Under the n = 2 case of Kirby's weak Schanuel conjecture, Diaz's conjecture is equivalent to: t² + π² is transcendental for every real t ≠ 0 with e^t algebraic

The hypothesis hWSC is the case $n = 2$ of Kirby's weak Schanuel conjecture (Kirby 2018, Conj. 1.5): if $\operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}(a, b, e^a, e^b) < 2$, then $ma + nb \in 2\pi i\mathbb{Z}$ for some integers $(m, n) \neq (0, 0)$. It is not the ``weak Schanuel'' of Calegari and Mazur, which is the algebraic independence of logarithms and would give Diaz's conjecture outright.

Under hWSC, Diaz's modulus conjecture (DiazModulus.DiazModulusConjecture) holds if and only if $t^2 + \pi^2$ is transcendental for every real $t \neq 0$ with $e^t$ algebraic. The forward implication holds without hWSC. So, under Kirby's weak form, the real half of (S) is not needed.

Proof. DiazModulus.leaf_iff_one says the single relation is equivalent to Diaz's implication for the $u$ off both axes with $e^u$ real and $\neq 1$; so Diaz's conjecture gives the relation unconditionally. Conversely, for a candidate $u$, DiazModulus.candidate_im_mem_pi_rat_of_weak_schanuel gives $\operatorname{Im} u = q\pi$ with $q \in \mathbb{Q}^\times$ and $\operatorname{Re} u \neq 0$. With $N$ the denominator of $q$, $v = Nu$ has $\operatorname{Im} v$ a non-zero multiple of $\pi$, so $e^v$ is real and $\neq 1$, and $|v| = N|u|$ is algebraic; leaf_iff_one makes $e^v$ transcendental, while $e^v = (e^u)^N$ is algebraic.

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