When $n = 2$, the points of $\mathbb{Z}[e^{\frac{2\pi i}{n}}]$ are the vertices of a regular tiling of $\mathbb{C}$ by unit squares. When $n = 4$, the points of $\mathbb{Z}[e^{\frac{2\pi i}{n}}]$ are the vertices of a tiling of $\mathbb{C}$ by right triangles. When $n = 3, 6$, the points of $\mathbb{Z}[e^{\frac{2\pi i}{n}}]$ are the vertices of a tiling of $\mathbb{C}$ by equilateral triangles. (NB: I'm not even sure that these remarks are correct. Someone please correct me if I'm wrong).
What about when $n \neq 2,3,4,6$? My guess would be that when $n = 2^m 3^k$ for some $m,k \in \mathbb{N}$, we still get a discrete set of points (what tiling would the points correspond to?). I would also guess that when $n \neq 2^m 3^k$ for some $m,k \in \mathbb{N}$, $\mathbb{Z}[e^{\frac{2\pi i}{n}}]$ is dense in $\mathbb{C}$. Are these guesses right? If not, what actually happens?
Edit
See the comments below this post. I made some mistakes in the first paragraph.
Claim: For $n\neq 1,2,3,4,6$, the cylotomic integers $\mathbb{Z}[\zeta_n]$ is dense in $\mathbb{C}$ (usual topology), where $\zeta_n=e^{2\pi i/n}$.
Proof: Note that, $2\cos(2\pi/n)-1=\zeta_n+\zeta_n^{n-1}-1\in\mathbb{R}^\times$, with absolute value $<1$ for these $n$. Combining with the rotation $\zeta_n\notin\mathbb{R}$ then yields the desired result.