Let $R=\mathbb{Z}(e^{2\pi i\over 3})$. Find $R^\times$.
Attempt:
Let $\zeta=e^{2\pi i\over3}$.
On
This follows from the following well-known lemma:
Lemma An element $x$ in a ring of integers is a unit if and only if $N(x)=\pm 1$.
Theorem: let $K=\Bbb Q(\sqrt{d})$ be an imaginary-quadratic number field and $d<0$ squarefree. Then the group of units is $\mathcal{O}_K^{\times}=\mu_K$, and this cyclic group is given as follows: $$ \mu_K= \begin{cases} \Bbb Z/4 =\{\pm 1,\pm i\} & \text{ if $d=-1$, } \\ \Bbb Z/6 =\{\pm 1,\pm \zeta,\pm \zeta^2\} & \text{ if $d=-3$, } \\ \Bbb Z/2 = \{\pm 1\} & \; \text{otherwise.} \end{cases} $$
Proof: An element $x=a+b\sqrt{d}$ is a unit in $\mathcal{O}_K$ if and only if $N(x)=\pm 1$. For $d\not\equiv 1 \mod 4$ we have $\mathcal{O}_K=\{a+b\sqrt{d} \mid a,b \in \Bbb Z\}$ and hence the norm condition is equivalent to the Diophantine equation $$ a^2-b^2d = \pm 1. $$ For $d\equiv 1 \mod 4$ we have $\mathcal{O}_K=\{a+b(1+\sqrt{d})/2 \mid a,b \in \Bbb Z\}$ and hence $x$ is a unit iff $$ (2a+b)^2-b^2d = \pm 4. $$ For $d<0$ these equations obviously have only finitely many solutions. Hence we have $\mathcal{O}_K^{\times}=\mu_K$. A primitive $m$-th root lies in $K$ if and only if $\Bbb Q(\zeta_m)\subseteq K$. Then we have $\phi(m)\mid [K:\Bbb Q]=2$, hence $m\mid 4$ or $m\mid 6$. This implies the result. Of course, we can solve the Diophantine equations now directly in each case.
Hints:
$R = \{ a+b\zeta :a,b \in \mathbb Z \}$ because $\zeta^2=-\zeta-1$.
$\alpha \in R^\times$ iff $N(\alpha)=1$, where the norm $N(\alpha)=\alpha \bar\alpha=a^{2}-ab+b^{2}$, for $\alpha=a+b\zeta$.
$a^{2}-ab+b^{2}=1$ implies $a,b \in \{ -1, 0, 1 \}$.