# A transcendental equation

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.

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Are you sure about min and not inf? – Kaster Feb 25 '13 at 21:11
@Kaster I haven't been very careful about "min" vs "inf". I would like to know what is the answer, however accurately it can be given. – user6818 Feb 25 '13 at 21:37
Answer is $z_0 > \frac \pi 2$, but this set doesn't have min. It has inf though. As for the difference, min is something that belongs to the set, i.e. reachable at some point, whereas inf not necessaryly, it can be outside of the set. Just like this interval -- it doesn't have min, because whatever value you choose as min you'll be able to find the value from the same interval which is less than that so called min. But inf, it is least possible number $m$ s.t. $m \le z, \forall z \in Z$. – Kaster Feb 25 '13 at 22:00

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 $\inf z_0 = \frac \pi2$.