If u is transcendental over a subfield F of ℂ and uū ∈ F, then a + bu + cū is transcendental over F for all a, b, c ∈ F with (b, c) ≠ (0, 0)

Let $F$ be a subfield of $\mathbb{C}$ and $u \in \mathbb{C}$ transcendental over $F$ with $u\bar u \in F$. Let $a, b, c \in F$ with $b \neq 0$ or $c \neq 0$. Then $a + bu + c\bar u$ is transcendental over $F$.

Proof. Put $e = a + bu + c\bar u$ and $\rho = u\bar u$, which is non-zero as $u \neq 0$. Multiplying by $u$ gives $bu^2 + (a - e)u + c\rho = 0$, so $u$ is a root of $bX^2 + (a - e)X + c\rho$, a polynomial over the intermediate field $F(e)$ that is non-zero because $b \neq 0$ or $c\rho \neq 0$. If $e$ were algebraic over $F$, then $F(e)$ would be algebraic over $F$, and $u$, algebraic over $F(e)$, would be algebraic over $F$, a contradiction.

Novelty. A standard fact of field theory; routine.