How to find the set of units of $\mathbb{Z}[w]$ Let $\mathbb{Z}[w]=\{a+bw : a,b \in \mathbb{Z}\}$ with $w=\frac{-1+i\sqrt 3}{2}$ and $w^2+w+1=0$.
I have proved so far that $\mathbb{Z}[w]$ is a subring of $\mathbb{C}$ and now I am tasked to find the set of units of this ring, $U(\mathbb{Z}[w])$.
To start with I said that we need to find two elements of the ring that multiply to give the one of the ring.
To this end suppose $(a+bw)(c+dw)=1$ with $a,b,c,d \in \mathbb{Z}$
Then 
$(a+bw)(c+dw)=1 \implies ac+ad(w)+bd(-w-1)=1 \implies (ac-ba)+(ad-bd)w=1=1+0w$
So we get that $ac-ba=1$ and $ad-bd=0$
From here I am unsure how to continue because the set should be like $U(\mathbb{Z}[w])=\{a+bw : a,b \in \mathbb{Z},\text{Extra restriction on a and b}\}$ but I have two equations that make reference to all of $a,b,c,d$.
Any help?
 A: As I noted in the comments, there are at least two errors in the algebra. Your method should be leading you to:
$$ac - bd=1\qquad ad+bc-bd=0$$
which is a more difficult system to deal with then the one you have posted, because there is no factoring.
Here is an alternative to setting $(a+b\omega)(c+d\omega)=1$. The norm maps this ring into $\mathbb{Z}_{\geq0}$: $$\mathrm{N}(a+b\omega)=(a+b\omega)(a+b\bar{\omega})=a^2-ab+b^2$$ You would have to have the norm equal $1$, since $1=\mathrm{N}(x x^{-1})=\mathrm{N}(x)\mathrm{N}(x^{-1})$.
If either $a=0$ or $b=0$, then you see that there is a solution with the other number being $1$ or $-1$. So here are four units: $1,-1,\omega,-\omega$.
If $a,b$ are nonzero with opposite signs, then clearly $a^2-ab+b^2>1$.
If $a>b>0$, then $a^2-ab+b^2=a(a-b)+b^2>1$.
If $b>a>0$, then $a^2-ab+b^2=a^2+b(b-a)>1$.
If $a=b>0$, then $a^2-ab+b^2=a(a-b)+b^2=b^2$ which could be $1$ if $b=1$. This gives units $1+\omega$.
And if both $a,b$ are negative with $a+b\omega$ being a unit, then $-(a+b\omega)$ would also be a unit, and we'd be in a case we've already examined. This allows one last unit: $-1-\omega$.
So there are six units.

Visually, this $\omega$ and $1$ form a lattice where it is clear the vertices never fall within the unit circle (except at $0$). So no number in the lattice with magnitude larger than $1$ could be a unit, since there is nothing with small magnitude to be its inverse. Only the six values on the lattice on the unit circle itself are possible units, and they all have inverses in the lattice.
A: You seem to have figured out the problem, so I am going to post an answer which uses a bit more machinery.  It looks like you're solving this problem in an abstract algebra class, but I would also consider this to be a number theory problem.  If you learn some number theory later, you might go back and solve this problem a different way, something like this: by the fundamental theorem of finitely generated abelian groups, every finitely generated abelian group $G$ is isomorphic to a finite direct sum $$\mathbb{Z} \times \cdots \times \mathbb{Z} \times H$$ where $H$ is a finite abelian group.  The number of $\mathbb{Z}$s in the direct sum is called the rank of $G$.
You can show that $A = \mathbb{Z}[w]$ is the integral closure of $\mathbb{Z}$ in the field $K = \mathbb{Q}(w)$.  Dirichlet's unit theorem then says that the units of $A$ form a finitely generated abelian group of rank $s-1$, where $s$ is half the number of complex embeddings of $K$ into $\mathbb{C}$.  
The only complex embeddings of $K$ into $\mathbb{C}$ are given by the identity mapping and the mapping $w \mapsto \overline{w}$.  So in fact $U(A)$ is a finitely generated abelian group of rank $0$.  This is the same as saying that the only units of $A$ are the roots of unity which lie in $K = \mathbb{Q}(w)$.
Clearly all $6$ sixth roots of unity lie in $K$, so then the question becomes whether there are any other roots of unity in $K$.  And you can see without too much trouble that the answer is no: you can first argue that $i, -i$ are not in $K$.  If this were the case, then also $\sqrt{3}$ would be in $K$, and therefore you would have $\mathbb{Q} \subsetneq \mathbb{Q}(\sqrt{3}) \subseteq K$.  Since $K$ and $\mathbb{Q}(\sqrt{3})$ both have the same degree over $K$, this implies $\mathbb{Q}(\sqrt{3}) = K$, which is impossible, because $\mathbb{Q}(\sqrt{3}) \subseteq \mathbb{R}$, while $K$ is not.  
The remaining roots of unity you need to exclude are primitive $5$th roots of unity, $7$th roots of unity, $8$th, and so on.  But none of these roots $\zeta$ can be in $K$, because $[K : \mathbb{Q}] = 2$, while $[\mathbb{Q}(\zeta) : \mathbb{Q}] > 2$.
