$f(exp(z)) = f(z) \cdot g(z) $ with closed form solutions?

Question

Are there closed form solutions for functions

$f(z),g(z)$

Such that $A)$

$$ \frac{f(exp(z))}{f(z)} = g(z) $$

$B)$ $g(z),f(z)$ are both analytic and nonconstant.

$C)$ $g(z)$ is analytic at $z$ Whenever $f(z) = 0$.

$D)$ the RHS is shorter (simplified).

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To clarify here is an analogue where $exp$ is replaced by $z^2$.

$$\frac{A(z^2)}{A(z)} = B(z)$$

$$ A(z) = z-1, B(z) = z + 1$$

Notice $A,B$ are analytic and nonconstant. And When $z = 1$ Then $A(z) = 0 $ and $B(z)$ is analytic.

Also $D)$ has been satified ; the RHS ( $B(z)$ ) is shorter / simplified version of the LHS.

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Also I do not accept the Abel function of $exp$ as a closed form.

I Will add tetration as a tag though, Because it relates.

Answer 1

EDIT: My answer works for a domain, but it appears the OP wants $f$ and $g$ to be entire functions. In this case, see Bumblebee's answer.

First, recognize that the composition of two analytic functions is analytic. This means that $f(\exp(z))$ is analytic as long as $f$ is.

Next, recognize that the quotient of two analytic functions is analytic whenever the denominator is non-zero. A proof can be found on pages 10-11 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch10.pdf

Finally, this implies that for a domain $\Omega$, as long as $f(\exp(z))$ and $f(z)$ are defined everywhere on $\Omega$ and $f$ is nowhere zero on $\Omega$, we can take: $$g(z) = \frac{f(\exp(z))}{f(z)}$$ This $g$ will be analytic, and non-constant so long as $f$ is non-constant. EDIT: this won't necessarily be non-constant, but it probably will be.

For example, take $\Omega = \{z: |z| < 1\}$ and $f(z) = z^2 + 1$. We get: $$g(z) = \frac{e^{2z} + 1}{z^2 + 1}$$ This pair satisfies.

Answer 2

Here I assume all functions are entire:

To make $g(z)$ analytic, we need both $f(z)$ and $f(\exp(z))$ have the same zeros or both have no zeros.
In the case where $f(z)$ has no zeros, it is of the form $\exp(F(z))$ for some entire function $F$ and then simply choose $g(z)=\exp(F(\exp(z))-F(z)).$

Answer 3

Let $f$ and $g$ be non-constant entire functions such that

$$ \frac{f(e^z)}{f(z)} = g(z) \quad \text{on} \ \mathbb{C}.$$

If $Z = \{z \in \mathbb{Z} : f(z) = 0\}$ denotes the zero set of $f$, then the above identity tells that $\exp(Z) \subseteq Z$. Now there are three possibilities:

1. $Z = \varnothing$. This case reduces to Bumblebee's answer.

2. If $Z$ is bounded, then $Z$ must be finite by the identity theorem. By partitioning $Z$ according to the orbits of $\exp$, we can find $(z_k)_{k=1}^{n} \subset \mathbb{C}$ and $(m_k)_{k=1}^{n} \subset \mathbb{Z}_{\geq 1}$ such that $z_k = \exp^{\circ m_k}(z_k)$ for each $k = 1,\cdots, n$ and

   $$Z = \bigcup_{k=1}^{n} \{ z_k, \exp(z_k), \cdots, \exp^{\circ(m_k-1)}(z_k) \} $$

   Of course, we may assume that all the elements appearing in the above formula distinct. Then there exists an entire function $F$ such that

   $$ f(z) = e^{F(z)} \prod_{k=1}^{n} \prod_{j=0}^{m_k-1} (z - \exp^{\circ j}(z_k)). \tag{*}$$

   Conversely, any $f$ of the form above induces an entire $g$ satisfying the functional equation.

3. If $Z$ if unbounded, then I suspect that $f$ cannot be written in terms of elementary functions, based on the guess that $\exp$-orbits in $Z$ will blow up so fast and such growth speed cannot be replicated by elementary functions. On the other hand, the behavior of orbits of $\exp$ seems quite complicated as demonstrated in this article, so I am not sure about this case.

To be honest, I have absolutely no idea about Case 3 except that there is indeed an entire function $f$ falling into Case 3 (which can be constructed by Weierstrass factorization theorem). So my bet is that $\text{(*)}$ will cover all the 'closed-form solutions of $f$ (by allowing the empty product so as to include Case 1 as a special case).