12/14/2009, 07:18 AM

Using the simplest conjugation function produces something of a mess; to simplify, I'll consider the conjugate F of exp(pi/2 x) instead; the fixed points are +/-I, so we can conjugate with w=z^2 + 1. The inverse function is z=sqrt(w-1); then I get: F(x) = exp(pi/2 * sqrt(x-1))^2 + 1 = exp(pi * sqrt(x-1)) + 1.

Note that this function has a branch point at 1, but the conjugation function moves [0, oo] to [1, oo], so we must have two different functions near zero, one for each choice of branch cut. They will presumably yield different regular iteration functions, which correspond to the two entire regular iteration functions for exp.

Note that this function has a branch point at 1, but the conjugation function moves [0, oo] to [1, oo], so we must have two different functions near zero, one for each choice of branch cut. They will presumably yield different regular iteration functions, which correspond to the two entire regular iteration functions for exp.