Find all entire $f$ such that $f(f(z))=z$.

Solution 1:

A general observation regarding the first two points:

If $f$ is an entire function such that $f(f(z)) = P(z)$ is a polynomial, then $f$ itself must be a polynomial, and hence $\deg P = \deg (f\circ f) = (\deg f)^2$ must be a square. That follows by Casorati-Weierstraß, for if $f$ had an essential singularity in $\infty$, there would be a sequence $z_n \to \infty$ with $f(z_n) \to 0$, and hence $f(f(z_n)) \to f(0) \neq \infty$.

That immediately rules out solutions to 2., and yields that solutions to 1. must be polynomials of degree $1$.

For part 3., if such an $f$ exists, it must be transcendental (otherwise $f\circ f$ would be a polynomial), so attain every $w \in \mathbb{C}$ with one exception infinitely often. Thus $\mathbb{C}\setminus \{0\} = f(f(\mathbb{C})) = f(\mathbb{C})$, and $f$ omits the value $0$. Since $\mathbb{C}$ is simply connected, $f$ has a logarithm, $f(z) = e^{g(z)}$. Now $f(f(z)) = e^z$ becomes

$$\exp \left(g(f(z))\right) = \exp(z) \iff \exp\left(g(f(z))-z\right)\equiv 1,$$

so $g(f(z)) - z \equiv 2\pi ik$ is constant, choosing $g$ accordingly, we have $g(f(z)) = z$. Hence $f$ is injective, which contradicts its transcendentality.

Solution 2:

Hellmuth Kneser found a solution to $f(f(x)) = e^x$ along the real line, and, since analytic, in a strip of varying width around the real axis in $\mathbb C.$

It does not extend to the entire plane. The obstruction in these problems is always the fixed points, $e^z = z.$ There are not any real fixpoints, but a countably infinite sequence in the plane, fairly easy to approximate. See SITE .

Evidently a full answer at https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root to part 3.

On part 2, for the real part of $z$ strictly positive, we can take the principal branch of logarithm and so define $f(z) = z^{\sqrt 2}.$ And $f(f(z)) = z^2.$ I think I am remembering that from a Robert Israel answer somewhere. As you are finding, this does not extend to the entire plane.

Solution 3:

Problem 1: As hinted by @DanielFisher, the automorphisms of the complex plane are the functions $f(z) = az+b$ with $a\ne0$. The condition $f(f(z))=z$ is equivalent to $a=1$ and $b=0$ or $a=-1$ and $b\in\mathbb{C}$ any complex number.

Problem 2: Notice that we must have $$f(z^2)=f(f(f(z)))=(f(z))^2$$ and thus $f(1)$ is either $0$ or $1$. Consider the set $A=\{z\in\mathbb{C}|\exists n\in\mathbb{N}:z^{2n}=1\}\subset S^1$ and is not discrete (consider the sequence $z_k=e^{\frac{\pi}{k} i}$, which converges to $1$). The condition above implies that $|f(z)|=|f(1)|$ for all $z\in A$. If $f(1)=0$, then the identity theorem implies that $f=0$, which is a contradiction. Thus we must have $f(1)=1$ and thus $f$ maps $S^1$ to itself. Consider the restriction of $f$ to $S^1$ and look at the degree. We have $\deg(f)^2=\deg(f\circ f)=\deg(z^2)=2$, which is impossible. Thus such a function $f$ cannot exist.

Problem 3: It is easy to see that $0$ is not contained in the image of $f$, and that we can look at $f$ as a surjective map $f:\mathbb{C}\rightarrow\mathbb{C}^*$.

Claim: $f:\mathbb{C}\to\mathbb{C}^*$ is a covering map.

Proof: $f\circ f=\exp$ is a covering map. Let $z_0\in\mathbb{C}^*$, then we have some open neighborhood $V$ of $z_0$ which is evenly covered by $\exp$, i.e. $\exp^{-1}(V)=\bigsqcup_{i\in I}U_i$ and $\exp:U_i\to V$ is a homeomorphism for all $i$. Since $f$ is holomorphic and thus an open map (by the open mapping theorem), and $f|_{U_i}$ is injective (else $f\circ f$ couldn't be injective), $f|_{U_i}$ is a homeomorphism onto its image. Now $$f^{-1}(V) = f\left(\bigsqcup_{i\in I}U_i\right) = \bigcup_{i\in I}f(U_i)$$ All those sets are open, we are left to show that they are disjoint. Assume there exist $i,j\in I$ such that $f(U_i)\cap f(U_j)\ne\emptyset$ and $f(U_i)\ne f(U_j)$. Let $B_{ij}=f(U_i)\cup f(U_j)$, then $f(B_{ij})=V$, which is connected (or it can be chosen so). Consider $F=f|_{B_{ij}}$, then $$X_i=\{z\in V|F^{-1}(z) \mathrm{\ contains\ } i \mathrm{\ elements}\}$$ for $i=1,2$ are disjoint and cover $V$, thus one of them must be empty. This, together with the fact that $f(f(U_i))=V$, is a contradiction. Thus $f$ is a covering map.

Assuming the claim above is true, we have that both $f$ and $\exp$ are universal coverings $\mathbb{C}\to\mathbb{C}^*$. Thus there must exist an automorphism $h:\mathbb{C}\to\mathbb{C}$ such that $f(z)=\exp(h(z))$. But as seen in problem 1, we must have $h(z)=az+b$ (with $a\ne0$), and thus $f(z)=e^{az+b}$, which is impossible. We conclude that there is no holomorphic function such that $f(f(z))=e^z$.

(My thanks to @DanielFischer for the great help furnished with this solution to the third problem!)