Both halves of the period-free split are non-empty

DiazModulus.diaz_of_exp_not_real_irrational_angle_period_free concerns the $u$ with $u \neq 0$, $|u|$ algebraic, $\operatorname{Im} e^{u} \neq 0$, $u$ on neither axis, $\operatorname{Im} u \notin \mathbb{Q}\pi$, and no non-zero rational $r$ making $\pi(\operatorname{Im} u + r\pi)$ algebraic. Its two children split this set according to whether $\pi \operatorname{Im} u$ is algebraic. Both parts are non-empty:

• $w_C = \sqrt{16 - \pi^{-2}} + i/\pi$ has $|w_C| = 4$ and $\pi \operatorname{Im} w_C = 1$, algebraic;
• $w_B = \sqrt{15} + i$ has $|w_B| = 4$ and $\pi \operatorname{Im} w_B = \pi$, transcendental.

Proof sketch. Both points have modulus $4$ and lie off the axes. $\operatorname{Im} u \in \mathbb{Q}\pi$ would give $1 = q\pi^{2}$ or $1 = q\pi$, impossible as $\pi$ is transcendental, and then $\sin(\operatorname{Im} u) \neq 0$ gives $\operatorname{Im} e^{u} \neq 0$. For a non-zero rational $r$, $\pi(\operatorname{Im} u + r\pi)$ is $1 + r\pi^{2}$ or $\pi + r\pi^{2}$; either being algebraic would make $\pi$ algebraic.

What it is for. Neither child is vacuous, so the split is a genuine case division: a proof of either child has to handle actual points.

Not claimed. Nothing about the transcendence of $e^{u}$ at these points.