If a·t² + b·π² is rational for a real logarithm t ≠ 0 of an algebraic number, with a ≠ 0 and b rational, then e^{iγ/π} is transcendental for every rational γ ≠ 0

Let $t \neq 0$ be real with $e^t$ algebraic, and let $a, b, c \in \mathbb{Q}$ with $a \neq 0$ and $a t^2 + b\pi^2 = c$. Then $e^{i\gamma/\pi}$ is transcendental for every $\gamma \in \mathbb{Q}^\times$.

The case $a = b = 1$ is DiazModulus.recip_pi_log_of_torsion_rational_modulus. The hypothesis cannot hold when $b = 0$, since $t$ would then be algebraic, against Hermite–Lindemann; the proof needs no case split. With $t = \log 2$, $a = -1/2$ and $b = 1/12$ it gives DiazModulus.dilog_half_irrational_or_exp_i_div_pi_transcendental.

Proof. Suppose $e^{i\gamma/\pi}$ is algebraic, so that $\lambda = i\gamma/\pi$ is a logarithm of an algebraic number. The relation gives $t^2/(i\pi) = -\frac{c}{a\gamma}\,\lambda + \frac{b}{a}\, i\pi$, a rational combination of the logarithms $\lambda$ and $i\pi$, hence again a logarithm of an algebraic number. The matrix $\begin{pmatrix} t & t^2/(i\pi) \\ i\pi & t\end{pmatrix}$ has determinant $0$, and its entries lie in a field of transcendence degree one, since $t$ is algebraic over $\mathbb{Q}(\pi)$. By the four exponentials theorem in transcendence degree one (DiazModulus.four_exponentials_trdeg_one), its rows or its columns are linearly dependent over $\mathbb{Q}$. Either way $\alpha t + \beta\, i\pi = 0$ with $\alpha, \beta \in \mathbb{Q}$ not both zero, which is impossible because $t$ is real and non-zero and $i\pi$ is purely imaginary.

Novelty. Not asserted. It is Brownawell's Corollary 5 (1974, p. 23) at $\eta = t/\pi$, followed by rational scaling. The statement for general $a, b$ was not found in the sources read.