Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)

[Ivars]

Infinite product-series can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...

Gottfried

Hmm. But if we have tetration of \( h(e^z) \) , then we can continue taking logarithms forever, reducing all heights? No.
Then my beloved \( h(e^{\pi/2}) = h(I^{1/I}) \) just ends where everything else:

\( \ln(\ln(\ln(....h(e^{\pi/2})) = \ln(\ln(\ln(....\pm..I))=0.318131505... \pm 1.337235701...I=-W(-1) \)(both branches), where \( W(z) \) is Lambert W function.

\( -W(-1)= \sum_{n=1}^{\infty} \frac{n^{(n-1)}} {n!} \)

So would than divergent products (and even some sums) similarly to divergent tetration end up in complex plane?

Ivars