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

[Ivars]

I am not sure what you mean. The method is about repeated applying of \( \log_{k,b} \) and not of \( \log_b \). You can not pull out the \( 2\pi i k \).

May be I am trying to say something useless, but I am sure, that in the region where h(z) does not converge, and may be even in the region where it does converge( e.g h(sqrt(2))=2 or 4 ) e.g h(e^pi/2) You can not limit yourself to only one root of -1, but You have to use both separately, or information is lost. So the real way to get accurate results is to study both rotations anticlokwise (i) and clockwise (-i) independently. I am not sure computer programs handle it that way. I will try once more:

Knoebel (1981) gave the following series for

h(z) = 1+ lnz + 3^2/3!*(lnz)^2 .........

Now, if we take h(e^(pi/2) we know that answers will be +i, -i.

ln(e^pi/2) = pi/2+- 2*pi*i*k -> I stress +-, k >0.

If You put this in series above( I know they do not coverge, but they might give h(z) analytic continuation), you will get :

h(z,k) = SUM (( (n^n-1)/n!) * ( pi/2+-2pi*I*k)^n-1 or
h(z,k) = SUM (((n^n-1)/n!)* pi/2^(n-1)* (1+-4*I*k)^n-1

Therefore, if we get h(z) = i when k = 1, than we must get h(z) =-i when with k =-1. If k=0, we have a special case, as we can not get i in the series if k=0.

For k=2 and -2 the values of h(z) again may be complex conjugates , but may be not.

For k=3 and -3 again the imaginary values will definitely differ by sign because values for 1 and - 1 were complex conjugates.

So basically for each pair of k +- You will get 2 values of h(z) which will be complex conjugates, but sometimes without imaginary part when they converge. Then there will seem to be only 1 value, a real number.But not all cases.

As a conjecture, perhaps h (sqrt2) will have one real value at from ln(sqrt2) and another from ln(-sqrt2).
h(cube root 2) may have 3 values, h(4th root of 2) may have 4 converging values etc.