Under what conditions on $f$, is $f(az)=g(a)f(z)$?

Question

Formal Statement

Given nonzero constant $a \in \mathbb{C}$, $|a|>0$ and $f:\mathbb{C} \to \mathbb{C}$, under what conditions on $f$ does the following hold?

\begin{equation} f\left(a z\right) = g\left(a\right) f\left(z\right) \end{equation}

for some $g:\mathbb{C} \to \mathbb{C}$.

Additional Thoughts

Obviously the above equation holds with $g(a)=a$ when $f$ is a linear function in $z$.

The property I seek seems initially similar to the separable property, though that property is typically applied to a multivariate function, and the decomposing functions are new, e.g. $p(x,y)=m(x)n(y)$. Upon closer inspection, this property is really quite different than my question.

1) Is there a general condition on my function $f$ that gives my equality ?

2) Is it possible to determine what the function $g$ is, given $f$ ?

EDIT I

1. My condition seems stricter than separability in that the same function $f$ appears on the RHS
2. However, my condition is weaker than separability in that a is a constant, rather than another complex variable

EDIT II

$a$ is indeed a genuine constant. In particular, I only hope to utilize this equation for two values of it:

3. $a=e^{i 2 \pi/3}$
4. $a=e^{i 4 \pi/3}$

Also, perhaps a better iteration of my question is this:

Given nonzero constant $a \in \mathbb{C}$, $|a|>0$ and $f:\mathbb{C} \to \mathbb{C}$, under what conditions on $f$ does the following hold?

\begin{equation} f\left(a z\right) = b f\left(z\right) \end{equation}

for some $b \in \mathbb{C}$.

Answer 1

Hint:

Reworking with logarithms and antilogarithms, you can write the equation as

$$f^*(a+z)=g^*(a)+f^*(z).$$

Then $g^*(a)=f^*(a)-f^*(0)$, implying

$$f^*(a+z)=f^*(a)-f^*(0)+f^*(z),$$ and with $f^+(z)=f^*(z)-f^*(0)$,

$$f^+(a+z)=f^+(a)+f^+(z).$$

If the equation only holds for a single $a$, then with $z=aw$ we have the recurrence

$$f^+(a(w+1))=f^+(aw)+f^+(a)$$ or

$$h(w+1)=h(w)+h(1)$$ for which the solution is easy.