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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#38
It is not that difficult to compute the branches of .
The branches of are simply the fixed points of . In this way you can compute the branches via
, where .
This is however valid only for as the numbering of the branches in this interval may be a bit different. For the branch 0 and -1 (which are non-real) join and for slightly smaller it splits again into two real branches.
The negative branches are simply the conjugates of the positive branches:.

Edit:
As it is so easy to compute it is perhaps more appropriate to compute from by instead of computing from .
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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 - 11/07/2007, 01:48 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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