Solving a functional equation in complex analysis

Is there any simple (general) way to find the complex solution of a functional-equation?

For example, it is given $f(z) = f\big(\frac{z^2+4}{3}\big)$, and the question is to find all the functions $f(z)$ witch satisfy the given condition.

In an another topic I have seen a interesting solution, where another equation ($f(2z) = f^2 (z)$) was solved by differencing, defining a new function, and using the identity theorem. This solution is elegant, but I think that it doesn't help in the case above, and in many other cases. So, what is "the usual way" (if it exists) of solving this?

$$f(z) = f\big(\frac{z^2+4}{3}\big)$$

I am happy to assume $f$ is an entire function, so that you can differentiate freely.

Answer

Let me assume you are interested in a functional equation of the form $f(z)=f(g(z))$ where $g(z)$ is some entire function and you want solutions $f(z)$ which are entire. In this case, for most choices of $g(z)$, there will be no nonconstant solutions. Indeed, suppose that there is some $a\in\mathbb{C}$ such that $g(a)=a$. If $|g'(a)|<1$, then if you start with a point $z_0$ near $a$ and iterate $g$, you will get closer and closer to $a$. Since $f$ must take the same value at all these points, we find that $f$ takes the same value at a set of points that accumulates at $a$, so by the identity theorem $f$ must be constant. On the other hand, if $|g'(a)|>1$, then $g$ has a holomorphic inverse $g^{-1}$ near $a$ such that $f(z)=f(g^{-1}(z))$ for all $z$ near $a$ and $|(g^{-1})'(a)|<1$, and we can apply the argument above with $g^{-1}$ to again find that $f$ must be constant.

So if $g$ has a fixed point $a$ at which $|g'(a)|\neq 1$, then any entire solution to $f(z)=f(g(z))$ must be constant. In particular, for instance, it is easy to verify that this condition holds for $g(z)=\frac{z^2+4}{3}$. If $g$ has a fixed point $a$ with $|g'(a)|=1$, I'm not sure whether you can always find a grand orbit of $g$ that accumulates at $a$; probably this is known and someone who knows complex dynamics better than me could answer that question.