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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#31
Ivars Wrote:Im (W) = +- ipi/2

then branches k=-1 and k=+1 are accordingly

-i(pi/2 - 2pi) = - 5/2pi i ; +i(pi/2+2pi) = + 5/2 pi*i and so on for all branches.

Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example
Code:
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
where the first argument denotes the branch.
The numbering of the branches is of course convention, but seems to match we the one given in your article (which seems to be a quite good article).
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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 bo198214 - 11/06/2007, 01:34 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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