$g(g(f(z)))=z$ and $\prod\limits_{f(x)=0}^{} g(x)=1$.

Question

My friend recently gave me this system of functional equations, asking me if I could find holomorphic $f,g: \mathbb{C} \to \mathbb{C}$ satisfying:

To which I promptly said, “no.” Thoughts? Hints?

How do you define the product? It may be an uncountable product. — Rushabh Mehta, Nov 28 '19 at 22:18
@DonThousand hmm, didn’t catch that—let’s say $f$ has finitely many zeros. If anyone wants to ignore that restriction and get creative, more power to them. — Descartes Before the Horse, Nov 28 '19 at 22:19

score 11 · Accepted Answer · edited Nov 28 '19 at 23:15

In general, given any two functions $p$ and $q$. If $p \circ q$ is injective, so does $q$.

Since the identity map $z \mapsto z$ is injective. The condition $g(g(f(z))) = z$ implies $f$ is injective. Since $f$ is holomorphic, $f(z)$ is a linear function, i.e. $f(z) = az + b$ with $a \ne 0$.

Now $g\circ g = f^{-1}$ is injective, so $g$ is also injective. This force $g(z)$ to be a linear function too, i.e. $g(z) = cz + d$ with $c \ne 0$.

Substitute this into the condition $g(g(f(z)) = z$, we can express $f(z)$ in terms of $c,d$.

$$f(z) = \frac{x}{c^2} - \frac{(c+1)d}{c^2}$$

$f(z)$ has a unique root at $z = (c+1)d$, In order for $\prod\limits_{f(z) = 0}g(z) = 1$, this forces $d = \frac{1}{c^2+c+1}$.

As a result,

$$f(z) = \frac{x}{c^2}-\frac{c+1}{c^2(c^2+c+1)},\quad g(z) = cx + \frac{1}{c^2+c+1}$$ is one family of solutions for the problem.

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