Qs on extension of continuous iterations from analytic functs to non-analytic

[tommy1729]

Idk if it's time to talk about continuous iterations for non-analytic functions that can still be defined on the whole complex plane, but these ideas did disastrously harm to my brain lol.

I worked out some cases last year, these are my results.

Consider any function g(x):R->R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as:
\[f(z)=ze^{ig(\|z\|)}\]
Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps:
\[\text{For real t, }f^t(z)=ze^{itg(\|z\|)}\]
Remember abs(e^(it))=1 for real t,
\[\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))\]
it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's.

Let h(z) be analytic with fixpoint 1.

g(z) = norm(z)
and h ' (1) = r * e^i

then 

h^[t](z) = h^[n] [ r^t f^[t] (h^[-n](z)) ] 

is that an analytic solution ? seems just the same as the regular fixpoint method ( koenings function ).

the idea is that the continu map is asymptotic to a linear map near the fixpoint 1 and that should be sufficient to make the whole formula in the limit analytic.

maybe a dumb question sorry.

And not suggesting to actually apply such formula's.

just for theory...

if not analytic somehow , complex-continu solutions to tetration ?? hmm 
been a while since we talked about that ...

 the disctinction between non-analytic C^oo and complex-continu fascinates me.

regards

tommy1729