Let $q \in \mathbb C$, $\|q\|=1$ and $q^n \neq 1, \forall n \in \mathbb N$. Show that $\{q^n: n \in \mathbb N\}$ is dense in $S^1$

Question

Let $q \in \mathbb C$, $|q|=1$ and $q^n \neq 1, \forall n \in \mathbb N$. Show that $\{q^n: n \in \mathbb N\}$ is dense in $S^1$.

My attempt: As $(q^n)$ is limited, there is subsequence $(q^{n_j})$ convergent. So given $\epsilon > 0, \exists j_0 \in \mathbb N, \quad |q^{n_j} - q^{n_k}| < \epsilon, \quad \forall j,k \geq j_0$. Supose that $n_j > n_k$, then $|q^{n_j} - q^{n_k}| = |q^{n_k}||q^{n_j - n_k} - 1| = |q^{n_j - n_k} - 1| < \epsilon, \quad \forall j, k \geq j_0$. So, there is a subsequence of $(q^n)$ that converges to $1$.

Using this fact, how can I ensure that for all $w \in S^1$ there is a subsequence of $(q^n)$ that converges to $w$?

Answer 1

Take a sequence $n_j$ such that $q^{n_j}\to1$. Given $\varepsilon>0$, take $j$ such that the angle between $q^{n_j}$ and $1$ is at most $\varepsilon$. Now consider the points $(q^{n_j})^n=q^{nn_j}$ for this $j$, which are at an angle at most $\varepsilon$ from each other. Hence, any point in the circle is at most at an angle $\varepsilon$ from some point in the sequence $q^{nn_j}$ (and so also $q^n$). Now let $\varepsilon\to0$.

Answer 2

We know $q=\exp({i\pi x})$ for some real $x$. Suppose $x=\frac a b\in\Bbb Q$.

Then $q^{2b}=\exp(2i\pi )^a=1$, contradiction.

Thus $x\in \Bbb R\setminus \Bbb Q$. The claim follows (why?).

Answer 3

We can assume $q=e^{i\cdot \alpha}, \alpha \in \mathbb{R}$.

If $\frac{2\pi}{\alpha}$ is irrational then, according to Kronecker's Approximation Theorem, $\left\{n + m \cdot \frac{2\pi}{\alpha} \mid n,m\in\mathbb{Z} \right\} $ is dense in $\mathbb{R}$. So must be $e^{i\cdot \alpha \cdot \left(n + m \cdot \frac{2\pi}{\alpha}\right)}=e^{i\cdot \alpha \cdot n}$ in $S^1$, because $f(x)=e^{i x \alpha}$ is periodic and continuous.

If $\frac{2\pi}{\alpha}$ is rational then $\frac{l}{m}\alpha=2\pi, l,m \in \mathbb{Z}$ or $\alpha = \frac{m}{l}2\pi$ and thus $$q^n=e^{i\cdot n\alpha}=e^{i\cdot n\frac{m}{l}2\pi}\overset{\text{div with remainder}}{=}e^{i\cdot (tl+r)\frac{m}{l}2\pi}=e^{i\cdot r\frac{m}{l}2\pi}e^{i\cdot tm2\pi}=e^{i\cdot r\frac{m}{l}2\pi}=q^r$$ where $0\leq r < l$, so $\left(q^n\right)_{n\in\mathbb{N}}$ will take only a finite number of values and won't be dense.