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

[jaydfox]

ln(sqrt2) gives : ln sgrt2 +- 2 pi k =1/2ln2 +- 2 pi I k
ln(-sqrt 2) gives: ln sqrt 2+-pi k= 1/2ln 2 +- pi I k but k is not 0 as negative logarithms can only be complex, never real.

Not to nitpick, but rather than saying k is not 0, you should really specify that k is odd (and inequality with 0 is then already implied). I tend to write it as, e.g., \( \ln\left(\sqrt{2}\right) \pm (2k+1)\pi i \) or ln(sqrt(2)) +- (2k+1)pi I, or if you really need to factor out k and put it at the end of the expression, then ln(sqrt(2)) + pi I +- 2 pi I k