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

[Ivars]

[quote=Gottfried]
Here is a more extended plot showing a wider range for beta.
The periodicity of alpha and s is not perfect; the same real s approximate the vertical axes at multiples of pi.

I omitted the curves for t (= a + b*i) here.
[\quote]

Thanks. Very interesting. But can You add t=a+b*i as well? It would be easier for me to place myself on this graph, when I see the exact relation with the smaller graph I hope I have understood.

Why log(log(s)) = alpha =0 at beta= pi/2+-n*pi? Is it exact relation? Would that mean that s= e there? what is IMAG( t) in these points? They seem to be roots of d^2 (log(log(s))/d(beta)^2=0 and d^2(alpha) /dbeta^2 = 0(except for n=0?) .
I rememmber on the small graph, Imag (t) had inflection points at b=pi/2 while alpha=real (t) had inflection points are integer beta=1. Does it continue, the Imag (t) and real(t) to oscillate around x axis with decreasing amplitude as beta gets bigger? How does their max amplitude decay with beta- what is the functional dependance?

Is there any information in a curve connecting points with the same slope in this graph on different branches? Where

d/(dbeta ) of (log(log(s) ) = const? ( and in d(alpha)/dbeta=const)

what is the value of alpha max in interval where beta<pi/2, so that alpha(beta=0) = ? And what is s at beta=0 ? a at beta =0 was 3, right?

Exciting

Best regards,

Ivars