If t² + π² is rational for a real t ≠ 0 with e^t algebraic, then e^{iγ/π} is transcendental for every rational γ ≠ 0

Let $t \neq 0$ be a real number with $e^{t}$ algebraic and $t^{2} + \pi^{2} = r \in \mathbb{Q}$. Then

is transcendental for every non-zero rational $\gamma$.

Two open statements meet here. Diaz's conjecture at $t + i\pi$ says that $t^{2} + \pi^{2}$ is transcendental. The statement (S) for real $\gamma$ (DiazModulus.recip_pi_not_log_real_gamma, Open) says that $e^{\gamma/(i\pi)} = e^{-i\gamma/\pi}$ is transcendental for every non-zero real algebraic $\gamma$; its smallest case is $e^{i/\pi}$ (compare DiazModulus.exp_i_div_pi_or_exp_i_pi_cube_transcendental). The hypothesis here is expected never to hold. What the theorem shows is that the two statements cannot both fail at rational data: a rational value of $t^{2} + \pi^{2}$ forces $e^{i\gamma/\pi}$ to be transcendental at every rational $\gamma \neq 0$.

Proof. By DiazModulus.diaz_number_forces_transcendence, $e^{r/(i\pi)} = e^{-ir/\pi}$ is transcendental. If $e^{i\gamma/\pi}$ were algebraic, write $r/\gamma = a/b$ with $a \in \mathbb{Z}$ and $b \geq 1$: then $(e^{ir/\pi})^{b} = (e^{i\gamma/\pi})^{a}$ is algebraic, so $e^{ir/\pi}$ and its inverse $e^{-ir/\pi}$ are algebraic.

Novelty. Not found in the sources read (among them Diaz 2004 and 2007, Roy–Waldschmidt 1995 and 1997, and Waldschmidt 1973, 1974 and 2005). The transcendence of $e^{-ir/\pi}$ is an instance of a classical dichotomy: by M. Waldschmidt, Nombres transcendants, Lecture Notes in Math. 402 (1974), p. 202, two $\mathbb{Q}$-independent logarithms $\ell_1, \ell_2$ are either algebraically independent, or $e^{\ell_1^{2}/\ell_2}$ is transcendental; take $\ell_1 = t$ and $\ell_2 = i\pi$. The scaling step is elementary, and DiazModulus.recip_pi_log_rational_line proves a stronger form of it.