If 1, v, v̄, g, vg are free and ḡ ∈ span(1, v, vg) for a Laurent g, their span is span(1, v, v̄, h, h̄) with h = G(v) + κ · conj(v G(v)), |κ| = 1

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $N \geq 0$, $F \in \overline{\mathbb{Q}}[X]$ and $g = F(v)/v^N$. Suppose $1, v, \bar v, g, vg$ are linearly independent over $\overline{\mathbb{Q}}$ and $\bar g \in \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, vg)$, and let $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, g, vg)$. Then there are $\kappa \in \overline{\mathbb{Q}}$ with $\kappa\bar\kappa = 1$ and $G \in \overline{\mathbb{Q}}[X]$ with $G(0) = 0$ such that $\deg G \geq 2$ or $v^2 \in W$, and $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, h, \bar h)$ for $h = G(v) + \kappa\,\overline{vG(v)}$.

Proof. Write $\bar g = a + bv + c\,vg$ with $a, b, c \in \overline{\mathbb{Q}}$. Conjugating once more and using $v\bar v = 1$ gives $(\bar a + \bar cb) + (\bar b + \bar ca)\bar v + (c\bar c - 1)g = 0$, so $c\bar c = 1$ and $a = -c\bar b$ by freeness. Then $g_1 = g - \bar b\bar v = (XF - \bar bX^N)(v)/v^{N+1}$ satisfies $\bar g_1 = c\,vg_1$, and DiazModulus.laurent_conj_twisted_symmetric_normal_form gives $G$ with $G(0) = 0$ and $g_1 \equiv h$ modulo $\mathrm{span}(1, \bar v)$, where $h = G(v) + \kappa\,\overline{vG(v)}$ and $\kappa = \bar c$; the identity $vh = \kappa\bar h$ then shows that $1, v, \bar v, g, vg$ and $1, v, \bar v, h, \bar h$ span the same space. If $\deg G \leq 1$, then $G = a_1X$: $a_1 = 0$ would put $g$ in $\mathrm{span}(1, \bar v)$, against freeness, and $a_1 \neq 0$ gives $\bar h = \bar a_1\bar v + ca_1v^2$, so $v^2 \in W$.

Novelty. Not found in the sources read in this form; routine. It is the Laurent form of DiazModulus.conj_equivariant_degree_one_space_eq_real_pencil_space.