If |u|² is algebraic and z is rational in u, the conjugation hull span(1, u, ū, z, z̄) contains u⁻¹ and consists of rational functions of u

Let $u \in \mathbb{C}$ be transcendental, that is, $u \notin \overline{\mathbb{Q}}$ (the algebraic numbers); spans and linear independence are over $\overline{\mathbb{Q}}$. Suppose $u\bar u = |u|^2$ is algebraic, let $z$ lie in $\overline{\mathbb{Q}}(u)$, the subfield of $\mathbb{C}$ generated by $\overline{\mathbb{Q}}$ and $u$, and let $W_z = \mathrm{span}(1, u, \bar u, z, \bar z)$. Then $u^{-1} \in W_z$, and every element of $W_z$ lies in $\overline{\mathbb{Q}}(u)$.

Proof. Since $u \neq 0$, $\rho = u\bar u$ is a non-zero algebraic number, so $u^{-1} = \rho^{-1}\bar u$ is an algebraic multiple of a generator of $W_z$. For $W_z \subseteq \overline{\mathbb{Q}}(u)$ it suffices that the five generators lie in $\overline{\mathbb{Q}}(u)$: $1$, $u$ and $z$ do, and so does $\bar u = \rho u^{-1}$. For $\bar z$: the preimage of $\overline{\mathbb{Q}}(u)$ under complex conjugation is a subfield of $\mathbb{C}$ containing $\overline{\mathbb{Q}}$ (conjugation maps algebraic numbers to algebraic numbers) and $u$ (as $\bar u \in \overline{\mathbb{Q}}(u)$), so it contains $\overline{\mathbb{Q}}(u)$; that is, conjugation maps $\overline{\mathbb{Q}}(u)$ into itself, and $\bar z \in \overline{\mathbb{Q}}(u)$. Of $u \notin \overline{\mathbb{Q}}$ only $u \neq 0$ is used.

Novelty. Not found in the sources read; routine.