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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#68
Ivars Wrote:Why log(log(s)) = alpha =0 at beta= pi/2+-n*pi? Is it exact relation? Would that mean that s= e there? what is IMAG( t) in these points? Ivars

It seems that if ln(ln(s))=0 than s=1/e (based on Eulers idea that ln(-1) = ln(1) + imaginary periodical branches = 0 + imaginary branches) or s= e. s=e is not solvable.

If s= 1/e, than s=exp(u/t) = 1/e -> u/t= -1
if u/t = -1 than u/exp(u) = -1 -> exp(u)/u = -1; epx(u) = -u

u = ln (-u) or u = - Omega = - 0,5671432........

so that -Omega= ln (Omega) , which is true.

but if u= - Omega = alpha+ i beta, than at points beta= npi/2

alpha= -Omega, Beta = npi/2 , i= +-i, or in this graph:

slog(slog(s)=0, s= 1/e, u= - Omega+- I*n*Pi/2

than t= exp((-Omega)+-I*n*pi/2) = exp(-omega)*exp(+-I*n*pi/2) = Omega*exp (+-I*n*pi/2) = Omega*(+-(I^n)) = +-I^n*Omega.

s= exp ((-Omega+-I*n*pi/2)/+-I^n*Omega) = exp(-+1/I^n+- I*n*pi/2/I^n*Omega) has values:

n=0

s= exp (-+1 +-0) = 1/e, e. This by the way implies that
h(1/e)= Omega
h(e) = - Omega

n=1 s complex

s= exp ( -+1/I +- pi/(2*Omega)) = exp(+-i+-pi/(2*Omega) =exp(+-i) exp(+-pi/2Omega) = exp(+-i) *15,95.., exp(+-i) *0,062682

exp(+-i) = i^(+-2/pi) = i^0,63662

n=2 s complex

s=exp(+-1+-2I*pi/(2*Omega))= exp(+-1+-I*pi/Omega) = exp(+-1)*exp(+-I* pi/Omega) = exp(+-1) *i^(+-2/Omega)

n=3 s Complex

s=exp((-+I+- 3pi/2*Omega)) = exp (-+I)*exp(+-3pi/(2*Omega))

n=4 s complex
s= exp(+-1+-4pi I/2*Omega) = exp(+-1)*exp (+-4pi/2Omega) = exp(+-1)*i^(+-4/Omega)

etc.

Or do I make a mistake somewhereSad

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 - 11/20/2007, 10:11 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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