Every unit in $\mathcal O_K$ is equal to a power of $\zeta$ times a real unit in $\mathcal O_K$, with $\zeta:=e^{2\pi\sqrt{-1}/p}$ and $K:=\mathbb Q(\zeta)$
The proof is below, but I don't understand some parts.
$1)$ Why is the multiplication of $u$ by $\zeta^j$ is the same as multiplying $u/\bar u$ by $\zeta^{2j}$ ?
$u/\bar u$ has the double argument of $u$ and ''length'' $1$ and if you multiply $u$ by $\zeta^j$ you just change the angle not the length, (and $u$ does not have necessarily length $1$), so what does it mean here, ''the same effect'' ?
$2)$ Why is $u\equiv\bar u\mod(1-\zeta)$ ?
The rest is OK, since in $K$ the roots of unity are $\pm1,\pm\zeta,\dots,\pm\zeta^{p-1}$ (already proved), so we just have to exclude $-1$, but I don't understand what modulo $1-\zeta$ means
Since Gerry Myerson answered the first question, here is the answer for the second question. If $a\equiv b\pmod{1-\zeta}$, it means we can write $a=b+x(1-\zeta)$ with $x\in\mathcal{O}_K$. This notion of equivalence is an equivalence relation. We have $1\equiv\zeta\pmod{1-\zeta}$, because $1=\zeta+(1-\zeta)$. From this follows that $\zeta\equiv\zeta^2\pmod{1-\zeta}$ and $1\equiv\zeta^2\pmod{1-\zeta}$. By induction you can show that $1\equiv\zeta^k\pmod{1-\zeta}$ for all integer $k$.
If we write $u=\sum_{m=0}^{p-1}a_m\zeta^m$, it follows that $$u=\sum_{m=0}^{p-1}a_m\zeta^m\equiv\sum_{m=0}^{p-1}a_m\equiv\sum_{m=0}^{p-1}a_m\zeta^{p-m}=\overline{u}\pmod{1-\zeta}.$$