In the plane spanned by a torsion-branch candidate and its conjugate only the two axes have algebraic modulus

Statement. Let $u$ satisfy $\Im u = k\pi$ with $k\in\mathbb{Z}\setminus\{0\}$ and let $u\bar u=\lVert u\rVert^{2}$ be algebraic. Then for rationals $a,b$,

In words: inside the real plane $\mathbb{Q}u\oplus\mathbb{Q}\bar u$, the only points with algebraic modulus are the rational multiples of $u$ and the rational multiples of $\bar u$. Every genuine mixture leaves the locus.

Source and attribution. All the mathematics of this section is Carlo Perassi's, in his manuscript C. Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz, unpublished manuscript, 15 August 2026, §Polar coordinates and the discreteness of the period. No novelty is claimed. This is part (a) of Theorem Period-plane classification (thm:period-plane), $V_u\cap\mathcal{D} =\mathbb{Q}^{\times}u\,\dot\cup\,\mathbb{Q}^{\times}\bar u$ for $\Im u\in\pi\mathbb{Q}$, formalised here in the slightly smaller range $\Im u\in\pi\mathbb{Z}$. Only the unconditional norm identity eq:pp-norm of that theorem's proof had been published, as Diaz.period_plane_norm; the classification it is there to serve had not. Nothing here is new.

Proof. The identity Diaz.period_plane_norm at $c=0$ gives $\lVert a u + b\bar u\rVert^{2}=(a+b)^{2}\lVert u\rVert^{2}-4ab\,(\Im u)^{2}$, and $(\Im u)^{2}=k^{2}\pi^{2}$. So the left-hand side differs from the algebraic number $(a+b)^{2}\lVert u\rVert^{2}$ by the rational multiple $4abk^{2}$ of $\pi^{2}$. If $ab\neq 0$ that multiple is non-zero and algebraicity of the left-hand side would make $\pi^{2}$ algebraic, against DiazModulus.pi_sq_transcendental; if $ab=0$ the term vanishes and what is left is algebraic.