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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#41
hej BO,
bo198214 Wrote:

How did You find this first value, exactly- where does b come in? Is it used as base of logarithm instead of e?
so log 1,b (z) (-1) is log with base b from -1?

Do I understand correctly?

So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?
Which in turn is calculated in each step via normal logarithms by formula:
log b (z) = (ln (z) + 2pik)/ ln (b)?

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 - 11/08/2007, 02:31 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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