At a generic point of the circle, no non-zero homogeneous polynomial over Q̄ vanishes at (u, ū, iπ)

Let $u \neq 0$ with $|u|^2$ algebraic, and suppose $u$ and $\pi$ are algebraically independent over $\overline{\mathbb{Q}}$ (a generic point of the circle $|z|^2 = |u|^2$). Then no non-zero homogeneous polynomial $P \in \overline{\mathbb{Q}}[X_0, X_1, X_2]$, of any degree, satisfies $P(u, \bar u, i\pi) = 0$.

The degree-2 case over $\mathbb{Q}$ is DiazModulus.generic_conj_pair_no_quadratic_relation.

Proof. With $\rho = |u|^2$ and $\bar u = \rho/u$, homogeneity gives $u^dP(u, \bar u, i\pi) = P(u^2, \rho, u\,i\pi)$. The monomial $X_0^aX_1^bX_2^c$ ($a + b + c = d$) goes to $\rho^b u^{2a+c}(i\pi)^c$, and distinct monomials go to distinct monomials in $u$ and $i\pi$, which are algebraically independent.

Novelty. None asserted: it is an instance of Waldschmidt's Proposition 12.13 ((i) ⇒ (iii)) in Diophantine Approximation on Linear Algebraic Groups, applied to $\overline{\mathbb{Q}} + \overline{\mathbb{Q}}\,\rho/u^2 + \overline{\mathbb{Q}}\,i\pi/u$. This instance was not found stated in the sources read. The contribution of this node is the formal proof.