Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))

[bo198214]

\( \log_{1,b}(-1)=6*I \)

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?

Yes, yes, yes to the last 3 questions. We defined \( \log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)} \) where \( b=e^{\frac{\pi}{2}} \) hence \( \log_{1,b}(z)=\frac{\log(z)+2\pi i}{\pi/2}=\frac{2}{\pi}\log(z)+4i \) and further \( \log_{1,b}(-1)=\frac{2}{\pi}\log(-1)+4i=\frac{2}{\pi}\pi i +4i=2i+4i=6i \).

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)?

Yes, yes