Determine all units in $\mathbb{Z}[\omega] := \{a+b\omega\mid a,b\in\mathbf{Z}\}$ where $\omega = \frac{-1 + i \sqrt{3}}{2}$ Determine all units in $\mathbb{Z}[\omega] := \{a+b\omega\mid a,b\in\mathbf{Z}\}$ where $\omega = \frac{-1 + i \sqrt{3}}{2}$
My Attempt:
$N(a + b\omega) = (a + b \omega)(a - b \omega) = a^2 + \omega^2 b^2$
I'm stuck here. Is my approach correct?
 A: The norm is the product of the conjugates, in the sense of algebraic extensions. The conjugate of $\omega$ is the other root of its minimal polynomial $x^2 + x + 1$ – it's $\omega^2$. So $$N(a + b \omega) = a^2 - ab + b^2.$$
Now to have $N(a + b \omega) = 1$, either $a = 0$, so we obtain $\pm \omega$ as units. Or $a \ne 0$, and we can write $a^2 - ab + b^2 = a^2(1 - x + x^2)$, where $x = \frac b a$. 
As $1 - x + x^2\ge \frac 34$, we can have $a^2 - ab + b^2 = 1$ only if $a^2 = 1$. 
If $a = 1$, $a^2 - ab + b^2 = 1 - b + b^2 = 1 \iff b = 0$ or $b = 1$. 
Similarly, if $a = -1$, we find $b = 0$ or $b = -1$. 
At the  end, we find the units are $\pm 1, \pm \omega, \pm \omega^2$. It's a cyclic group, generated by $-\omega$ or $-\omega^2$.
A: No, your second equality is wrong. See this :
I note $A := \mathbf{Z}[\omega] := \{a+b\omega\;|a,b\in\mathbf{Z}\}$ where $\omega = \frac{-1 + i \sqrt{3}}{2}$. Consider $N : A \to \mathbf{N}$ defined by $N(z) := z\overline{z}$ for all $z\in A$, where $\overline{z}$ denote $z$'s complex conjugate. This definition shows that $N$ is multiplicative, as complex conjugation is multiplicative. Now let $z$ be a unit in $A$, so that we have a $z'$ in $A$ such that $zz' =1$. Then $N(zz') = N(1)=1$, but $N(zz') = N(z)N(z')$ as $N$ is multiplicative. So that $N(z)N(z') = 1$. But $N$ takes values in $\mathbf{N}$, so there is not choice : $N(z) = 1$. If you write $z = a+b\omega$ with $a,b\in\mathbf{Z}$, this is equivalent to say that $a^2 + ab(\omega + \overline{\omega}) + b^2 |\omega|^2 = 1$, that is, to $a^2 - ab + b^2 = 1$, with $a,b\in\mathbf{Z}$. Knowing this equation, what can you say now about $a$ and $b$ ?
