exp^[0.5](x + 2 pi i) - exp^[0.5](x)

[sheldonison]

Im thinking about exp^[0.5](x + 2 pi i) - exp^[0.5](x).

I have little time.

But some ideas.

regards

tommy1729

Hmmm can it be 2*pi*i periodic? Like exp?
I can only compute that from my c++
tet(ate(x)+0.5) ....
tet(ate(x)+1) --> exp(x) is 2*pi*i periodic, hmmm

If the cut lines are drawn to the right from the logarithmic singularity at -0.36237+/-pi i, then exp^{0.5}(z) is 2pi i periodic. In this plot, the singularity at L,L* is drawn vertically, away from the real axis, but it is so slight that it is still not really visible, even where it intersects the pi i cut lines. This graph goes from +/-10 real, +/-10 imag, with grid lines every 5 units.