${\theta}$ is an integer. Now $x = e^{it({\theta}+1/2)} = e^{i\pi(2m+1)/(2{\theta}+1)}$ or $-e^{-it({\theta}+1/2)} = -e^{-i\pi(2m+1)/(2{\theta}+1)}$, so $x$ is a $2(2{\theta}+1)$-th root of unity. But we claim that $x$ is a $2(2{\theta}+1)$-th root of unity only when ${\theta}= 1$. To see this, note that ${\left|{\mathrm{Re}}(x)\right|} = 1/2{\theta}$, while any $2(2{\theta}+1)$-th root of unity has a real part of larger absolute value. (The root smallest real part in absolute value is ${\omega}=
e^{i\pi({\theta}+1)/(2{\theta}+1)}$, which has ${\left|{\mathrm{Re}}({\omega})\right|} = \sin(\tfrac{\pi}{2(2{\theta}+1)}) > 1/2{\theta}$ for ${\theta}> 1$.) In the case ${\theta}= 1$, we get the $3 \times 3$ Latin square graph, for which uniform mixing does occur.

If ${\theta}$ is not an integer, then ${\theta}= (\sqrt{n}-1)/2$ for integer $n$, so ${\theta}$ and consequently $x$ are both algebraic. We use the Gelfand-Schneider theorem, as fomulated in Theorem \[gelfond\_schneider\]: since $\theta+1/2$ is irrational and $e^{it}$ is an $n$-th root of unity (and so algebraic) it follows that $e^{it({\theta}+1/2)}$ is transcendental, a contradiction. So there are no solutions if ${\theta}$ is not an integer.

[[**Case**]{} $x = \tfrac{1 \pm \sqrt{(2{\theta}+1)(2{\theta}+3)}i}{2({\theta}+1)}$:]{} This case is similar: we find $ce^{ikt} = 1$. Thus $t = \frac{\pi(2m+1)}{k+1/2} = \tfrac{2\pi(2m+1)}{n}$, for some integer $m$. So $t({\theta}+1/2) = (2\pi m)/(2{\theta}+1)$, and $x = e^{it({\theta}+1/2)}$ or $-e^{-it({\theta}+1/2)}$ 