Unit-modulus elements of the dyadic cyclotomic ring

For nonzero $z$, write its canonical form $z=a\mathrm{/}{2}^r$ using [F1]. If $r>0$, then $a\bar{a}=2^{2r}$, so $2\mid a\bar{a}$ in $O$ and [F2] gives $a=\sqrt{2}{a}_1$ for some $a_1\in O$. Whenever $a_j\in O$ has $a_j{\bar{a}}_j=2^m$ with $m>0$, the same argument gives $a_j=\sqrt{2}{a}_{j+1}$ and $a_{j+1}{\bar{a}}_{j+1}=2^{m-1}$. Applying this for $m=2r,2r-1,\dots ,1$ yields $a=(\sqrt{2}{)}^{2r}{a}_{2r}=2^r{a}_{2r}$. All canonical coefficients of $a$ are then even, contradicting the minimality of $r$. Thus $r=0$ and $z\in O$.

Write $z=a+b\omega +c{\omega }^2+d{\omega }^3$ with integer coefficients. Direct multiplication gives $z\bar{z}=P+\sqrt{2}Q=1$, where $P=a^2+b^2+c^2+d^2$ and $Q=a(b-d)+c(b+d)$ by [F2]. The numbers $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$: if $\sqrt{2}=m\mathrm{/}n$ in lowest terms, then $m^2=2n^2$ forces $m$ even and then $n$ even, a contradiction. Thus $P=1$ and $Q=0$. Four integer squares sum to $1$ only when exactly one coefficient is $1$ or $-1$ and the others are zero. Hence $z\in \{\pm 1,\pm \omega ,\pm {\omega }^2,\pm {\omega }^3\}=\{{\omega }^j:0\le j<8\}$.