{k}{T} = \omega\\ &\Leftrightarrow& &\omega\in\mathbb{Q} \end{align*}