Let $w$ be a primitve third root of unity. Find the units of $A=\{a+bw, a,b \in \mathbb{Z}\}$ What I have so far: if $x \in \mathbb{C}$, then $N(x)=\bar{x}x$ is multiplicative ($N(xy) = N(x)N(y)$). So $N$ restricted to $A$ is also multiplicative.
if $a+bw \in A$, then it's easy to see that $N(a+bi)=(a+bw)(a+b\bar{w})= (...) = a^2-ab+b^2$ (I'm using that $\bar{w}=w^2$ and that $w+w^2+1=0$)
If $x$ is a unit of $A$, $\exists\ y$ which satisfies $xy=1 \Rightarrow N(xy)=N(x)N(y) = 1$. It must be that, if $x=a+bw$ , $|a^2-ab+b^2| = 1$.
$a^2-ab+b^2=(a-b)^2 + ab$, so if $(a-b)^2=0$ then $a=b$, then $a^2 = 1$, so we have $a=b=\pm 1$. It's easy to find x^-1 in this case (and in other similar ones, where a=0, for example).
My problem is then we have $(a-b)^2 >0$. I'm stuck. Any hints?

Your way can be pushed through. I prefer to use
$$2(a^2-ab+b^2)=(a-b)^2+a^2+b^2.$$
So exactly two of $a$, $b$, and $a-b$ are equal to $\pm 1$ and the other is equal to $0$. Now the solutions should not be hard to find.
