When is $a(z) = b(c(z)) $?

Question

Let $a(z)$ be a given transcendental entire function.

When is $a(z)=b(c(z))$ where $b,c$ are also transcendental entire functions ?

How to find such $b,c$ ?

In particular when $a$ is given by a closed form for the n'th derivative or as An integral transform or product ( Weierstrass ).

Im aware of the connections to invariants and carleman matrices , but despite that Im stuck.

Answer 1

Well, I can give some examples where such a representation does not exist. Consider $a(z) = \exp(z)$. Using Picard's theorem, $b$ takes the value $0$ at most once. WLOG we may assume that, if it does take the value $0$, it does so at $0$. Thus there are two cases:

1. $b(z) = \exp(g(z))$ for some nonconstant entire function $g$
2. $b(z) = z \exp(g(z))$ for some nonconstant entire function $g$.

In case (1), $\exp(z) = b(c(z)) = \exp(g(c(z)))$, so $z = g(c(z)) + 2 \pi i n$ for some constant $n$, but this is impossible since the left side is one-to-one and the right is not.

In case (2), $\exp(z) = c(z) \exp(g(c(z)))$, so $c(z) = \exp(z - g(c(z))) = \exp(h(z))$ where $$ h(z) = z - g(\exp(h(z))$$ is entire. Now since $z = h(z) + g(\exp(h(z)))$ is one-to-one, $h$ must be one-to-one, thus $h(z) = s + t z$ for some constants $s, t$ with $t \ne 0$, and then $g(\exp(s+tz)) = z - (s + t z)$ which is impossible because the right side is either one-to-one or constant, and the left side is not.