= Solution
Assume that $\zeta$ is a <primitive root of unity>[primitive] fifth root of unity. The <cyclotomic field> $K=\mathbb Q(\zeta)$ has degree four and signature $(r_1,r_2)=(0,2)$. The <Dirichlet unit theorem> therefore gives free rank
$$
r_1+r_2-1=1.
$$
The <roots of unity in a rational cyclotomic field> are the ten elements $\mathord\pm\zeta^j$, so
$$
\mathcal O_K^\times\cong C_{10}\times\mathbb Z.
$$
For explicit units, $-\zeta$ has order ten, while
$$
N_{K/\mathbb Q}(1+\zeta)
=\prod_{j=1}^4(1+\zeta^j)
=\Phi_5(-1)=1.
$$
Hence $1+\zeta$ is a unit. It has infinite order because in the standard complex embedding
$$
|1+\zeta|=2\cos(\pi/5)>1,
$$
whereas every root of unity has modulus one. Its image in the free $\mathbb Z$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequently
$$
H=\langle-\zeta,1+\zeta\rangle
$$
has finite index in $\mathcal O_K^\times$, as described by the <unit group of the fifth cyclotomic field>.
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