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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#26
Ivars Wrote:so h(e^-pi/2) = -i-2pik
or h(e^-pi/2 = +i+2pik

...

Or do I make a mistake somewhere?

Yes Wink

Different branches are not necessarily apart. Though this is true for the logarithm, it is for example not true for or .
To compute (a closed form of) the power series coefficients of at say -1 depending on the branch, seems a rather laborious task.
I am not sure how Mathematica and Maple compute the Lambert W function outside the convergence radius of its standard power series.
Perhaps they dont use a power series expansion but an iteration formula.
For example the different branches of can be computed by infinitely applying the logarithm: (corresponding with the branches of the logarithm) or as we know already by infinitely applying the exponential for the range .

In this post I showed the dependency of the fixed points of from . The fixed points of are the different branches of .
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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 - 11/04/2007, 04:31 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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