I’m thinking about the number $x=r \alpha e^{2i\pi \theta}$ where $$ r = \sqrt{\frac{\sqrt{2}}{3-\sqrt{3}}},\quad \alpha = 1-\frac{\sqrt{3}}{2} + \frac{i}{2},\quad \theta = \frac{1}{48}. $$ Each one is algebraic over $\mathbb{Q}(i)$ and hence so is the product $x=r\alpha e^{2i\pi/48}$. We also have the relation $$ r^2 e^{4i\pi\theta}(1+i\alpha^2) = 1. $$
Question: Is it possible that $x$ satisfies a degree $2$ polynomial over $\mathbb{Q}(i)$? Can I say for certain that it is not degree $2$ over $\mathbb{Q}(i)$?
Context: I'm interested in studying the dynamical behaviour of certain diagonal orbits in $\operatorname{SL}_2(\mathbb{C})/\operatorname{SL}_2(\mathbb{Z}[i])$. The behaviour of an orbit generated by an upper triangular unipotent matrix is known to be related to the continued fraction expansion of the top right entry. In particular the 'quadraticness' of such an entry would have dynamical implications - See, for example, the beginning remarks of section 5 in these notes by Hensley which say that quadratic numbers have expansions which are eventually periodic.
Edit: Thanks so much for your answers. For anyone wondering about where these constants come from, $r$ is the constant appearing Minkowski's theorem on minimal values of two binary complex linear forms (evaluated on $\mathbb{Z}[i]$). See Satz LXII on page 218 of Minkowski's Diophantische Approximationen.
There are some other (other than qudraticness) algebraic criteria which guarantee my desired orbit behaviour so I will try and explore those now.
The minimal polynomial of $x$ over $\mathbb Q$ is $$9x^8+9x^4+6x^2+1$$ which splits in $\mathbb Q(i)$, but the splitting proves that $x$ has algebraic degree $4$ there, not $2$. But then $x^2$ is degree $2$ in $\mathbb Q(i)$: $$x^4+ix^2+\frac i3=0$$