An algebraic $\pi\,\operatorname{Im} u$ forces an algebraic multiple of $1/(i\pi)$ into $\mathcal{L}$

Let $u \in \mathbb{C}$ with $e^{u}$ algebraic, and suppose $\operatorname{Im} u$ is not a rational multiple of $\pi$ but $\pi \cdot \operatorname{Im} u$ is algebraic. Then there is an algebraic $\gamma \neq 0$ with $e^{\gamma/(i\pi)}$ algebraic.

This is the exact analogue, for the degenerate case $r = 0$, of the published DiazModulus.recip_pi_log_of_period_aligned, which carries the same conclusion under the period-aligned hypothesis $\exists r \in \mathbb{Q}^{\times}$ with $\pi(\operatorname{Im} u + r\pi) \in \overline{\mathbb{Q}}$.

The proof is shorter than the aligned one, because no period translate is involved: with $\nu = (u - \bar u)/2 = i \operatorname{Im} u$ one has $e^{2\nu} = e^{u}/\overline{e^{u}}$, so $e^{\nu}$ is algebraic, and $-\pi \operatorname{Im} u$ is the required $\gamma$ since $\gamma/(i\pi) = \nu$.

Together with DiazModulus.recip_pi_not_log this closes the leaf DiazModulus.diaz_of_exp_not_real_irrational_angle_period_free_pi_im_algebraic.