I have read that limit points of $z^n$, where $z\in\mathbb{C}$,$n\in\mathbb{N}$, and $|z| = 1$ ; depend on rationality of $\theta$ ,where $z = \mathrm{e}^{i\pi\theta}$.
In fact, If $\theta$ is rational, then the set of limit points is finite and otherwise the set of limit points is $\mathbb{R}$
Can someone explain this more detailed? I don't get the deduction.
Let's define $f(\theta)=e^{i \pi \theta}$. When $\theta=\frac{p}{q}$, then $f^{n}(\theta)=f^{n \mod 2q}(\theta)$, so you have finitely many limit points on the unit circle. If $\theta$ is irrational, then you can prove that the set $\{f^n(\theta):n \in \mathbb Z\}$ is dense on the unit circle (e.g. by the usual $\epsilon$ argument, you show that for any $z$ on the unit circle, there exists $N \in \mathbb Z$ such that $|z-f^N(\theta)|< \epsilon$).