zero's of exp^[1/2](x) ?

[sheldonison]

.... The cutpoint for L is not even visible at 0.318+1.34i, and only becomes visible around -1+1.34i. This gives somewhat of an idea as to how mild the singularity for the half iterate at L is, with a magnitude of a little less than 1 part in 100,000 in the immediate vicinity of L itself.
- Sheldon

I'm fascinated by the "mild" singularity of exp^[1/2](L). L is the fixed point for base e, or ~0.318+1.337*I. It seems that the limiting value for exp^[1/2](L)=L from all directions in the complex plane. The branch singularity discontinuity goes to zero in the vicinity of L. Here is a parametric contour graph, where the inner red circle is a circular radius L+0.5*exp(i*z). The inner green circle is a radius of 1. The middle red circle is a radius of 1.5, and the outer green circle is a radius of 2. The outer red circle is a radius of 2.5. The branch cut-point jump is visible in the graph at radius 1.5, 2, and 2.5. The starting point for the circle is L-radius, going counter clockwise, which matches Mike's exp^[1/2] graph cutpoints.
   
This chart plots the absolute value of the cutpoint branch delta, after circling around the singularity, from radius 0.5 to radius 2.5. At a radius of 0.5, the singularity has a magnitude of 0.00038, which is pretty small. We are comparing exp^[1/2](L+r*exp(-Pi*i)) to exp^[1/2](L+r*exp(+Pi*i)), a difference of one rotation around the singularity.
   
This chart plots the absolute value of the singularity, from radius 0.02 to radius 0.5. At a radius of 0.02, the singularity has a magnitude of 1.4E-11, which is very small. So, at the singularity itself, it is likely that exp^[1/2](L)=L, from all directions. Mathmetically, circling counterclockwise around the singularity corresponds to adding approximately one Period to the slog. slog(L-r) becomes slog(L-r)+Period, where Period is the Period of the superfunction before the Riemann mapping.
- Sheldon