At a candidate, $e^{u^{2}/\bar{u}}$ is transcendental

At a candidate, the third angular exponential is transcendental.

Let $u$ be a candidate for a counterexample to Diaz's modulus conjecture: $u \neq 0$, $|u|$ algebraic, and $e^{u}$ algebraic. Then $e^{u^{2}/\bar{u}}$ is transcendental.

The argument. Put $c = u\bar{u} = |u|^{2}$, which is non-zero and algebraic, and $t = u^{2}/c$. Feed the six exponentials theorem — the mission's DiazModulus.six_exponentials — the two families $x = (u, \bar{u})$ and $y = (1, t, t^{-1})$. Their product matrix is

because $u\,t^{-1} = c/u = \bar{u}$ and $\bar{u}\,t = u$. Here $u^{3}/c = u^{2}/\bar{u}$ is the exponent in the statement, and $\bar{u}^{2}/u$ is its complex conjugate.

Both independence hypotheses come from Hermite–Lindemann, which makes $u$ transcendental (the mission's DiazModulus.hermite_lindemann_holds, applied to the algebraic value $e^{u}$). A rational relation between $u$ and $\bar{u}$ would make $u/\bar{u} = u^{2}/c$ rational, hence $u^{2}$ and then $u$ algebraic. A rational relation among $1, t, t^{-1}$, multiplied by $t$, is a non-zero rational polynomial of degree at most $2$ vanishing at $t$; but $t$ is transcendental too, since $u^{2} = ct$ would otherwise be algebraic.

Four of the six entries are $u$ or $\bar{u}$, whose exponentials are algebraic by the candidate hypothesis together with the stability of algebraicity under complex conjugation. So if $e^{u^{2}/\bar{u}}$ were algebraic, its conjugate $e^{\bar{u}^{2}/u}$ would be as well, all six would be algebraic, and the six exponentials theorem would be contradicted.

Attribution — this is not new, and what is known is stronger. Guy Diaz, Produits et quotients de combinaisons linéaires de logarithmes de nombres algébriques : conjectures et résultats partiels, J. Théor. Nombres Bordeaux 19 (2007), 373–391, proves in théorème 7(1), p. 390: for $u, v \in \mathbb{C}$ with $(1, u, \bar{u})$ and $(v, \bar{v})$ both $\bar{\mathbb{Q}}$-linearly independent, $\{v,\ vu,\ v\bar{u}\} \not\subset \tilde{\mathcal{L}}$. Taking $v := u$ and his $u := u/\bar{u}$ turns this into $\{u,\ u^{2}/\bar{u},\ \bar{u}\} \not\subset \tilde{\mathcal{L}}$. For a candidate both of his hypotheses hold, by the same Hermite–Lindemann argument recalled above, so his theorem applies and yields strictly more than the statement here: it places $u^{2}/\bar{u}$ outside $\tilde{\mathcal{L}}$, the $\bar{\mathbb{Q}}$-span of $\{1\} \cup \mathcal{L}$, which properly contains $\mathcal{L}$ (it contains $1$, and $1 \notin \mathcal{L}$ because $e$ is transcendental); and his hypotheses do not require $|u|$ to be algebraic. Diaz derives it from the strong six exponentials theorem.

No novelty is claimed for this node. The one thing observed here is about implementation, not about transcendence: for the weaker conclusion — membership in $\mathcal{L}$ rather than in $\tilde{\mathcal{L}}$ — the ordinary six exponentials theorem already suffices, so the node closes from material the mission already carries, with no appeal to the strong form.

Role in the mission. This is the first genuine consequence drawn from six_exponentials here. It does not settle the conjecture at $u$ — that would need $e^{u}$ itself to be transcendental — but it records one more constraint every candidate must satisfy: a candidate cannot have all three of $u$, $\bar{u}$ and $u^{2}/\bar{u}$ in $\mathcal{L}$.