(1, v + βv̄) and (1, v̄ + eβv, v̄² + ζv + ev²) form a 2×3 configuration in span(1, v, v̄, v̄² + ev², βv̄³ + ζv² + ev³)

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $e, \beta, \zeta \in \overline{\mathbb{Q}}$ be arbitrary and $S = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, \bar v^2 + ev^2, \beta\bar v^3 + \zeta v^2 + ev^3)$. Then $1, v + \beta\bar v$ are linearly independent over $\overline{\mathbb{Q}}$, so are $1, \bar v + e\beta v, \bar v^2 + \zeta v + ev^2$, and each of the six products of a member of the first family with a member of the second lies in $S$.

With $v = u/s$ these are the configurations of class (Q) in DiazModulus.laurent_conj_hull_two_by_three_configuration_iff.

Proof. Using only $v\bar v = 1$, the six products are $1$, $\bar v + e\beta v$, $(\bar v^2 + ev^2) + \zeta v$, $v + \beta\bar v$, $\beta(\bar v^2 + ev^2) + (1 + e\beta^2)$ and $(\beta\bar v^3 + \zeta v^2 + ev^3) + \bar v + \beta\zeta + e\beta v$, each an explicit $\overline{\mathbb{Q}}$-combination of the five generators. For freeness, a $\overline{\mathbb{Q}}$-relation multiplied by $v$ (first family) or by $v^2$ (second family) becomes a polynomial identity of degree at most $4$ at the transcendental $v$; for the second family the coefficients at $1, v, v^2$ are the three coefficients of the relation, since the orders $0, -1, -2$ are distinct.

Novelty. In print: for $v = u/s$, $|e| = |\beta| = 1$ and $\zeta = -\beta e^2\bar\zeta$, this is the configuration $(x_1, x_2) \otimes (1, y, \bar y)$ to which the proof of Diaz (2004, Th. 2, p. 538) applies the strong six exponentials theorem, at $x = (1, (u + \beta\bar u)/s)$; also Diaz (2007, Th. 4(1), p. 385). Here $e, \beta, \zeta$ are arbitrary.