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.

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    $\begingroup$ Qualifying exam for what? What tools do you have at your disposal? $\endgroup$
    – Martin R
    Sep 21, 2021 at 7:44
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    $\begingroup$ If $f$ is constant, $f$ even has to equal $0$ everywhere since then $f(f(z))=f'(z)=0$. $\endgroup$
    – md2perpe
    Sep 21, 2021 at 8:51
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    $\begingroup$ @MartinR Phd Qualifying exam for your information. I can assume any theorem from a first course in complex analysis. $\endgroup$
    – biden
    Sep 21, 2021 at 14:11

2 Answers 2


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

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    $\begingroup$ 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. $\endgroup$
    – TomTom314
    Sep 21, 2021 at 12:28
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    $\begingroup$ 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. $\endgroup$
    – Gerd
    Sep 21, 2021 at 12:42
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Sep 21, 2021 at 13:04

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