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

Ivars Wrote:Hej Gotfried,

Gottfried Wrote: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.


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

Quote: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 Helms, Kassel

Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/15/2007, 06:16 AM
RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM
RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM
RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM

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