Real roots plot of the modified bessel function Could anyone point me a program so i can calculate the roots of
$$ K_{ia}(2 \pi)=0 $$
here $ K_{ia}(x) $ is the modified Bessel function of second kind with (pure complex)index 'k' :D
My conjecture of exponential potential means that the solutions are $ s=2a $ with 
$$ \zeta (1/2+is)=0. $$

I can quantify somewhat Raymond's suggestion that the zeros of $K_{ia}(2\pi)$ are much more regular than the zeros of $\zeta(1/2+i2a)$.  The calculations below are rough and I didn't verify the details, so this is perhaps more of a comment than an answer.
The Bessel function in question has the integral representation
$$
K_{ia}(2\pi) = \int_0^\infty e^{-2\pi \cosh t} \cos(at)\,dt.
$$
Naively applying the saddle point method to this integral indicates that, for large $a$,
$$
\begin{align}
K_{ia}(2\pi) &\approx \frac{e^{-a\pi/2} \sqrt{\pi}}{(a^2-4\pi^2)^{1/4}} \Bigl( \sin f(a) + \cos f(a) \Bigr) \\
&=\frac{e^{-a\pi/2} \sqrt{2\pi}}{(a^2-4\pi^2)^{1/4}} \cos\left( f(a) - \frac{\pi}{4} \right),
\end{align}
$$
where
$$
f(a) = \frac{a}{2} \log\left(\frac{a^2-2\pi^2+a\sqrt{a^2-4\pi^2}}{2\pi^2}\right) - \sqrt{a^2-4\pi^2}.
$$
Here's a plot of $K_{ia}(2\pi)$ in blue versus this approximation in red (both scaled by a factor of $e^{a\pi/2}\sqrt{a}$ as in Raymond's graph).

The approximation has zeros whenever
$$
f(a) = \frac{3\pi}{4} + n\pi, \tag{$*$}
$$
and since
$$
f(a) \approx a\log(a/\pi) - a
$$
for large $a$ we expect that solutions of $(*)$ for large $n$ satisfy
$$
a \approx \frac{3\pi/4+n\pi}{W\Bigl((3\pi/4+n\pi)/e\Bigr)},
$$
where $W$ is the Lambert W function.
Here's a plot of $e^{a\pi/2} \sqrt{a} K_{ia}(2\pi)$ in blue with these approximate zeros in red.

