For a real Laurent polynomial f in u/s of width k ≥ 3, or 2 with f(−λ̄) ≠ 0: W(λ, f) is conjugation-stable with a 2×3 configuration invisible to Theorem B

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and let $s \in \overline{\mathbb{Q}}$ be real with $s^2 = u\bar u$ (so $\bar u = s^2/u$; with $v = u/s$, conjugation acts as $v \mapsto 1/v$ on coefficients). Let $\lambda \in \overline{\mathbb{Q}}$ with $|\lambda| = 1$, and let $f = \sum_{|j| \le k} c_j (u/s)^j$ with $c_j \in \overline{\mathbb{Q}}$, $c_{-j} = \bar c_j$ and $c_k \neq 0$ (so $f$ is real). Let $W(\lambda, f)$ be the \overline{\mathbb{Q}}-span of $1, u, \bar u, f, f/(s + \lambda u)$, and write $f(-\bar\lambda)$ for $\sum_j c_j(-\bar\lambda)^j$, the value at $v = -\bar\lambda$ (the pole of $f/(s + \lambda u)$). Suppose the five numbers are free, and that $k \geq 3$, or $k = 2$ and $f(-\bar\lambda) \neq 0$. Put $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$. Then $W(\lambda, f)$ is stable under conjugation and carries a $2 \times 3$ configuration, and for no $w \in W(\lambda, f)$ outside $H_0$ does $H_0 + \overline{\mathbb{Q}}w$ carry one.

With DiazModulus.laurent_real_pencil_space_contains_square_of_root this settles visibility for Laurent $f$: $W(\lambda, f)$ is visible exactly when $k = 2$ and $f(-\bar\lambda) = 0$.

Proof. $f$ is real: conjugation sends $u/s$ to $(u/s)^{-1}$ and $c_j$ to $c_{-j}$, so reindexing $j \mapsto -j$ returns the sum. Then DiazModulus.conj_stable_real_pencil_space_carries_two_by_three_configuration gives conjugation stability and a configuration, and Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff) with exchange against DiazModulus.laurent_real_pencil_space_excludes_squares_and_reciprocals rules out every $H_0 + \overline{\mathbb{Q}}w$ with $w \notin H_0$.

Novelty. Not found in the sources read; it is short.