$1$, $u$ and $\bar u$ are linearly independent over the base field

Let $K \subseteq \mathbb{C}$ be a subfield and $u \in \mathbb{C}$ transcendental over $K$ with $\rho := u\bar u \in K$. If $a, b, c \in K$ satisfy

then $a = b = c = 0$.

Why. Substituting $\bar u = \rho/u$ and clearing the denominator turns the relation into $b\,u^{2} + a\,u + c\rho = 0$, a quadratic over $K$. If any of $a,b,c$ were non-zero, that polynomial would be non-zero — its coefficients are $b$, $a$ and $c\rho$, and $\rho \neq 0$ — so $u$ would be algebraic over $K$, contrary to hypothesis.

Role. This is one of the two clauses of the stability lemma of the accompanying note; the other, conjugation-stability of the hull, is Diaz.conj_mem_hull. Together they say that the span $W_u$ of $\{1, u, \bar u\}$ over $K$ is a three-dimensional conjugation-stable space, which is the setting in which the rank and coefficient statements about the matrix $H$ are formulated.