What is the minimum real value of $z_0$ for which the following equation has a non-zero solution in $z$?
$$\tan(z) = - \frac{1}{\sqrt{ \left(\dfrac{z_0}{z}\right)^2-1 } }$$
It seems that $z_0 = \frac{\pi}{2}$ is a the threshold but I am unable to prove this.
Plot both functions $y = \frac {\tan z}z$ and $y = -\frac 1{\sqrt{z_0^2-z^2}}$ (to better understand both functions' domains). Also let's consider $z>0$ domain, since both functions are even.
So you can see, that second function has vertical asymptote at $x = z_0$, which means it never goes beyond that line. Also, it's always negative. At the same time, $\tan z$ (as well as $\frac {\tan z}z$) is positive, and change its sign only when $z > \frac \pi2$, so the only way to those two function to intersect is when $y = -\frac 1{\sqrt{z_0^2-z^2}}$ function's asymptote lies beyond $\frac \pi2$.
So the answer is - $z_0$ to those two lines defined above to intersect is set $z_0 > \frac \pi2$. This set doesn't have minimum value, but $\inf z_0 = \frac \pi2$.