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

[sheldonison]

As to your original question,

I found a much simpler way of thinking about it. For some cases, we conjecture that
\( f(z) =\text{slog}(\text{sexp}(z)) = z + \theta(z) \), where theta(z) is a well defined 1-cyclic function as imag(z) increases, but has singularities at the real axis.

So, z2=slog(sexp(z1)). Now look at the sequence of points. If these are all equal to zero, then we can conjecture that we have a 1-cyclic solution, since sexp(z) goes to the fixed point of L as real(z) goes to minus infinity.
sexp(z1-1)-sexp(z2-1)
sexp(z1-2)-sexp(z2-2)
....
sexp(z1-n)-sexp(z2-n)

If they're not all equal, than there will be some lowest "n", for which \( \text{sexp}(z1-n) + 2m\pi i = \text{sexp}(z2-n) \)

At the point z1-n, we solve the function
\( f(z) =\text{slog}(\text{sexp}(z)+2m\pi i) \), which looks nothing like the 1-cyclic solution, due to the mismatching branches. The solution is only well behaved locally, near \( f(z) =\text{slog}(\text{sexp}(z)+2m\pi i) \), where f(z) doesn't vary much at all as z increases to some medium sized imaginary number, and then z increases by a Period, all while f(z) hardly changes at all. This is clearly a very different function mostly uninteresting solution, than \( f(z) =\text{slog}(\text{sexp}(z)) \approx z + \theta(z) \).
- Sheldon