For u ∉ Q̄ with uū = ρ algebraic and z = u/(u² − a) (aā ≠ ρ²) or u/(u² − a)² (aā = ρ²), a ∈ Q̄ non-zero, the space Q̄ + Q̄u + Q̄ū + Q̄z + Q̄z̄ carries a 2×3 configuration

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $\rho = u\bar u$ algebraic, let $a \in \overline{\mathbb{Q}}$, $a \neq 0$, and let $z$ be $u/(u^2 - a)$ if $a\bar a \neq \rho^2$, or $u/(u^2 - a)^2$ if $a\bar a = \rho^2$. Put $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ and $W = H_0 + \overline{\mathbb{Q}}z + \overline{\mathbb{Q}}\bar z$. Then there are $x_1, x_2$, linearly independent over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, linearly independent over $\overline{\mathbb{Q}}$, with all six products $x_iy_j$ in $W$.

The configuration is $x = (1, u^2)$, $y = (b, bu^2, bu^4)$, with $b = 1/(u(u^2 - a)(u^2 - a'))$ and $a' = \rho^2/\bar a$ in the first case and $b = 1/(u(u^2 - a)^2)$ in the second: $W$ contains the progression $b, bu^2, bu^4, bu^6$. Since $\widetilde{\mathcal{L}}$ is closed under complex conjugation, a hypothesis $z \in \widetilde{\mathcal{L}}$ brings $\bar z$ with it, so $W$ is the space Roy's strong six exponentials theorem sees at a candidate; DiazModulus.circle_point_extension_two_by_three_configuration_iff (Theorem B) treats the four-dimensional spaces $H_0 + \overline{\mathbb{Q}}w$ only. The statement does not involve $e^u$, so it keeps its content if Diaz's conjecture holds.

Proof. In the first case $a' = \rho^2/\bar a$ is algebraic and differs from $a$, and $\bar z = -(\rho/\bar a)\,u/(u^2 - a')$; in the second, $\rho^2/\bar a = a$ and $\bar z = (\rho/\bar a^2)\,u^3/(u^2 - a)^2$. With $u^{-1} = \bar u/\rho$, partial fractions in $v = u^2$ write each of $b, bu^2, bu^4, bu^6$ as a combination of $u$, $u^{-1}$, $z$ and $\bar z$ with algebraic coefficients; for instance $(a - a')\,bu^2 = z - u/(u^2 - a')$ in the first case and $bu^2 = z$ in the second. The six products $x_iy_j$ are $b, bu^2, bu^4, bu^2, bu^4, bu^6$. Both families are free because $u$ is transcendental over $\overline{\mathbb{Q}}$: $s + tu^2 = 0$ and, after dividing by $b \neq 0$, $g_0 + g_1u^2 + g_2u^4 = 0$ force all coefficients to vanish.

Novelty. Not asserted for the configuration: rescaled, it is the configuration $(u, \bar u) \times (1, U, \bar U)$, $U = 1/(u^2 - a)$, in the proofs of Diaz (2004, Th. 2) and Diaz (2007, Th. 6(1) and 7(1)), and the progression of Fischler (2001, Lemma 6.1). That no four-dimensional subspace of $W$ containing $H_0$ carries one is DiazModulus.circle_point_conjugate_pair_configuration_invisible_to_four_dimensional_extensions.