For g = G(v) + κ · conj(v G(v)) with |κ| = 1 and deg G ≥ 1, (1, v) and (v̄, 1, g) form a 2×3 configuration in span(1, v, v̄, g, ḡ)

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $\kappa \in \overline{\mathbb{Q}}$ with $\kappa\bar\kappa = 1$, let $G \in \overline{\mathbb{Q}}[X]$ with $\deg G \geq 1$, and put $g = G(v) + \kappa\,\overline{vG(v)}$. Then $1, v$ are linearly independent over $\overline{\mathbb{Q}}$, so are $\bar v, 1, g$, and each of the six products of a member of $(1, v)$ with a member of $(\bar v, 1, g)$ lies in $\mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, g, \bar g)$.

Proof. The products are $\bar v, 1, g$, $v\bar v = 1$, $v$ and $vg = \kappa\bar g$, the last identity using only $v\bar v = 1$ and $\kappa\bar\kappa = 1$. The family $(1, v)$ is free as $v \notin \overline{\mathbb{Q}}$. For $(\bar v, 1, g)$, with $n = \deg G$, $v^n\overline{G(v)}$ is the value at $v$ of the reflected conjugate polynomial, of degree at most $n$, so a relation $g_0\bar v + g_1 + g_2g = 0$ times $v^{n+1}$ is a polynomial identity at the transcendental $v$ whose coefficient at $X^{2n+1}$ is $g_2$ times the leading coefficient of $G$. So $g_2 = 0$, and then $g_0\bar v + g_1 = 0$ gives $g_0 = g_1 = 0$.

Novelty. Not found in the sources read in this form; routine. It is the configuration of DiazModulus.conj_stable_real_pencil_space_carries_two_by_three_configuration, written for Laurent data.