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

Suppose $f:\mathbb{C}\to \mathbb{C}$ is entire.

1. If $f(f(z))=z$, find all such $f$.
2. Can we find $f$ such that $f(f(z))=z^2$?
3. How about $f(f(z))=e^z$?

Ideas: For #1, we can show that $f$ must be a bijection, since $f$ failing to be either injective or surjective quickly leads to a contradiction.

My intuition tells me that $z\mapsto -z$ should be the only such (non-identity) function.

We know $f'(z)\neq 0$ for all $z$ (or else there would be an area where $f$ were not injective). Therefore $1/f'(z)$ is an entire function.

Taking derivatives of both sides, we get $f'(f(z))f'(z) = 1$, or $f'(f(z))=1/f'(z)$.

$1/f(z)$ should be analytic everywhere, except for a simple pole at $z=f(0)$.

Let $\max_{|z|=1}|f(z)-f(0)|=R_1$, then Schwarz's lemma implies that $f'(0)\leq R_1$. Let $\max_{|z|=1}|f(z+f(0))|=R_2$, then Schwarz's lemma implies $f'(f(0))\leq R_2$. But $f'(f(0)) = 1/f'(0)$, so $$\frac1R_2\leq f'(0)\leq R_1.$$

Trying to put all this together...

Update: I've got #1. Consider the singularity of $f$ at infinity. If it is removable, then $f$ is a constant by Liouville's theorem. If it is a pole, then $f$ is a polynomial. If it is an essential singularity, then by the Big Picard theorem, $f$ cannot be injective since $f$ will take every value, except perhaps one, in a neighborhood of infinity.

Therefore $f$ is a polynomial, and since $f'\neq 0$, we have $f=az+b$ for $a\neq 0$. Now we have $a(az+b)+b = z$, giving that the involutions are $z\mapsto z$ and $z\mapsto -z+c$ for $c\in \mathbb{C}$.

Update 2: I think I've got #2 as follows: Suppose $f$ entire such that $f(f(z))=z^2$. Since $z\mapsto z^2$ is surjective, we must have $f$ surjective.

Now $2z=f'(f(z))f'(z)$. So we know that $f'(f(z))f'(z)$ is zero only at $z=0$. Either $f'(f(0))=0$ or $f'(0)=0$. Suppose the former holds and not the latter. That is impossible, since then $f(f(z))$ would have zero derivative at two points, zero and $f(0)$. So we know $f'(0)=0$.

Since $f$ is surjective, some point must map to zero. It cannot be a nonzero point $a$, because then $f(f(z))$ would have zero derivative at $a$.

This implies that zero is the unique point mapping under $f$ to zero. Now $$(z^2)'' = 2 = f''(f(z))f'(z) + f'(f(z))f''(z)$$ and the right hand side is zero at $z=0$, a contradiction.

Update 3: Here's what I've got for #3 so far:

Note $e^z$ is nowhere zero. So if $f:w\mapsto 0$ then $w\not\in \text{Rg}(f)$. But only $0$ is not in $\text{Rg}(f)$. So only zero could possibly map to zero, but it does not, because $f(f(0))\neq 0$. So $f$ is nonzero.

$f'$ is also nonzero since $$(e^z)'=e^z=f'(f(z))f'(z).$$

$f$ should have an essential singularity at infinity.

• Your intuition for the first part is in error, there are more such functions. – Daniel Fischer Dec 31 '13 at 17:55
• First thing, what are the automorphisms of $\mathbb{C}$? Once you have those, it's easy to find the involutions among them. – Daniel Fischer Dec 31 '13 at 17:59
• @danielfischer Whoops, forgot the identity. – Eric Auld Dec 31 '13 at 18:01
• There are still more. – Daniel Fischer Dec 31 '13 at 18:02
• @DanielFischer OK, I've got #1 I think – Eric Auld Dec 31 '13 at 20:46

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.

• Very good point! – Eric Auld Dec 31 '13 at 22:38
• Nice! I am trying to get a proof of non-existence for $f(f(z))=z^2$ using degree theory or homotopy. I will post if I get something. – Daniel Robert-Nicoud Jan 1 '14 at 2:19
• I wrote my own answer to problem $2$. I would like to have your opinion on its correctness, if you have the time to take a look at it. Thanks. – Daniel Robert-Nicoud Jan 1 '14 at 16:03
• If solutions to 1 are linear polynomials, then they are of the form $ax+b$. Then you have $a^2x+ab+b=x\quad\forall x$. So for $m\neq n$, you get $a^2(m-n)=m-n$ by $f(m)-f(n)$. This gives $a=\pm1$. $a=1$ gives $b=0$ and $a=-1$ gives $0=0$. So the only such functions must be $x=x$ or $-x+b\quad\forall b$.$\qquad$ – DynamoBlaze Jul 16 '18 at 19:39

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.

• I made a pdf of Kneser's 1950 paper, but zakuski will not let me ssh today. Or as in a song: There was I, waiting at the church, Waiting at the church, Waiting at the church; When I found he'd left me in the lurch, Lor, how it did upset me! All at once, he sent me round a note Here's the very note, This is what he wrote: "Can't get away to marry you today, My wife, won't let me!" – Will Jagy Jan 1 '14 at 0:30

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!)

• In 1., we have $z = a^2z + (a+1)b$, so $a^2 = 1$ and $(a+1)b = 0$. If $a = -1$, then $b$ can be arbitrary. 2. is correct. After having seen $f(1) = 1$, we could also use that then $f(0) = 0$, so $f(z) = c_kz^k+ \dotsc$ and $f(f(z)) = c_k^{k+1} z^{k^2} + \dotsc$ to obtain the contradiction. – Daniel Fischer Jan 1 '14 at 16:15
• @DanielFischer Thanks, I will correct $1$. I am onto something for $3$. Care to discuss it in chat? – Daniel Robert-Nicoud Jan 1 '14 at 16:18
• Main chat room? – Daniel Fischer Jan 1 '14 at 16:27
• Truth to tell, I created this: chat.stackexchange.com/rooms/12256/complex-analysis but if you prefer, main chat room is good for me. – Daniel Robert-Nicoud Jan 1 '14 at 16:28