Solving the functional equation $ f(z) = f\left(\frac{z^2+4}{3}\right) $ 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\left(\frac{z^2+4}{3}\right) $, 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 differentiating, 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?
$$ f(z) = f\left(\frac{z^2+4}{3}\right) $$
I am happy to assume $f$ is an entire function, so that you can differentiate freely.
 A: 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.
