Let $a, b \in \mathbb C$, $x,\theta \in \mathbb R$, $x>0$
I'm looking for a (strict, if possible) condition on $t > 0$ such that the number $|\xi(t)| = |1 - \frac{2at}{x^2}\sin^2(\frac{\theta}{2})+bt|$ will be no greater than $1$ for all $\theta$.
What I tried:
Writing $a = a_R + ia_I, b = b_R + ib_I$ we thus have
$|\xi(t)|^2 = |1 - \frac{2(a_R + ia_I)t}{x^2}\sin^2(\frac{\theta}{2})+(b_R + ib_I)t|^2 = |(1-\frac{2a_Rt}{x^2}\sin^2(\frac{\theta}{2})+b_Rt)+i(-\frac{2a_It}{x^2}\sin^2(\frac{\theta}{2})+b_It)|^2=(1-\frac{2a_Rt}{x^2}\sin^2(\frac{\theta}{2})+b_Rt)^2+(-\frac{2a_It}{x^2}\sin^2(\frac{\theta}{2})+b_It)^2 = \dots$
I got to a very ugly term with $t$ and $t^2$ and there's no obvious way to proceed.
When $\theta\in\Bbb R$, the range $R$ of $\zeta(t)$ is a segment with the endpoints $1+bt$ and $1 - \frac{2at}{x^2}+bt$. Since the unit disk $D$ is convex, the segment $R$ is contained in $D$ iff its endpoints are contained in $D$, that is iff $|1+bt|\le 1$ and $|1 - \frac{2at}{x^2}+bt|\le 1$.