Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))

[bo198214]

However, as e^-pi/2 > e^-e,

h(e^-pi/2) = 0,474541.......= (2/pi)*W(pi/2)

So there is no symmetry and i will not be a result of analytic continuation of infinite tetration of anything, while -i is.

This depends on how you analytically continue h, on which path.
W (and hence h) has a singularity at \( -\frac{1}{e} \), so if you continue on different paths around \( -\frac{1}{e} \), you possibly get different results (similar to the analytic continuation of the logarithm with singularity at 0).
Surely on another path/branch
\( h(e^{-\pi/2})=i \).

Edit: Perhaps its not at all "Surely", was just my conjecture.
Edit2: Now its not a conjecture anymore but regarded as nonsense XD, instead \( h(e^{\pi/2})=i \) on another branch.