$e^{it(k-{\theta})} = -1$ and $e^{it}$ is a $2(k-{\theta}) = 4{\theta}^2$-th root of unity while $e^{it(k+1/2)}$ is an $8{\theta}^2$-th root of unity. Since none of the primitive $m$-th roots of unity of algebraic degree $4$ have real part $\tfrac{2{\theta}^2+2{\theta}+1}{\sqrt{n}(2{\theta}({\theta}+1)}$ for any ${\theta}$, there are no solutions.

For the complement of $X$, the characteristic equations are: $$\begin{aligned}
\sqrt{n}e^{it((n-k-1)+1/2)} &= \frac{-2{\theta}^2-2{\theta}-1 \pm i(2{\theta}+1)\sqrt{4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)} \\
\sqrt{n}e^{it({\theta}+1/2)} &= \frac{2{\theta}^2+2{\theta}+1 \pm i(2{\theta}+1)\sqrt{4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)} \end{aligned}$$ Then $e^{it(n-k+{\theta})} = 1$, so $e^{it}$ is a $(n-k+{\theta}) = (2{\theta}^2+4{\theta}+2)$-th root of unity. Again no solutions exist.

Conference graphs
-----------------

The final family of graphs that arise in Chan’s theorem are known as conference graphs. A conference graph has parameters $((2{\theta}+1)^2,2{\theta}^2+2{\theta}, {\theta}^2+{\theta}-1,{\theta}^2+{\theta})$ with ${\theta}+ \tau = -1$. Chan shows that $y = \bar{x}$. As in the previous case, the characteristic equations reduce to , , and with $c = e^{it/2}$. Then equations minus reduce to ${\mathrm{Im}}(x) = {\mathrm{Im}}(e^{it({\theta}+1/2)})$ and therefore $x =
e^{it({\theta}+1/2)}$ or $x = -e^{-it({\theta}+1/2)}$. There are two possible values of $(x,y)$, namely $$x \in \left\{\frac{-1 \pm \sqrt{(2{\theta}+1)(2{\theta}-1)}i}{2{\theta}},  
        \frac{1 \pm \sqrt{(2{\theta}+1)(2{\theta}+3)}i}{2({\theta}+1)}\right\}, \; y = \bar{x}.$$

[[**Case**]{} $x = \tfrac{-1 \pm \sqrt{(2{\theta}+1)(2{\theta}-1)}i}{2{\theta}}$:]{} In this case reduces to $ce^{ikt} = -1$. Thus $t = \frac{\pi(2m+1)}{k+1/2} =
\tfrac{2\pi(2m+1)}{n}$, for some integer $m$.

Suppose 