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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#54
Ivars Wrote:Hej Gottfried,

Could there be another symmetrical curve for b=imag(t) ? So that at point 0;0 You will have 2 curves crossing each other?

Then You could take away the left 2 quadrants of the graph totally- I can not understand why they are needed as it seems to me - maybe wrongly- that there beta=-pi/2 also belongs to h(e^pi/2) which can not be true?

Ivars

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
Gottfried Helms, Kassel
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Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/14/2007, 08:57 PM
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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