Fixed point question for branches

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Let $A(z)$ be a real-entire function.

$$ A(z) = a_0 + a_1 z + a_2 z^2 + \cdots $$

Also for real $x$ we have

$$ A’(x) > 0 .$$

Let $f$ be the functional inverse: $ f(A(z)) = A(f(z)) = z $. More specific $f$ is the main branch inverse.

An example could be $A(z) = \exp(z) $ and $f(z) = \ln(z) $ (not $ \ln(z) + 2 \pi i $ !!).

Ok that was just an example.

So now the question.

Consider the (fixed point) equation

$$ f(z) = z .$$

I'm looking for examples where there are NO solutions for any $z$ on the Riemann Sphere (so complex or complex infinity) or a proof that none exist.

We know for instance that there is always a solution to $ \ln(z) = z \ln(b) $ (the logarithm with base $b$ always has a fixed point). So that is NOT an example.

I also wonder about a description for all these without a solution.

I thought about Riemann surfaces intersecting with the identity function but without result.

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  • There is no such thing as "the main branch".

  • Let $g(z)=A(z)-z$. If $g(c) =0$ then let $n$ be the order of the zero of $A(z)-c$ at $c$, there is a branch of $f$ such that $f(z)^n$ is analytic and $f(A(z))=z$ near $c$ $$A(c)=c\implies c=f(c)$$

  • That $A'(x)>0$ for $x\in\Bbb{R}$ means that there is a branch of $f$ analytic on $A(\Bbb{R})$ such that $f(A(x))=x$. But there is no canonical way to continue analytically this function. It will have some branch points at every $s$ such that $A(z_1)=A(z_2)=s$, moreover the branch points may move and appear disappear when rotating around another one, the set of all such branch points doesn't need to be isolated. But locally it is.

  • Iff $g$ doesn't vanish then $g(z)=\exp(h(z))$ and $A(z)=\exp(h(z))+z$ where $h$ is entire. In that case whatever the branch $f(z)-z$ doesn't vanish.

  • All the branches of $f$ are the analytic continuations of the same function, analytic continuations along different curves.