Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#46
Ivars Wrote:So we have just limit log b log b log b log b ... log b (+-i*2pik) to be concerned with for branches in numerator. Right?
...
So we have lim k, n-> infintylog b log b log b ...... log b ( +-i*2pik/(pi/2+- i*2pi/k))

So the simplest starting value is then +-i*2pik/(pi/2+-i*2pik)?
Are these k in numerator and denominator the same or running independently ?

I am not sure what you mean. The method is about repeated applying of \( \log_{k,b} \) and not of \( \log_b \). You can not pull out the \( 2\pi i k \).
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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 - 11/12/2007, 08:23 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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