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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
Gottfried Wrote:The plot shows this (in principle, it is far too granular!) by colors for imag(s) where t = x + I*y. Where the color is black, purely real values occur for s, because black means imag(s)~0. The real value of s is not visible from that plot, but I showed in my other graph, that it ranges
I've just got another image with a higher resolution and stronger rescaling of imag(s)

The black's are still areas in the plot, and I'd like to see them as lines.
The closer the argument real(t) is at zero, the more solutions for real(s)>e^(1/e) are available, and also the absolute values real(s) increase strongly(which again cannot be seen in the graph)
The tick-marks for x and y-axes are not correct in the plot; the range is the right half of the complex half-unit-square.

I'd like to understand, whether the imaginary nulls of the function f(x)=x^(1/x) form continuous lines...


[update] plot removed, improved plot see next post
Gottfried Helms, Kassel

Messages In This Thread
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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