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

[Ivars]

Thanks again,

So we choose one branch of logarithm with k=-1-> ln(-1) = i*pi and by that got to value of W(-pi/2) on the branch which gives = ip/2 and by that got value of h(e^pi/2) = - i

But, if I understand correctly ( probably not) the picture 4. in the article http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf if
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.

but h(e^pi/2) = - W(-pi/2)/ ln(e^pi/2) so

as ln(e^pi/2) = pi/2 +- 2k*pi *sqrt(-1)

h(e^pi/2) = - (+- i(p/2+-2pi*k))/ (pi/2 +-2 k*pi * i)=

= (-+i *pi/2-+i *2*k*pi) / pi/2+- i* 2 k*pi) = pi/2 (-+i-+i*4k)/ pi/2*(1+-i*4K) = (-+i-+i*4* k)/ (1+-i*4*k)

So every time we take -i - 4*k*i in upper part, we get 1+4*k*i in denominator, and vice versa. Now what would be the period of such ratio?

h(e^pi/2) = -i-4k*i/(1+4ki)

if k=0, h(e^pi/2) = -i
if k=1, h(e^pi/2) = (-i - 4i)/(1+4i) = -5i/(1+4i) = 1/ (i/5-4/5)
if k=2 h((e^pi/2)= (-i - 8i)/(1+8i) = (-9i)/(1+8i) = 1/(i/9-8/9)
if k=3 h(e^pi/2) = (-i-12i)/(1+12i) = (-13i)/(1+12i)= 1/(i/13-12/13)

if k=-1 h=(e^pi/2) = (-i+4i)/ (1-4i) = 3i/(1-4i)= 1/(1/3i-4/3)
if k=-2 h(e^pi/2) = (-i+8i)/(1-8i) = 7i/(1-8i) = 1/(1/7i-8/7)
if k=-3 h(e^pi/2) = (-i+12i)/(1-12i) = 11i/(1-12i) = 1/(1/11i-12/11)
if k=-4 h(e^pi/2) = (-i+16i)/(1-16i) = 15i /(1-16i)= 1/(1/15i)-16/15)

we can do the same calculations starting with i+4ki in upper part, 1-4ki in lower:

h(e^pi/2) = i+i 4k / (1-4ki)

if k=0, h(e^pi/2) = i
if k=1 h(e^pi/2) = (i+4i)/(1-4i) = 5i/(1-4i) = 1/(-i/5-4/5)
if k=-1 h(e^pi/2) = (i-4i)/(1+4i) = -3i/(1+4i) = 1/(-1/3i-4/3)

seems like a complex conjugates to first 2 sets of values.

I do not know about periodicity, there seems to be none , but these might be other fixed points for e^pi/2?

Or have I made a mistake again

Best regards,
And thank You once more for time spent advising me.

Ivars