I met this problem preparing for my qualifying exam:
Find all entire function $f$ such that $f(f(z))=f'(z)$ for all $z$.
My guess is that $f$ is a constant so I may need Liouville's theorem somewhere, but I don't see how.
Any help is appreciated.
Entire function satisfying $f(f(z))=f'(z)$
analysiscomplex numberscomplex-analysis
Best Answer
Let’s consider a more general situation. Not all details are included, but elementary properties of holomorphic functions suffice for those, such as the open mapping theorem and Casorati-Weierstrass. Suppose that $f$ and $g$ are entire and $$f’(z)=g(f(z)).$$ Then for all higher derivatives $f^{(n)}$ there is an entire $g_n$ such that $$f^{(n)}(z) = g_n(f(z)).$$ This follows by induction. As a consequence, $f$ has the same series expansion at $z_0$ and $z_1$ whenever $f(z_0) = f(z_1)$. In particular $z_0-z_1$ is a period of $f$ for all such pairs.
The group $P$ of periods of $f$ is either $\{0\}$, $\mathbb C$, or $\mathbb Z 2 \pi \mathrm i\alpha$ for some complex $\alpha \neq 0$. If $P = \{0\}$ then $f$ is injective and therefore a polynomial of degree one and $g$ is a constant. If $P = \mathbb C$ then $f$ is a constant and $g=0$. In the remaining case $$h(z) = f(\alpha \log(z))$$ is well-defined, non-constant, and injective on $\mathbb C^{\ast}$. Therefore either $h(z)$ or $h(z^{-1})$ is a polynomial of degree one. Then $$f(z) = h(e^{\alpha^{-1}z})$$ and $g$ is a polynomial of degree one. Now the only of all these options where $g=f$ is when $f=0$.
It is instructive to follow the same argument for meromorphic $f$ and see where the differences appear. Note for example that $f(z)=\tan(z)$ satisfies $f’(z) = f(z)^2 + 1$ but no quadratic $g$ could appear for entire functions.