= Solution
Put $L=\mathbb Q(\omega)$ and $F=\mathbb Q(\omega+\omega^{-1})$. The <maximal real subfield of the fifth cyclotomic field> is $F=\mathbb Q(\sqrt5)$. The <roots of unity in a rational cyclotomic field> show that the roots of unity in $L$ are exactly $\{\mathord\pm\omega^a\}$.
Let $u\in\mathcal O_L^\times$. For every embedding $\sigma:L\to\mathbb C$,
$$
\left|\sigma\left(\frac{u}{\bar u}\right)\right|=1.
$$
Thus the <kernel of the logarithmic unit embedding> implies that $u/\bar u$ is a root of unity. Since conjugation acts trivially modulo the prime element $1-\omega$, we have $u/\bar u\equiv1\pmod{1-\omega}$. Among the tenth roots of unity this excludes the elements $-\omega^a$, so $u/\bar u\in\langle\omega\rangle$. Squaring is a bijection on this group of order five, so choose a root of unity $\eta$ with $\eta/\bar\eta=u/\bar u$. Then $v=u/\eta$ is fixed by conjugation and is therefore a unit of $F$.
The <units of the quadratic field Q square root of five> are $\mathord\pm\phi^b$, where $\phi=(1+\sqrt5)/2$. Absorbing the sign and $\eta$ into $\mathord\pm\omega^a$ proves
$$
\boxed{\mathcal O_L^\times
=\left\{\mathord\pm\omega^a
\left(\frac{1+\sqrt5}{2}\right)^b:a,b\in\mathbb Z\right\}.}
$$
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