Functional equation $ f(z-g(z))=g(z)$

Question

Does anyone know if there is a way to determine the possible solutions to the equation

$f(z-g(z))=g(z)$

For example, one for which the solutions behave as

$f(z), g(z) \sim \frac{1}{z}, \qquad z\rightarrow \infty $

plus subleading terms in $z$.

Answer 1

Here is how you can produce a lot of examples.

Let $h(z)$ be any onto function. Define $$g(z)=z-h(z)$$ and hence $z-g(z)=h(z)$.

Therefore, the functional equation becomes $f \circ h = g$.

Since $h$ is onto, we can find some $h'$ such that $h \circ h'=Id$.

This gives the solution $$g(z)=z-h(z) \\ f=g \circ h'$$

There are very likely many other solutions.

Note In this construction, if we ask for $g$ to be entire, then $g$ is linear. Then $g(z)=az+b$ with $a \neq 1$, $h(z)=(1-a)z-b$ and $h'(z)=\frac{1}{1-a}(z-b)$.

Then $$f(z)=\frac{a}{1-a}(z-b)+b=\frac{a}{1-a}z+\frac{b}{1-a}$$