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

[Ivars]

Hello, Bo

Ivars, have a look at the imaginary part of the Lambert W function (picture from Mathworld)


You see there two branches for the negative real values \( <-\frac{1}{e} \).
For one branch is \( W(-\frac{\pi}{2})=i\frac{\pi}{2} \) and for the other branch is \( W(-\frac{\pi}{2})=-i\frac{\pi}{2} \).

This is clear now, very good explanation. Especially the picures are now clear. Thanks a lot. I hope You are not tired yet
Next Item how to find power series expansion along any branch and the formulas You show in previous post I am afraid I am not ready yet to even comment, but I will take notice, and return to the series expressions of +- i.

So, depending on branch,
h(e^pi/2) will be either -i if W(-pi/2) = +ipi/2 or
h(e^pi/2) = +i W(-pi/2) = - ipi/2.


That means that even if h^(e^pi/2) can have values i+pi*k, but they can not be reachable simultaneously because as soon as we fix branch (if we are moving along it, figuratively speeking), we are doomed to change of sign for i:

so h(e^pi/2) = -i-2pik
or h(e^pi/2 = +i+2pik

Euler would have written h(e^pi/2)=sgrt(-1)+- 2pik and W(-pi/2) = sgrt(-1) *pi/2 +-2piK (which he did without mentioning +-2pik specifically because he limited himself to angle<=pi in the work Gotfried mentioned). Leaving question open.

In Eulers works, sgrt(-1) has argument: pi/2+- 2pi; +- 4pi; +- 6pi etc.

Or do I make a mistake somewhere?

Regards,

Ivars