slog_b(sexp_b(z)) How does it look like ?

[sheldonison]

\( \text{slog}(\text{sexp}(z))=z+\theta_n(z)+n\text{period} \), where theta is a 1-cyclic function, which goes to zero as \( \Im(z) \) goes to infinity, and theta is uniquely determined by n.
- Sheldon

Second :

Why L as ' invariant ' ? Why not 2pi i ?

afterall iterations of exp(z) + 2pi i seems to suggest branches that differ by 2 pi i !?

Tommy,

There's a lot of misconceptions in your post; you might want to go back and look at the threads where we went over the Kneser construction, which starts with the formal Schroeder equation. When you use the formal Schroeder equation to generate f^n, you get a solution that has a period of \( \frac{2\pi i}{\ln(\lambda)} \). For base e, the period~=4.447+1.058i. A case you are familiar with is iterating 2sinh, where \( \lambda=2; \; \lambda \) is the derivative of f(L+z) at the fixed point, where f(L)=L. For \( f^{oz} \) for iterating \( f=b^z, \; \lambda=\ln(L) \). For base e, L~=0.318+1.337i. For base e, \( \lambda=\ln(L)=L \), though for base bases lambda<>L.

For the inverse Abel function generated from the formal Schroeder function, then iterating \( f^{o z} = L+ S(\lambda^z) \approx L+\lambda^z \), where \( S=x+a_2 x^2 + a_3 x^3... \) is the formal inverse Schroeder function. For our case for b>e^(1/e), \( f(z)=b^z \), and f^z approaches L as imag(z) increases, and as real(z) decreases. For our case, f^z is not real valued at the real axis.

To get a real valued function, we do a Riemann/theta mapping to generate sexp(z). Because theta(z) very quickly decays to a constant as imag(z) increases, you can see the pseudo periodicity for sexp(z) as imag(z) increases in Dimitrii's complex plane plots, or my plot of the inverse Abel function and sexp. This link/picture also tries to show the incredible complexity of the singularity at z=0 for Abel(sexp(z)), for Kneser's sexp(z) solution, which Tommy alludes to in some of his comments. This last plot also shows the contours of theta(z)+z, which would be similar for slog(sexp(z)), n>0; see the n=1 period graph below.

The program I have posted to calculate Kneser's sexp(z) solution starts with this inverse Abel function, and converges to the inverse Abel function as imag(z) increases. Finally, here is the contour graph for n=1, slog(sexp(-1.5+i1E-6..0.5+i1E-6))=z+theta1(z)+period, where the solution is 1-period away from the origin, for base e, for z just above the real axis, where \( \Im(z)=i10^{-6} \). As you can see, the function is not "almost always flat", and instead resembles the theta(z) mapping for Kneser's solution, just as I predicted.
   
- hope this helps, Sheldon Levenstein