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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#65
Gottfried Wrote:But, wait a moment. If you want I and -I in your formula for any variable x (or in my case u), what else are you doing as considering
sqrt(-1)_1 = 0 + b*I
and sqrt(-1)_2 = 0 - b*I
when you ask for considering both complex roots of -1 ?
So in my negative and positive beta I'm just doing that, what you demand; it is the negative and positive b in the above example.
So since it seems to me, that I'm already doing what you ask for, this is the source of my not-understanding of your concern.

Now I understand- and that is fine- I was just distracted by that (-pi/2) on X axis giving the same value for s=e^pi/2=4,81 (graphically it looks so ) . So that -pi/2 is just the same b=pi/2 multiplied by -1 which could be looked upon as taking the other root of (-1). OK!

Quote:Your demand for considering -1*I and +1*I as possible roots of -1 is perfectly modeled here - at least as far I can see.

Good! Now I think I will be finally able to place myself in the graph...
On other hand, then it is just a coincidence that having b= pi/2 the result for IMAG(t) = +-i coincides with h(e^pi/2) = +- i .

But please could You show a plot with few more branches of IMAG(t) on it? I am very interested what patterns You described look like now when I seem to understand where on the plot what is placedSmile

Best regards,

Ivars
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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 Ivars - 11/16/2007, 03:17 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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