Question Let $R$ be a subring of $\mathbb{C}$ and the group of units $\mathcal{U}(R)$ is finite. Show $\mathcal{U}(R)$ is cyclic.
My Idea is to let an element of $R$ be $z\in\mathbb{C}$ so if $z\in\mathcal{U}(R)$ then $\exists w\in\mathcal{U}(R)$ such that $zw=1$. Then $z^n\in\mathcal{U}(R)$ with inverse $w^n$. But $\mathcal{U}(R)$ is finite so we must have $|z|=1$ otherwise $z^n\ne z^m$ for all $m\ne n$ as $|z|^n\ne |z|^m$ if $|z|\ne1$. Therefore elements of $\mathcal{U}(R)$ must have the form $e^{i\theta}$ for some $\theta\in[0,2\pi)$.
My Problem is that I do not really know how to proceed to show $\mathcal{U}(R)$ is cyclic. In fact it is intuitively clear that those elements of $\mathcal{U}(R)$ should be the roots of unity for some $n$. But I am not sure how to prove this rigorously.
Extra Problems There are some extra problems given: calculate the order of the following groups (i) $\mathcal{U}(\mathbb{Z}_3[X])$; (ii) $\mathcal{U}(\mathbb{Z}_{4}[X])$ and (iii) $\mathcal{U}(\mathbb{Z}_{2016}[X])$.
I know the first one's solution. Since $\mathbb{Z}_3$ is a field hence a domain so $\mathcal{U}(\mathbb{Z}_3)=\mathcal{U}(\mathbb{Z}_3[X])$ so the order is $2$. But what about the next two?
$\mathcal{U}(R)$ is a finite multiplicative subgroup of a field $\mathbb{C}$. Any such group is cyclic.
For the extra problems, consider the polynomials $f_n = 2X^n+1 \in \mathbb{Z}_4[X]$ and $g_n = 1008X^n+1 \in \mathbb{Z}_{2016}[X]$. They're all units since $f_n^2 = 1$ and $g_n^2 =1$. Hence $\mathcal{U}(\mathbb{Z}_{4}[X])$ and $\mathcal{U}(\mathbb{Z}_{2016}[X])$ are both infinite.