For every u ∈ ℂ, z lies in span_ℚ ℒ + ℚu + ℚū exactly when z = c + pu + qū with c ∈ ℒ and p, q ∈ ℚ, and the Q̄-span of 1 and this space is ℒ̃ + Q̄u + Q̄ū

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 $u \in \mathbb{C}$ be arbitrary and $\mathcal{L}_u = \mathrm{span}_{\mathbb{Q}}(\mathcal{L}) + \mathrm{span}_{\mathbb{Q}}(u, \bar u)$. Then:

• $z \in \mathcal{L}_u$ if and only if $z = c + pu + q\bar u$ for some $c \in \mathcal{L}$ and $p, q \in \mathbb{Q}$;
• $\mathrm{span}_{\overline{\mathbb{Q}}}(\{1\} \cup \mathcal{L}_u) = \widetilde{\mathcal{L}} + \mathrm{span}_{\overline{\mathbb{Q}}}(u, \bar u)$.

Proof. $\mathcal{L}$ is already a $\mathbb{Q}$-subspace of $\mathbb{C}$: it contains $0$, it is closed under addition since $e^{a+b} = e^ae^b$, and under rational multiples by Diaz.exp_ratMul_isAlgebraic; so by span induction $\mathrm{span}_{\mathbb{Q}}(\mathcal{L}) \subseteq \mathcal{L}$, and membership in the sum unfolds to $z = c + pu + q\bar u$. For the second part, $1$ and every $c + pu + q\bar u$ lie in $\widetilde{\mathcal{L}} + \mathrm{span}_{\overline{\mathbb{Q}}}(u, \bar u)$, as rationals are algebraic. Conversely $\widetilde{\mathcal{L}} = \mathrm{span}_{\overline{\mathbb{Q}}}(\{1\} \cup \mathcal{L})$ lies in $\mathrm{span}_{\overline{\mathbb{Q}}}(\{1\} \cup \mathcal{L}_u)$ because $\mathcal{L} \subseteq \mathcal{L}_u$, and so do $u$ and $\bar u$, which lie in $\mathcal{L}_u$.

Novelty. Elementary; routine.