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

[Gottfried]

Hi Ivars -

Hej Gotfried,

Hmm; I don't know absolutely. See my derivations in the article
function; it seems pretty straightforward and I don't think, there are other solutions - but...

t is t=exp(u); so even if we use periodical u (including multiple 2*Pi*I*k) there is only one t according to my considerations. Don't know, whether there are other solutions possible - but if you should see another solution, I'd like to know this.

Gottfried

Why not:

Define complex number u Euler style , as
u=a+sqrt(-1)b so u= a+-ib where +i and -i must be used separately but together- they are both in sqrt(-1) always. And stick to it throughout the derivations.

[b]Then red point coordinates will be:

u= 0+-ipi/2

so this describes the curves in the 2'nd and third quadrant: if I use beta or -beta. So this is dealt with.

t=0+-i

??? t = exp(u) (whatever u) and this is unique, since exp is multiple->unique function (no two values/branches for function-result)

s=e^pi/2

Yes. Since you have selected one beta (+ or -) you have t and s uniquely determined.
My only problem was, whether possibly there is another way/formula to force the result s being purely real - I don't know, whether my considerations are exhaustive here.

And that will correspond to h(e^pi/2) having values +i, -i in at least in first 2 branches. So on right quadrants will have 2 curves instead of 1. May be symmetric, may be not- it has to be calculated according to Your methods. On left quadrants I am not sure as I can not really understand what values in left quadrants correspond to. .

Best seems to me, you would compute some examples, show this/plot this and nail down the actual problem. I'm still feeling insure whether I'm getting your caveats correctly.

Gottfried