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

[bo198214]

Im (W) = +- ipi/2

then branches k=-1 and k=+1 are accordingly

-i(pi/2 - 2pi) = - 5/2pi i ; +i(pi/2+2pi) = + 5/2 pi*i and so on for all branches.

Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example

Code:

LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I

where the first argument denotes the branch.
The numbering of the branches is of course convention, but seems to match we the one given in your article (which seems to be a quite good article).