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show/hide this revision's text 2 Cleaned up a bit.

I might as well add another two interpretations of this group from the theory of algebraic tori - this should make the isomorphism between $SO_2(\mathbb{Q})$ and $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$ mentioned in another answer look a bit more natural. It The group you seek is the group of rational points of the norm torus of the field extension $\mathbb{Q}[i]/\mathbb{Q}$, and it is also the group of rational points of the quotient of the Weil restriction (along the same extension) of the multiplicative group by the counit subgroup.

Recall that restriction of scalars (also known as Weil restriction) of the multiplicative group $\mathbb{G}_m$ along $\mathbb{Q}[i]/\mathbb{Q}$ produces a two dimensional algebraic torus, that in particular is a subgroup of $GL_{2,\mathbb{Q}}$. The restriction of determinant to this subgroup is called the norm map, and its kernel (i.e., the intersection with $SL_2$) is the norm torus. Oddly enough, base change to $\mathbb{R}$ yields $U(1)$, the The norm torus is equal to $SO_2$ as an algebraic subgroup of the extension $\mathbb{C}/\mathbb{R}$.SL_2$ - this is special to $\mathbb{Q}[i]$ among quadratic extensions, as other quadratic fields have norms that are not isomorphic to the standard diagonal quadratic form.

Since restriction of scalars is right adjoint to base change, you get a counit homomorphism $\mathbb{G}_{m, \mathbb{Q}} \to \operatorname{Res}_{\mathbb{Q}}^{\mathbb{Q}[i]} \mathbb{G}_{m, \mathbb{Q}[i]}$ of group schemes. The resulting quotient is naturally isomorphic to the norm an anisotropic rank one torus , and I think this is most easily whose rational points are in natural bijection with elements of $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$.

An isomorphism between these two descriptions can be seen by passing to character lattices as Galois modules (this passage is an antiequivalence)antiequivalence - see SGA3 Exp. 9). The character lattice of $\mathbb{G}_m$ is a copy of $\mathbb{Z}$ with trivial action, and restriction of scalars corresponds to taking a tensor product with the group ring of the Galois group (i.e., it is the induced representation - there is also a way to think of this as a pushforward of étale sheaves)sheaves on the corresponding spectra of fields). The norm torus construction corresponds to taking the quotient by the invariant sublattice, and the counit quotient corresponds to taking the Galois anti-invariant sublattice. As abstract Galois modules, both character lattices are free of rank one with isotropy group equal to the absolute Galois group of $\mathbb{Q}[i]$.

You can show that there are elements of infinite order by noting that $m$th roots of unity have degree $\phi(m)$ for all $m \geq 1$, so you only have to find a point that is neither a 4th root of unity nor a 6th root of unity. Any nondegenerate pythagorean Pythagorean triple will suffice to produce a suitable rotation, and there is a neat characterization of such triples using cohomology given in this MathOverflow answer.
show/hide this revision's text 1

I might as well add another two interpretations of this group from the theory of algebraic tori. It is the group of rational points of the norm torus of the field extension $\mathbb{Q}[i]/\mathbb{Q}$, and it is also the group of rational points of the quotient of the Weil restriction (along the same extension) of the multiplicative group by the counit subgroup.

Recall that restriction of scalars (also known as Weil restriction) of the multiplicative group $\mathbb{G}_m$ along $\mathbb{Q}[i]/\mathbb{Q}$ produces a two dimensional algebraic torus, that in particular is a subgroup of $GL_{2,\mathbb{Q}}$. The restriction of determinant to this subgroup is called the norm map, and its kernel (i.e., the intersection with $SL_2$) is the norm torus. Oddly enough, base change to $\mathbb{R}$ yields $U(1)$, the norm torus of the extension $\mathbb{C}/\mathbb{R}$.

Since restriction of scalars is right adjoint to base change, you get a counit homomorphism $\mathbb{G}_{m, \mathbb{Q}} \to \operatorname{Res}_{\mathbb{Q}}^{\mathbb{Q}[i]} \mathbb{G}_{m, \mathbb{Q}[i]}$ of group schemes. The resulting quotient is naturally isomorphic to the norm torus, and I think this is most easily seen by passing to character lattices as Galois modules (this passage is an antiequivalence). The character lattice of $\mathbb{G}_m$ is a copy of $\mathbb{Z}$ with trivial action, and restriction of scalars corresponds to taking a tensor product with the group ring of the Galois group (i.e., it is the induced representation - there is also a way to think of this as a pushforward of étale sheaves). The norm torus construction corresponds to taking the quotient by the invariant sublattice, and the counit quotient corresponds to taking the Galois anti-invariant sublattice.

You can show that there are elements of infinite order by noting that $m$th roots of unity have degree $\phi(m)$ for all $m \geq 1$, so you only have to find a point that is neither a 4th root of unity nor a 6th root of unity. Any nondegenerate pythagorean triple will suffice to produce a suitable rotation.