# 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.

• If you differentiate $f$ you are assuming that $f$ is analytic. This is a very strong condition that you have not included in your problem. – ajotatxe Jul 12 '16 at 17:30
• Of course, this condition isn't given in my problem (but I think that it was in the similar topic here), so I probably would lose some of the solutions by taking that $f$ is a holomorphic function. (I tried to derive looking for some solutions, I understand that it is not necessary the whole solutions set) – Icarus 369 Jul 12 '16 at 17:44
• A typical functional equation of this sort will have no nonconstant analytic solutions, so it makes a huge difference whether you require that. – Eric Wofsey Jul 12 '16 at 18:13
• Assume that $f$ is an entire function (holomorphic over the whole complex plane), so the differentiation is possible and allowed. – Icarus 369 Jul 12 '16 at 18:24

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.
• Thank you, your answer was very helpful - most of the equations that I had to solve are of the type $f(z) = f(g(z))$. I still have only one question, what to do if there is a situation $f(h(z)) \pm f(z)= g(z)$? For example, if we have $f(z) - f(z^2) = z$, what is the usual aproach? I tried something (again with derivinig), but my logic looks circular, so I would be grateful for an advice. – Icarus 369 Jul 13 '16 at 7:28
• If $h(0)=0$, one technique will probably often be useful is looking at the Taylor series of $f$. You can write each side of the equation as a power series, and then equate them term by term, which will give you a bunch of equations relating the Taylor coefficients of $f$. If it is particularly simple to write down the Taylor series of $g(z)$ and $h(z)$ (as in your example), these equations will be easy to understand. – Eric Wofsey Jul 13 '16 at 7:35