The field generated by ℒ̃ is countable, and every circle uū = ρ > 0 contains a point transcendental over it

Here $\mathcal{L}$ is the set of logarithms of algebraic numbers (the $z \in \mathbb{C}$ with $e^z$ algebraic), and $\widetilde{\mathcal{L}}$ is the $\overline{\mathbb{Q}}$-vector space spanned by $1$ and $\mathcal{L}$. Let $F$ be the subfield of $\mathbb{C}$ generated by $\widetilde{\mathcal{L}}$, that is, by $\overline{\mathbb{Q}}$ and $\mathcal{L}$. Let $\rho > 0$ be a real number. Then $F$ is countable, and there is $u \in \mathbb{C}$ with $u\bar u = \rho$ that is transcendental over $F$.

Proof. $\mathcal{L}$ is countable: it is the union, over the countably many algebraic numbers $a$, of the fibres $\exp^{-1}(a)$, each a translate of $2\pi i\mathbb{Z}$. As $\overline{\mathbb{Q}}$ is countable, $\widetilde{\mathcal{L}}$, the image of the finitely supported functions from $\{1\} \cup \mathcal{L}$ to $\overline{\mathbb{Q}}$ under linear combination, is countable, and the subfield $F$ it generates has cardinality at most $\max(\#\widetilde{\mathcal{L}}, \aleph_0) = \aleph_0$. The circle point is the one that DiazModulus.exists_noncandidate_transcendental_on_circle gives for this countable field, with its third conclusion ($e^u$ transcendental) dropped.

Novelty. A countability argument; routine.