Some slog stuff

[sheldonison]

... Unless the functional equations no longer hold there.

Consider that slog(0)=-1, slog(1)=0
given the 2pi i periodicity, then
slog(2pi i)=-1; slog(1+2pi i)=0
slog(2npi i)=-1; slog(1+2npi i)=0 for any value of n

sexp is the inverse of slog. Therefore, somewhere on the sexp(z) Riemann surface, as you circle around the singularity at -2:
sexp(-1)=0; sexp(0)=1
sexp(-1)=2pi i; sexp(0)=1+2pi i
sexp(-1)=2npi i; sexp(0)=1+2npi i for any value of n

Well approximately yes.
Analytic continuation forces perfect periodicity to extend everywhere.

Everywhere until you hit the singularity. The singularity is at L,L*. To the left of those singularities, slog(z) is both analytic and exactly 2pi i periodic. To the right, slog is no longer 2pi i periodic. A similar 2pi i periodic would be \( \sqrt{1-\exp(z)}\; \); it is exactly 2pi i periodic in the left half of the complex plane, but not the right half.