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

[bo198214]

Logarithm itself has 2 branches such that log (-1) = +i *pi or - i*pi.

The logarithm has a branch for each integer \( k \), that is \( \ln(-1)=i\pi+2\pi i k=(2k+1)i\pi \) for \( k=-1 \) you get \( \ln(-1)=-i\pi \). This is because the logarithm is the inverse of the exponential and \( e^{i\pi + 2\pi i k} = -1 \).

So its not really about chosing the sign of \( i \) but about chosing the \( k \).

h(e^pi/2) = +-i +- 2pik.

Again, this is not true. If we have a fixed point \( a \) of \( b^z \) then \( a+2\pi k \) (or \( a+2\pi i k \)) is usually not again a fixed point (and hence not a branch of \( h \)).
For example \( b=e^{\frac{\pi}{2}} \) then \( b^i=i \) but \( b^{i+2\pi}=ib^{2\pi}\neq i \) also \( b^{i+2\pi i}\neq i \).

PS: the name is sqrt and not sgrt