[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

[tommy1729]

This reduces into a problem with riemann mappings.

We can write tommy's equation as:

\[
f(g(f^{-1}(z)) = g(z)\\
\]

These equations are only solvable if \(g : D \to D\). Then we need that \(f : D \to D\). The only domain \(g = \exp\) takes to itself is \(\mathbb{C}\). Trying to create conjugations on the entire complex plane is a null effort. There exists no conjugations in this case. The cases where you can create conjugations on all of \(\mathbb{C}\), always require a smaller domain \(D\) that satisfies this.

If \(D\) were simply connected and biholomorphic to the unit disk; then we can write the solutions for:

\[
f(g(f^{-1}(z))) = h(z)\\
\]

Very simply for all \(f,g,h\). But as \(\exp(z)\) only fixes the complex plane, this isn't possible; unless you introduce a lot of branching talk.

So, Daniel is right in saying that \(z \mapsto z\) is the trivial solution, but it's also the only solution on \(\mathbb{C}\). To get a different solution, we can take something like Kneser's:

\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\
\]

But this is not holomorphic on \(\mathbb{C}\). And we get a chicken and the egg situation... Do we solve tetration, or do we solve this conjugation first?

I'm not sure how helpful it is to look at this first; rather than looking at iterates which appear more natural; but I could be wrong...

Regards, James

Minor silly comment : exp maps C to C \ {0} 

But yeah that may be an issue.

also quote :

\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\
\]

end quote

I think you meant

\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + (-1)^{1/2})\\
\]

\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + i)\\
\]

or maybe

\[
f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2 \pi i)\\
\]

so complex ? right ?

afterall f(f(z)) = z 

and the periodicity in the complex direction would make f and exp commute potentially.

 
Or maybe im missing something.
( my mistake )


regards

tommy1729