For u algebraically independent of π on a circle of algebraic radius, no configuration of the strong six exponentials theorem lies over Q̄ + Q̄u + Q̄ū + Q̄iπ

The strong six exponentials theorem cannot refute a generic candidate, even with $i\pi$.

Let $u \neq 0$ with $|u|^{2}$ algebraic, and suppose that $u$ and $i\pi$ are algebraically independent over $\overline{\mathbb{Q}}$. Then there are no $x_1, x_2$, linearly independent over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, linearly independent over $\overline{\mathbb{Q}}$, such that all six products $x_i y_j$ lie in the $\overline{\mathbb{Q}}$-span of $1, u, \bar u, i\pi$.

Roy's strong six exponentials theorem says that for such $x$ and $y$ one of the six products lies outside $\overline{\mathbb{Q}} + \overline{\mathbb{Q}}\mathcal{L}$. A candidate for Diaz's conjecture certifies $u$, $\bar u$ and $i\pi$ as logarithms of algebraic numbers, and its algebraic modulus makes the constant available. This node shows that these four dimensions never hold a configuration, so the theorem gives nothing against a generic candidate. It extends DiazModulus.sixExponentials_cannot_refute_candidate from $1, u, \bar u$ to $1, u, \bar u, i\pi$, where counting dimensions no longer suffices: the proof uses that every quadratic relation among $1, u, \bar u, i\pi$ is a multiple of the norm (DiazModulus.generic_quadratic_relation_is_norm). G. Diaz (2007) remarks that the strong six exponentials theorem has produced no example of a $\lambda$ with $|\lambda| \notin \overline{\mathbb{Q}} + \overline{\mathbb{Q}}\mathcal{L}$; on a generic candidate's own data, this node is one reason.

Novelty. None claimed. Elementary given the norm lemma; not found in Diaz (2007), Waldschmidt (2005) or Roy–Waldschmidt (1997).