\tfrac{1}{2{\theta}^2} + O(\tfrac{1}{{\theta}^4}), \\
t &= \pm  \tfrac{\pi}{2{\theta}} \pm \tfrac{1}{2{\theta}^3} + O(\tfrac{1}{{\theta}^5}),\\
\arg(c) & = t  \pm  \tfrac{\pi}{2} \pm \tfrac{1}{2{\theta}} + O(\tfrac{1}{{\theta}^2}), \\ 
\arg(ce^{i(2{\theta}^2-2)t}) &= \arg(c) + (2{\theta}^2-2)t \\
& = \pm  \tfrac{\pi}{2} + \tfrac{1}{{\theta}}[\pm \tfrac{1}{2} \pm 1 \pm \pi/2] + O(\tfrac{1}{{\theta}^2}).\end{aligned}$$ Meanwhile from equation , $$\arg(ce^{i(2{\theta}^2-2)t}) = \pm  \tfrac{\pi}{2} - \tfrac{1}{{\theta}} + O(\tfrac{1}{{\theta}^2}).$$ Since the two expressions are not equal, no solution exists. Thus there is no uniform mixing in the symplectic-type graphs or their complements.

Regular conference matrix type graphs
-------------------------------------

We treat the fourth family of graphs from Chan’s theorem. These have parameters $(n,k,{\lambda},\mu) = (4{\theta}^2+4{\theta}+2,2{\theta}^2+ {\theta}, {\theta}^2-1,{\theta}^2)$ and ${\theta}+ \tau = -1$.

Chan found $x = \pm i$ or $x = \tfrac{-1 \pm i\sqrt{(4{\theta}^2({\theta}+1)^2-1}}{2{\theta}({\theta}+1)}$ and in either case, the $y = \bar{x}$. The characteristic equations reduce to: $$\begin{aligned}
ce^{ikt} \sqrt{n} & = 1 + xk + y(n-k-1) \label{eqn:flatsrg4.1} \\
ce^{i{\theta}t} \sqrt{n} & = 1 + x{\theta}+ \bar{x}(-{\theta}-1) \label{eqn:flatsrg4.2} \\
ce^{i(-{\theta}-1) t} \sqrt{n} & = 1 + x(-{\theta}-1) + \bar{x}{\theta}\label{eqn:flatsrg4.3} .\end{aligned}$$ From plus we have $$\sqrt{n}c(e^{i{\theta}t} + e^{i(-{\theta}-1)t}) = 2 - 2{\mathrm{Re}}(x),$$ so $ce^{i{\theta}t} + ce^{i(-{\theta}-1)t}$ is real. Thus either $ce^{i{\theta}t} = -ce^{i (-{\theta}-1)t}$ or $ce^{i{\theta}t} = \overline{ce^{i (-{\theta}-1)t}}$. But in the first case $x = y 