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

[Ivars]

I may be messing up the thread, on other hand, this must be trivial (if it is correct) -I just wanted to check:

W(2ln2)=ln2
W(3ln3)=ln3
W(4ln4)=ln4
W(nln(n))=ln n , n>=1

So h(1/4) = h((1/2)^2) = W( -ln (1/2)^2)/-ln(1/2)^2=W(2ln2)/2ln2= 1/2

h(1/27) = h((1/3)^3) = 1/3

h(1/256) = h(1/4)^4) = h((1/2)^2^2^2) = 1/4

h(1/3125)=h(1/5)^5=1/5

h(1/n^n)=1/n

Which is the same as taking n-th root of n^n.

so that :

h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series

h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.

We also have W(-ln2/2) = -ln2

So h (2^1/2) = W(-ln2/2)/-1/2ln2 = -ln2/(-1/2ln2)= 2 and simmetry ends at e^1/e

Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.

so h(x^1/x) = 1/h(1/x^x). h(1/x^x) converges for all -<1/e x<1/e so all positive integers can be produced as

n=1/(h(1/n^n))), n>=1 , fractions
1/n=h(1/n^n))

In that sense h(1/n^n) works opposite integers -when n is in denominator, h(1/n^n) is in numerator.

so ln n = Integral (1/n) = integral h((1/n^n))

we can also make:

m/n = h(1/n^n) / h(1/m^m)

practically all integer numbers and rational fractions can be substituted by h of something, also functions like ln , Especially, h(e^pi/2) = +- i .......... but i^4=1 so in the end

1= (+h(e^pi/2))^4, -1 = (+h(e^pi/2))^2

2= (+h(e^pi/2))^4+ (h(e^pi/2))^4

So even i can be replaced by h(e^pi/2) so that

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

My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works?

You can take any infinite series and replace integer n with 1/h(1/n^n) for a starter and see what happens.