For z ≡ a v̄³ + b v̄² + c v² + d v³ mod span(1, v, v̄), the hull is in the real pencil class exactly when |a c̄ − b d̄| = ||a|² − |d|²| ≠ 0

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $a, b, c, d \in \overline{\mathbb{Q}}$ and let $z \in \mathbb{C}$ with $z - (a\bar v^3 + b\bar v^2 + cv^2 + dv^3) \in \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v)$, and let $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, z, \bar z)$ (no assumption is made on $\dim W$). Then $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, g, \bar g)$ with $g = G(v) + \kappa\,\overline{vG(v)}$, for some $\kappa \in \overline{\mathbb{Q}}$ with $\kappa\bar\kappa = 1$ and some $G \in \overline{\mathbb{Q}}[X]$ with $G(0) = 0$ and $\deg G \geq 2$, if and only if

Proof. Here $W = \mathrm{span}(1, v, \bar v, m, \bar m)$ with $m = a\bar v^3 + b\bar v^2 + cv^2 + dv^3$; write $p_{12} = a\bar c - b\bar d$ and $p_{14} = a\bar a - d\bar d$ (the dimension of $W$ is not needed: $p_{14} \neq 0$ already makes $m, \bar m$ free modulo $\mathrm{span}(1, v, \bar v)$). If $g = x_0 + x_1v + x_2\bar v + xm + y\bar m$ and $f_{-3}, f_{-2}, f_2, f_3$ are the coefficients of $xm + y\bar m$, then $G(v) - (x_0 + x_1v + f_2v^2 + f_3v^3) = \bar v\,\overline{Q(v)}$ for $Q = \bar x_2 + \bar f_{-2}X + \bar f_{-3}X^2 - \bar\kappa G$; multiplied by $v^{n+1}$, $n = \deg Q$, the left side has order at least $n + 1$ and the right side degree at most $n$, so both vanish, as $v$ is transcendental. Hence $G = x_0 + x_1X + f_2X^2 + f_3X^3$ and $\bar\kappa G = \bar x_2 + \bar f_{-2}X + \bar f_{-3}X^2$, so $f_3 = 0$, $f_{-3} = \kappa\bar f_2$ and $f_2 \neq 0$ as $\deg G \geq 2$; with $D = x\bar x - y\bar y$ the Plücker identities give $Dp_{14} = f_2\bar f_2 \neq 0$ and $Dp_{12} = \kappa\bar f_2^2$, so $p_{14} \neq 0$ and $p_{12}\bar p_{12} = p_{14}^2$. Conversely $\kappa = p_{14}/p_{12}$ has $\kappa\bar\kappa = 1$; with $G = G_1X + \bar p_{12}X^2$, $G_1 = \kappa\,\overline{\bar ab - d\bar c}$, one has $g = G_1v + (\bar am - d\bar m)$ and $\bar g = \bar G_1\bar v + (a\bar m - \bar dm)$, and the determinant $a\bar a - d\bar d = p_{14} \neq 0$ of this change of basis gives $W = \mathrm{span}(1, v, \bar v, g, \bar g)$.

Novelty. Not found in the sources read; routine.