= 1$, which is not a solution. We conclude that or $ce^{i{\theta}t} = \overline{ce^{i (-{\theta}-1)t}}= \bar{c}e^{i({\theta}+1)t}$ and therefore $c = e^{it/2}$. The same analysis shows that $c = e^{it/2}$ for the complement of $G$ as well.

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[[**Case**]{} $x = i$, $y = -i$:]{} The characteristic equations reduce to: $$\begin{aligned}
\sqrt{n}e^{it(k+1/2)} &= 1 -i(2{\theta}+1) \\
\sqrt{n}e^{it({\theta}+1/2)} &= 1 +i(2{\theta}+1) \label{eqn:algdeg4} \end{aligned}$$ It follows that $e^{it(k+{\theta}+1)} = 1$, so $e^{it}$ is a $(k+{\theta}+1) = (2{\theta}^2+2{\theta}+1)$-th root of unity and $e^{it({\theta}+1/2)}$ is a $(4{\theta}^2+4{\theta}+2)$-th root of unity. But by , noting that $\sqrt{n}$ is not an integer, $e^{it({\theta}+1/2)}$ has algebraic degree exactly $4$. Since none of the primitive $m$-th roots of unity of algebraic degree $4$ have real part $1/\sqrt{n}$ for any ${\theta}$, there are no solutions.

For the complement of $X$, the characteristic equations reduce to $$\begin{aligned}
\sqrt{n}e^{it((n-k-1)+1/2)} &= 1 -i(2{\theta}+1) \\
\sqrt{n}e^{it({\theta}+1/2)} &= 1 -i(2{\theta}+1) \label{eqn:algdeg4.2} \end{aligned}$$ So $e^{it(n-k-1-{\theta})} = 1$ and $e^{it}$ is a $n-k-1-{\theta}= (2{\theta}^2+2{\theta}+1)$-th root of unity. Again there are no solutions.

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[[**Case**]{} $x = \tfrac{-1 \pm i\sqrt{(4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)}$:]{} This case is similar. The characteristic equations reduce to: $$\begin{aligned}
\sqrt{n}e^{it(k+1/2)} &= \frac{-2{\theta}^2-2{\theta}-1 \pm i(2{\theta}+1)\sqrt{4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)} \\
\sqrt{n}e^{it({\theta}+1/2)} &= \frac{2{\theta}^2+2{\theta}+1 \mp i(2{\theta}+1)\sqrt{4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)} \end{aligned}$$ Thus $-e^{it({\theta}+1/2)} = e^{it(k+1/2)}$ and so 