I apologize, I forgot to add an additional element of algebra. I forgot that you are requiring \(f(f(z)) = z\). For that you have to go a little tricky. I'm going to assume that \(\Im(y) \neq 0\) and that \(\overline{y} = - y\).
I meant that:
\[
f(z) = \text{sexp}_K\left( \text{slog}_K(z)+ y\right)\\
\]
Is holomorphic on \(\mathbb{C} / S\), where \(S\) is a set of singularities and branches. But:
\[
f(\exp(z)) = \exp(f(z))\\
\]
To get the function you want we have to be way more clever. Sorry, my eyes must've glazed over before. I think a viable solution will require using the use of the conjugation function; So if I take:
\[
F(z) = f(\overline{f^{-1}(\overline{z})}) \neq z\\
\]
This is a holomorphic function because Sexp is conjugate symmetric. But then this is also satisfying:
\[
F(F(z)) = f(\overline{f^{-1}(\overline{f(\overline{f^{-1}(\overline{z})}))}}) =z\\
\]
Unfortunately, I think this is still just Daniel's trivial solution \(z\), just repackaged in a fancy way; I'm sure you can find more like this. Also, I may have fucked up some switch ups, lmao. This should be \(z\)... I think, lmao.
I apologize; don't think I have much to add to this.
Well considering the infinite composition method/beta method is nowhere holomorphic for \(b = e\); I'm not surprised it's nowhere equivalent to the Carlemann method. Even when you choose arbitrary solutions; the beta method is unequivalent--which was kind of the point. Though it can be holomorphic, and if it was holomorphic it was unique; Carlemann always chooses the Schroder or Kneser approach. Which is regular iteration or crescent iteration.
I apologize, long day. Sorry If I'm not making any sense. I do think the switch from upper half plane to lower half plane is the trick to this problem (when using Kneser).
I meant that:
\[
f(z) = \text{sexp}_K\left( \text{slog}_K(z)+ y\right)\\
\]
Is holomorphic on \(\mathbb{C} / S\), where \(S\) is a set of singularities and branches. But:
\[
f(\exp(z)) = \exp(f(z))\\
\]
To get the function you want we have to be way more clever. Sorry, my eyes must've glazed over before. I think a viable solution will require using the use of the conjugation function; So if I take:
\[
F(z) = f(\overline{f^{-1}(\overline{z})}) \neq z\\
\]
This is a holomorphic function because Sexp is conjugate symmetric. But then this is also satisfying:
\[
F(F(z)) = f(\overline{f^{-1}(\overline{f(\overline{f^{-1}(\overline{z})}))}}) =z\\
\]
Unfortunately, I think this is still just Daniel's trivial solution \(z\), just repackaged in a fancy way; I'm sure you can find more like this. Also, I may have fucked up some switch ups, lmao. This should be \(z\)... I think, lmao.
I apologize; don't think I have much to add to this.
Well considering the infinite composition method/beta method is nowhere holomorphic for \(b = e\); I'm not surprised it's nowhere equivalent to the Carlemann method. Even when you choose arbitrary solutions; the beta method is unequivalent--which was kind of the point. Though it can be holomorphic, and if it was holomorphic it was unique; Carlemann always chooses the Schroder or Kneser approach. Which is regular iteration or crescent iteration.
I apologize, long day. Sorry If I'm not making any sense. I do think the switch from upper half plane to lower half plane is the trick to this problem (when using Kneser).