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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
Ivars Wrote:
Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis).
So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.


I do not take +- beta, I take +- I . beta is still 0<beta<pi.

But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 .
And then see, what value the real part must have such that the resulting s is purely real.
And while the graph shows only u = alpha -pi*I .. u=alpha + pi*I there is no problem to extend the graph to u = alpha + (beta + 2*k*pi)*I - there is a rough periodicity with a distortion then. Just plot it in other ranges.

Quote:In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant.

The question is, can we even use beta = - pi/2 as that would imply

Yes, just look in the 3'rd quadrant; the required value for alpha is on the blue curve, and the resulting values for real and imag of t are on the magenta curves, and the resulting value for s is on the green curve.

Quote:h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i.

??? hmmm....
Regards -

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 - 11/16/2007, 09:09 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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