A period-aligned candidate makes an algebraic multiple of 1/(iπ) a logarithm

A period-aligned counterexample to Diaz's modulus conjecture would make an algebraic multiple of $1/(i\pi)$ a logarithm of an algebraic number.

Statement. Let $u\in\mathbb C$ with $\Im u\notin\pi\mathbb Q$ and $e^{u}\in\overline{\mathbb Q}$, and suppose $\pi(\Im u+r\pi)\in\overline{\mathbb Q}$ for some $r\in\mathbb Q^{\times}$. Then there is $\gamma\in\overline{\mathbb Q}^{\times}$ with $e^{\gamma/(i\pi)}\in\overline{\mathbb Q}$, that is, $\gamma/(i\pi)\in\mathcal L$.

This is not a decomposition. It is a route lemma: there is no reduction edge to submit for it, and it neither follows from nor implies the mission's target. It is published because it is what makes DiazModulus.diaz_of_exp_not_real_irrational_angle_period_aligned attackable. That child follows at once from

equivalently "$1/\pi$ is not, up to an algebraic factor, a logarithm of an algebraic number" — a statement about $\pi$ alone, implied by the strong four exponentials conjecture (take $x=(1,\nu)$, $y=(1,i\pi)$), open, and far more special than the leaf.

Proof. Put $\nu=i(\Im u+r\pi)$. It is non-zero because $\Im u\notin\pi\mathbb Q$. Then $(i\pi)\,\nu=-\pi(\Im u+r\pi)\in\overline{\mathbb Q}^{\times}$, which is $\gamma$ up to sign, and $\nu=\gamma/(i\pi)$. It remains to see $\nu\in\mathcal L$: with $d=\operatorname{den} r$ and $c=\operatorname{num} r$ one has $2d\,\nu=d\,(u-\bar u)+c\,(2\pi i)$, hence

which is algebraic; and a complex number whose $2d$-th power is algebraic is algebraic.

What is not used. No transcendence input at all: not Hermite–Lindemann, not the transcendence of $\pi$, not six exponentials. Nor is $|u|$ algebraic, nor $u\neq0$, nor $e^{u}\notin\mathbb R$ used — the leaf's remaining clauses are irrelevant to this implication, which is why the statement carries only three hypotheses.

Novelty. Elementary; possibly folklore, not found in the sources consulted (Diaz 2007 and this mission's notes). No claim of novelty is made.