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

Gottfried Wrote:Ivars -

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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