If g = F(v)/v^N satisfies ḡ = γ v g with |γ| = |v| = 1, then g ≡ G(v) + γ̄ · conj(v G(v)) modulo span(1, v̄) for some polynomial G with G(0) = 0

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $N \geq 0$, $F \in \overline{\mathbb{Q}}[X]$ and $g = F(v)/v^N$, and let $\gamma \in \overline{\mathbb{Q}}$ with $\gamma\bar\gamma = 1$ and $\bar g = \gamma vg$. Then there is $G \in \overline{\mathbb{Q}}[X]$ with $G(0) = 0$ such that

Proof. Divide $F = A + X^NB$ with $\deg A < N$, and put $f_0 = B(0)$ and $G = B - f_0$, so $G(0) = 0$ and $g = A(v)/v^N + f_0 + G(v)$. As $\bar v = 1/v$, a conjugated polynomial value times a power of $v$ is the value of the reflected conjugate polynomial, so $\overline{A(v)/v^N} = A'(v)$ with $A'(0) = 0$. Multiplying $\bar g = \gamma vg$ by $v^{N+K}$, $K = \deg G$, gives a polynomial identity at the transcendental $v$, and comparing degrees yields $\overline{A(v)/v^N} = \gamma vG(v) + \gamma f_0v$. Conjugating back, $g - (G(v) + \bar\gamma\,\overline{vG(v)}) = f_0 + \bar\gamma\bar f_0\bar v$.

Novelty. Not found in the sources read; routine.