06/15/2014, 08:10 PM
(06/15/2014, 07:35 PM)sheldonison Wrote: how do we prove that \( \theta(z) \) has to be entire? Is there the possibility that theta(z) has singularities, but sexp(z+theta(z)) is still analytic? Those are the two issues I've struggled whenever I think about theta(z) mappings and uniqueness.
Well if sexp is analytic and z+theta(z) is not analytic we get the composition of an analytic function of a function with a singularity.
Now that has to remain a function with a singularity UNLESS MAYBE the singularities of z+theta(z) get cancelled by functional inverse.
Thus z+theta(z) must have slog singularities.
If I recall correctly the slog singularities are similar to the sexp singularities : of type ln^[m](x) , m>0.
Now locally that gives sexp(ln^[m](x)).
Now sexp(complex oo) IS a singularity as can be clearly seen from sexp(-n) , sexp(+ oo i) and sexp(+oo).
Hence sexp(ln^[m](0)) remains a singularity.
examples : exp(sqrt(x)), ln(x)^2.
regards
tommy1729
