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

[Gottfried]

The plot shows this (in principle, it is far too granular!) by colors for imag(s) where t = x + I*y. Where the color is black, purely real values occur for s, because black means imag(s)~0. The real value of s is not visible from that plot, but I showed in my other graph, that it ranges

I've just got another image with a higher resolution and stronger rescaling of imag(s)
\( u = asinh(imag(s)) \hspace{24} y_{row,col} = u_{row,col} / \sqrt{max(abs(u_{,col})) } \)

The black's are still areas in the plot, and I'd like to see them as lines.
The closer the argument real(t) is at zero, the more solutions for real(s)>e^(1/e) are available, and also the absolute values real(s) increase strongly(which again cannot be seen in the graph)
The tick-marks for x and y-axes are not correct in the plot; the range is the right half of the complex half-unit-square.

I'd like to understand, whether the imaginary nulls of the function f(x)=x^(1/x) form continuous lines...

Gottfried


[update] plot removed, improved plot see next post