Consider the function $f : \mathbb{R} \to \mathbb{C}$ defined by $f(\theta) = e^{i \theta} + e^{i \xi \theta}$ for all $\theta \in \mathbb{R}$.
Let $f[\mathbb{R}] \subset \mathbb{C}$ denote the image of $f$; that is $f[\mathbb{R}] := \left\{f(\theta) : \theta \in \mathbb{R} \right\}$.
Define $S := \left\{ z \in \mathbb{C} : \lvert z \rvert \leq 2\right\}$.
Proposition. If $\xi$ is irrational, then $f[\mathbb{R}]$ is dense in $S$.
I have worked through and shown that $\xi$ being rational is both a necessary and sufficient condition for $f$ being periodic. And I can understand, intuitively, that if $f$ fails to be periodic (due to the assumed irrationality of $\xi$), then its image must be dense in $S$.
But I'm unsure how to go about proving this rigorously. In particular, whether it would be more fruitful to try and show:
a) For any $\varepsilon > 0$ and $s \in S$, there exists $z \in f[\mathbb{R}]$ such that $\lVert s - z \rVert < \varepsilon$, or
b) For any $s \in S$, there exists a sequence $(s_n)$ in $f[\mathbb{R}]$ such that $\lim\limits_{n \to \infty} s_n = s$, or
c) Use another approach I haven't considered above.
Any advice would be greatly appreciated.
It is a well-known fact that the map $$\phi : \mathbb R \to S^1 \times S^1, \phi(t) = (e^{it}, e^{i\xi t})$$ with an irrational $\xi$ is a smooth injection having a dense image $\phi(\mathbb R)$. This has been addressed several times in this forum.
We claim that the map $$\sigma : S^1 \times S^1 \to S, \sigma (z,w) = z + w$$ is a surjection.
Once we know this, it is easy to see that $f(\mathbb R)$ is dense in $S$. Since $f = \sigma \circ \phi$, we get $$S = \sigma (S^1 \times S^1) = \sigma (\overline{\phi((\mathbb R)}) \subset \overline{\sigma (\phi((\mathbb R))} = \overline{f(\mathbb R)} .$$
Let us prove the above claim.
Let $p \in S$ and $S_p = \{z \in \mathbb C \mid \lvert z - p \rvert = 1\}$ be the circle with center $p$ and radius $1$. The intersection $D_p = S_p \cap S^1$ is non-empty: For $\lvert p \rvert = 2$, we get $D_p = \{ \frac p 2 \}$, for $0 < \lvert p \rvert < 2$ its contains exactly two points and for $p = 0$ we get $D_p = S^1$.
Pick $z \in D_p \subset S^1$ and set $w = p - z$. Since $z \in S_p$, we have $\lvert p - z \rvert = 1$, thus $w \in S^1$. Then $\sigma(z,w) = z + w = p$.