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

[Ivars]

hej bo198214

-i = power tower (e^pi/2) = h (e^pi/h2)

But be aware that this is mathematically wrong!
Correct is:
power tower (e^(pi/2)) = oo

If you mean something different for example an infinite valued function or a fixed point of x^x then you have to explicitely state that.

Thanks, I understood that ; I will try more carefully:

Analytic continuation of h(z) defined as = -W(-ln(z))/ln(z) at point z=e^(pi/2) is - i.
also, in this case:

h(e^pi/2)^i = e^(pi/2)
h(e^pi/2)^-i= e^(-pi/2) as

-i^i=e^(pi/2) = 4,810477.. -i^-i = e^(-pi/2)=0,207879
i^i = e^(-pi/2)= 0,207879.. i^-i = e^(pi/2) = 4,810477

and

i*ln h(e^pi/2) = pi/2

ln h(e^pi/2) = -i pi/2 = pi/2*(h(e^pi/2)

(h(e^pi/2))^2 = -i^2 = e^-ipi = i^6

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.

Best regards,

Ivars