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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
Gottfried Wrote: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.

What I mean is that any time You define an arbitrary complex number u you have to consider both roots of (-1) and only later get rid of one if that does not add any information. Usually it does not, so people already initially dismiss the idea of 2 roots of sqrt(-1). Euler did not, however, and that is why he never made mistakes.

Which means that beta is 1, but sqrt(-1) is -i and +i simultaneously not because of beta, but because of 2 values of sqrt(-1) which can occur simultaneously for any given beta.

If You take beta = - 1 , You get u=a-sgrt(-1) but sqrt (-1) again taken as usual is +-i, so You have a-+i instead of a+- i as in case of b=1. Not that it matters, but it matters in your graphs, as by using +-i instead of +- beta You will get 2 values of IMAG(t) = +i and -i in the right side of Y axis, meaning 2 symmetric curves there.

Quote: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.

Could You please plot few examples on the current plot just to get the idea-it is very interesting to see the character of those branches- I am really sorry, I have only excel as a tool and even that I use for business tables not for graphing functions. So please...

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.

I think, may be wrongly, that beta in Your case corresponds to the power of e in tetration of (e^beta). If not, please correct me.

So I think if beta = pi/2 it is Ok, as h(e^pi/2) diverges and has 2 values +I , -I and ONLY then other branches, so e^pi/2 is a real value of s that corresponds to complex to real transformation as s>e^1/e. You should be able to see both I and - I at beta = pi/2.

But if beta = -pi/2 something is wrong as (e^-pi/2) = 0,20..and
(e^-pi/2)^(e^-pi/2)^(e^-pi/2) ....... converges, and is 0,47..... which means that is real to real case where beta should have been 0, not -pi/2- there should not have been such branch on a graph in this quadrant.

So for me it looks that if You just fold the graph along Y so that left side superimposes on right side You will have the right graph+ have left side free to plot convergent values of h(e^beta) where e^beta< e^1/e so beta < 1/e (including negative beta).

May be I misunderstand something still in the role of each variable, but, if You will be able to explain it to me, others will easily understandSmile

Best regards,


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, 12:06 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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