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

[Ivars]

hej Bo,

Thanks again. But: if you calculate W (-pi/2) via iteration of logarithm, it will have the same 2pik difference between different branches at point -pi/2.

Logarithm itself has 2 branches such that log (-1) = +i *pi or - i*pi. So we are in the same position again- if we choose one branch, we end up with W(-pi/2) = +ipi/2, and h(e^pi/2) = - i ; if another, W(-pi/2) = - ipi/2 and h(e^pi/2) = + i , but it does not mean h(e^pi/2) = +- i at the same time.

Which means that value of h(e^pi/2) depends on choice we make about usage of +i or -i on the way.

Now e^pi/2 is real, positive and has one value. The principal value of h(e^pi/2) is -i or i depending on our choice of +i or - i earlier in the process of calculations, so it is dependant on the direction we choose to go around singularity -1/e in W or negative numbers in x= ln (y/x).

Now if the value of power tower is dependant on the choice of +i or - i along the way, we have to keep for further calculations values:

h(e^pi/2) = +-i +- 2pik. This look very similar to situation with ln(-1) = +-i*pi+- 2pik but the order is different.If we take ln(-1)=+ip, we get h(e^pi/2) = - i

So ln(-1) = -(pi*h(e^pi/2) -+2pik).

I do not see any other way to differentiate between sgrt(-1) and h(e^pi/2) except that they seem to have opposite order of signs -if we choose +i somewhere, we get - h(e^pi/2) and if we choose - i we get +h(e^pi/2), or one could write

h(e^pi/2)= -(sgrt(-1) +2pik)

A mistake again?