What is meant by a number to be a root of unity? I am proving first case of Fermat's last theorem for regular primes by following Marcus' book "Number Fields". I have to prove following statement:

If $\varepsilon$ is a unit in $\mathbb{Z}[\omega]$ where $\omega = e^{2\pi i/p}$ then $\varepsilon$ is a power of $\omega$ times a real unit.

I have been able to prove that $\varepsilon/\overline{\varepsilon}$ is a root of unity. Now from here how do I conclude that it's equal to $u\omega^k$ where $u$ is a real unit and $k\in \mathbb{Z}$. 
Also, in an expository article I saw that, in fact,  $\varepsilon/\overline{\varepsilon} = \omega^{2r}$ for some $r\in \mathbb{Z}$. How to prove this assertion?
 A: (I am following Ireland and Rosen, Chapter 17, Section 11, Lemma 1)
I assume $p>2$.
A root of unity in $\mathbb{Z}[\omega]$ is of the form $\pm\omega^h$ for some integer $h$, so you have $\varepsilon/\overline{\varepsilon}=\delta\omega^h$ with $\delta\in\{+1,-1\}$.
If $\delta=+1$, choose $k\in\mathbb{Z}$ so that $h\equiv 2k\pmod{p}$, where $p$ is your regular prime. Then you have $\omega^{-k}\varepsilon=\omega^{k}\overline{\varepsilon}=\overline{\omega^{-k}\varepsilon}$ so $u:=\omega^{-k}\varepsilon$ is real and it is a unit and $\varepsilon=u\omega^k$.
You now have to show that $\delta=-1$ is actually impossible. For this, set $\lambda:=1-\omega$. Note that $\omega^j\equiv 1\pmod{\lambda}$ for every $j$, thus you have $\varepsilon\equiv\overline{\varepsilon}\pmod{\lambda}$ (write it as a linear combination of powers of $\omega$ and note that $\overline\omega$ is still a power of $\omega$). But if it were $\delta=-1$, then you would also have $\varepsilon\equiv-\varepsilon\pmod{\lambda}$, so $\lambda\,|\,2\varepsilon$ and, since $\varepsilon$ is a unit, $\lambda\,|\,2$, which is absurd (to see this, use for example the fact that $N(\lambda)=p$).
