11/14/2007, 08:57 PM

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