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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#64
Ivars Wrote:
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.

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.

Quote: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.
How comes that you think this? I say: h(x) has many solutions t_0, t_1, t_2, ... So what is the form of the different t? Then I say, t_0,t_1,t_2 must be an exponential of u_0,u_1,u_2,... . complex, multivalued like your example of sqrt(-1). Now: what is the complex form of all these t's? They must be the exponential of another number u_0,u_1,u_2,... multivalued again. Now what is the form of the u_0,u_1,u_2,..., which are complex numbers alpha_0 + beta_0*I, alpha_1 +beta_1*I, maybe each second beta is just the negative signed other beta, which would satisfy your request to consider positive and negative complex roots.

Then I find, that if I select one imaginary part beta*I, to get a real s, the real part alpha must be of a certain value. I can insert beta_0*I, -beta_0*I and so on, compute the required alpha and always get one real s.
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.

Quote: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.

My process goes three steps:
beta*I -> u -> exp(u) -> exp(u/exp(u)) = s = real(s)
and beta is not in that way related to s as you assume here. (It is not s = e^beta, for instance)

Regards -

Gottfried
Gottfried Helms, Kassel
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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 Gottfried - 11/16/2007, 02:02 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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