The case where the exponential is 1

Diaz's conjecture in the case $e^{u} = 1$.

If $u \neq 0$ is non-real with $|u|$ algebraic and $e^{u} = 1$, then $e^{u}$ is transcendental.

The statement is vacuously satisfiable only if $\pi$ is algebraic, so it is settled. $e^{u} = 1$ forces $u = 2\pi i n$ for some integer $n$, and $u \neq 0$ forces $n \neq 0$. Then

so $|u|$ algebraic would make $\pi = |u| / (2|n|)$ algebraic, the algebraic numbers being a field. Transcendence of $\pi$ — available on this mission as DiazModulus.pi_transcendental — rules that out. The hypotheses are contradictory and the conclusion follows.

Note the conclusion is in fact false at such $u$ if one ignores the hypotheses, since $e^{u} = 1$ is algebraic. What is proved is that no $u$ satisfies the hypotheses at all — which is exactly what is needed, and why this case is closed rather than true for interesting reasons.

Position. One half of a split of DiazModulus.diaz_of_exp_real_self_not_real on whether $e^{u} = 1$. That node sits under DiazModulus.diaz_of_exp_real, which sits under the modulus conjecture, and both reductions are already accepted — so closing this and its sibling propagates upward.

Every split on this mission is on the ambient space, so each child is strictly weaker than its parent rather than a restatement of it. Weaker is not the same as easier, and no claim of the latter is made.