# Entire function satisfying $f(f(z))=f'(z)$

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.

• Qualifying exam for what? What tools do you have at your disposal? Sep 21, 2021 at 7:44
• If $f$ is constant, $f$ even has to equal $0$ everywhere since then $f(f(z))=f'(z)=0$. Sep 21, 2021 at 8:51
• @MartinR Phd Qualifying exam for your information. I can assume any theorem from a first course in complex analysis. Sep 21, 2021 at 14:11

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.

Let $$f \in H(\mathbb{C})$$ be a solution of (1) $$f(f(z))=f'(z)$$. Assume that $$f'$$ is not the zero function. We have $$f'(f(z))f'(z)=f''(z)$$. Thus $$f'(f(z)) = \frac{f''(z)}{f'(z)}$$ is a entire function. By the argument principle $$f'$$ has no zeros. Thus $$f''$$ has no zeros. If $$f(\mathbb{C})=\mathbb{C}$$ then (1) yields that $$f'$$ has zeros, a contradiction. Thus $$f$$ omits one value (Little Picard), call it $$w$$. Set $$g(z)=f(z)-w$$. Now $$g(z)\cdot g'(z)\cdot g''(z)$$ has no zero. By a theorem of Polya and Saxer $$g$$ is of the form $$g(z)=\exp(az+b)$$, $$a \not= 0$$. Thus $$f(z)=\exp(az+b)+w$$. Such an $$f$$ cannot solve (1). Thus $$f'=0$$ and (1) forces $$f=0$$.

For the Thm of Polya and Saxer see for example

R. B. BURCKEL, An Introduction to Classical Complex Analysis. Vol. 1. Basel-Boston 1979; p.433

• argument principle... I don't get it. I see only: (roots of f') is a subset of (roots of f''). Polya & Saxer... could you provide an idea how this theorem works? It doesn't look like a common known theorem - at least for me. Sep 21, 2021 at 12:28
• Argument principle says: If $h$ is an entire function and $\gamma(t)=R\exp(it)$ $(t \in [0,2\pi])$ then $\frac{1}{2\pi i} \int_\gamma h'(z)/h(z) dz$ is the the number of the zeros of $h$ in $\{z:|z|<R\}$. In the case above this is applied to $h=f'$ and since the integral is $0$ in this case there are no zeros of $f'$ (since $R$ can be arbitrarily big). The Polya Saxer Thm. is indeed a bit deep; its from the area of value distribution of entire functions (such as Little Picard). In fact I couldn't find an elementary proof.
– Gerd
Sep 21, 2021 at 12:42
• The fact that $f'$ has no zeroes is easy by induction too keeping differentiating the original relation ($f'(a)=0$ then $f^{(n)}(a)=0, n\ge 1$ so $f$ constant etc), then $f$ has no zeroes too from the relation $f(f(z))=f'(z)$ and a bit of entire function theory (if $f$ has zeroes it cannot be surjective as $f'$ would have zeroes, but then it omits only one other value and the preimage of zero has infinitely many points, so $f(f(z))$ has zeroes etc) and $f''$ doesn't have zeroes by differentiation; from there I do not see an elementary argument either at least so far Sep 21, 2021 at 13:04