$f(f(z)) = z, f' (f'(z)) = z $?

Question

Consider an analytic function $f(z)$ defined over a nonempty subset of the complex plane such that $f(f(z)) = z$.

(Also called analytic involutions)

One wonders when we also have $f'(f'(z)) = z$ where $'$ denotes the derivative with respect to $z$.

Since $f$ is its own functional inverse we can rewrite $f'(f'(z))$ (by the so called inverse function rule) as

$$f'(f'(z)) = \frac{1}{f'(f(f'(z)))}[1]$$

But it is unclear how to proceed.

Another idea is this , $w(h(z)) = h(w(z)) = z$ and

$$f(z) = h(1 - w(z))$$ $$f'(z) = h'(1- w(z))w'(z) = \frac{h'(1 - w(z))}{h'(w(z))}$$

$$f'(f(z)) = \frac{h'(1 - w(f))}{h'(w(f))}= \frac{h'(1 - (1-w(z)))}{h'(1-w(z))}=\frac{h'(w(z))}{h'(1-w(z))} = \frac{1}{f'(z)} [2]$$

But I am not getting much closer to a solution, if one exists.

Combining both $[1],[2]$ by plugging in $f'(z)$ I get

$$f'(f(f'(z)) = \frac{1}{f'(f'(z))}$$

But that is just equivalent to $[1]$.

Also $[2]$ was trivial from $f(f(z) = z$ implies $f'(f(z)) f'(z) = 1$

So I am running in circles and going nowhere with this.

How to solve such problems ??

edit

Just to be clear, I am not asking for functions $f$ that are counterxamples ! Most $f$ do not satisfy this for sure. Im looking for the ones that do have this property, if any. And how to find them or prove it.

edit2

I believe no " nice " solutions exist.

Let me explain; with nice I mean no complicated branch cuts and yet valid for all $z$. Preferably meromorphic everywhere.

Then we get apparantly that to satisfy $f(f(z)) = z$ , we must have that $f(z)$ must be well approximated by a polynomial of degree $0$ or degree $1$ in a large region of the riemann sphere.

For instance Linear Fractional Transforms approximate polynomials of degree $0$ or degree $1$ far away from their pole in a large part of the complex plane.

But this forces $f'(z)$ to be well approximated by a constant function.

And then $f'(f'(z))$ cannot be close to $z$ in most cases.

so we have an issue.

Notice that if we have an invariant $f(z) = f(g(z))$ then $f(f(z)) = z$ cannot hold for most $z$ since then we have $f(f(z)) = f(f(g(z))) = g(z)$ what is a paradox for $g(z) \neq z$.

So a degree $2$ or higher polynomial for $f$ fails, as does any other functions that makes copies of the complex plane.

And the same applies to $f'(z)$ hence another issue.

Therefore I do not believe in a nice solution.

Proving this formally, finding less nice solutions or proving nonexistance is however the goal here.

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Answer 1

Let $U$ be the domain of $f$, if you only require that $f(f(z))=z\,\forall z,f(z)\in U$ and $f'(f'(z))=z\,\forall z,f'(z)\in U$ then there's an example with the domains $V_1,V_2$ non empty but disjoint ($V_1=\{z\mid z,f(z)\in U\},V_2=\{z\mid z,f'(z)\in U\}$), otherwise if they're empty this problem is trivial. Let $U=U_1\cup U_2$ where $U_1,U_2$ are open and simply connected but disjoint from each other. Then the idea is that we make $$f(U_1)\subseteq U_1,f(U_2)\subseteq\mathbb{C}\setminus U_2;f'(U_2)\subseteq U_2,f'(U_1)\subseteq\mathbb{C}\setminus U_1$$ So $V_1\subseteq U_1$ and $V_2\subseteq U_2$, therefore $V_1\cap V_2=\emptyset$.

An example: pick $U_1=\{z\mid |z|<0.5\},U_2=\{z\mid |z-2|<0.5\}$. Then set $$f(z)=z\,\forall\,z\in U_1\implies f'(z)=1\,\forall\,z\in U_1$$ $$f'(z)=2-z\,\forall\,z\in U_2\implies f(z)=2z-\frac{z^2}2+c\,\forall\,z\in U_2$$ For some $c$, for convenience pick $c=4$. It's easy to check that $f(U_1)=U_1,f'(U_2)=U_2,f'(U_1)=\{1\}\subseteq\mathbb{C}\setminus U_1$. While $$|f(z)-2|=\left|2z-\frac{z^2}2+2\right|=\left|-\frac{(z-2)^2}2+4\right|\ge 4-\frac{0.5^2}2>3$$ $$\implies f(U_2)\subset\mathbb{C}\setminus\{z\mid |z-2|\le 3\}\subset\mathbb{C}\setminus U_2$$ Therefore $V_1=U_1,V_2=U_2$. And $$f(f(z))=z\,\forall\,z\in V_1,f'(f'(z))=2-(2-z)=z\,\forall\,z\in V_2$$

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If you require that it holds for all $z\in U$ which will make $f: U\to U,f': U\to U$ then it's trickier but still possible. It's hard to explain but I found a solution: $$f_0(z)=-\frac1{2z},f_1(z)=i\sqrt{-2z},f_2(z)=\frac{z^2}2$$ $$U_0=\left\{z\mid\left|z+i\sqrt{\frac35}\right|<\frac1{\sqrt{10}}\right\},U_1=f_0'(U_0),U_2=f_1(U_1)=\frac1{U_0}$$ $$f=\begin{cases}f_0(z)&\text{if }z\in U_0\\f_1(z)&\text{if }z\in U_1\\f_2(z)&\text{if }z\in U_2\end{cases}$$ P.S. The principal branch is chosen. Since the domain does not contain any part of the positive axis, the constructed function is therefore analytic.