For $n\geq 2$ I need to find all entire functions $f:\mathbb{C} \rightarrow \mathbb{C}$ Such that $f(z^n)=f(z)^n$ for all complex numbers $z$. I've tried expanding it's series at 0 and have found some relations with the coefficients, but none seem to finish the problem. I couldn't conclude using Liouville's theorem neither.. My gut feeling tells me that the only the functions of the form $z^k$ will work, but I am a bit stuck.

  • $\begingroup$ An idea would be to try to use Cauchy's integral formula at $0$ on the one hand and to calculate derivatives of $g_n (z) = f(z^n)$ from the power rule and chain rule on the other hand. Comparing the two might give you something. $\endgroup$ Oct 22, 2023 at 19:00
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    $\begingroup$ $f\left( x \right) = 0$ and $f\left( x \right) = e^{2 \cdot \frac{k}{n - 1} \cdot \pi \cdot i}$ where $k \in \mathbb{Z}$ are also solutions. $\endgroup$ Oct 22, 2023 at 19:01
  • $\begingroup$ To elaborate my idea (which may be fruitless) I would play a game like $g_k^{(n)} (z) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)^k}{z^{n + 1}} dz$. It feels like it should be possible to go somewhere from there by comparing with the answer you get calculating $g_k^{(n)}(z)$ explicitly. $\endgroup$ Oct 22, 2023 at 19:04
  • $\begingroup$ @CharlesHudgins The equation you have just provided makes no sense. The LHS depends on $z$ whereas the RHS does not. $\endgroup$
    – K.defaoite
    Oct 22, 2023 at 19:07
  • $\begingroup$ My mistake. I meant to write $g_k^{(n)}(0)$ in each case. $\endgroup$ Oct 22, 2023 at 19:08

5 Answers 5


This answer uses the Weierstrass factorization theorem.

First, prove that if $f$ is non zero, then $f$ has no non zero root. Consider, proving this for yourself before reading on.

Suppose that $f$ has a non-zero root $z_0$ [Edit: that is also not equal to 1], then ${z_0}^{1/n}$ is also a root. Iterating that argument, every member of the family ${z_0}^{1/n^p}$, for $p$ integer, is a root. That family has an accumulation point at 1 so $f(z)=0$ for all $z$ which contradicts the initial hypothesis.

[Edit: Now, consider the case where $f$ has a root that is equal to 1. Then, a $n$-th root of unity is also a root. As $n>1$, $f$ would have a root that is not equal to 1 nor zero which was proved to not be possible in the previous paragraph.]

Thus, according to the Weierstrass factorization theorem, if $f$ is non zero then,

$$f(z)=z^m e^{g(z)}$$

where $g(z)$ is an entire function

Thus, $g$ verifies


for some integer $k$

Taking the series expansion of $g$ with coefficients $a_l$, one finds that for $l>0$

$$a_l=na_{n l}$$

Iterating that relationship we have

$$a_l=n^p a_{ln^p}$$

for $p$ integer. $r a_r \rightarrow 0$ for $r\rightarrow\infty$ as the series representation of $g'(1)$ converges. Thus, $n^p a_{ln^p}$ also converges to $0$ for large $p$ and so $g$ is constant. It is then clear that


where $$\Theta^n=\Theta$$

  • $\begingroup$ The finitely many zeros case of Weierstrass factorization is "easier" than the general case: once you know $f(z)/z^m$ is a map $\mathbb{C} \to \mathbb{C}^*$, it lifts to the universal cover, ie is of the form $e^{g(z)}$. $\endgroup$
    – ronno
    Oct 23, 2023 at 8:49
  • $\begingroup$ @ronno Hi sorry I am a physicist and so I might sometimes have a bit of trouble filling in gaps. I am not sure which theorem you are using to infer directly that $f(z)/z^m$ is of the form $e^{g(z)}$ but I am interested in knowing if you happen to have the name of the theorem. $\endgroup$ Oct 25, 2023 at 4:21
  • $\begingroup$ Couldn't $f$ have a zero at $1$? After all, the set $\left\lbrace 1^{n^p} : p \in \mathbb{Z} \right\rbrace$ has no accumulation point. $\endgroup$ Oct 25, 2023 at 5:48
  • $\begingroup$ @BrunoKrams darn, yep you are right that I did not take into account the case where $f$ has a zero at 1. I had also considered looking at the Taylor expansion around non zero roots to check if the expansion around the root is different on both sides of the equality but It would involve considering zeroes with multiplicity and even if it worked it was not as pretty as what I see now to be an incomplete accumulation point argument. $\endgroup$ Oct 25, 2023 at 6:11
  • $\begingroup$ @userrandrand $\mathbb{C}$ is simply connected, so any map $\mathbb{C} \to \mathbb{C}^\times$ must factor through the universal cover of $\mathbb{C}^\times$, namely $\exp : \mathbb{C} \to \mathbb{C}^\times$. $\endgroup$
    – ronno
    Oct 25, 2023 at 8:48

Let $n\geq 2$ and consider an entire function $f:\mathbb{C}\to\mathbb{C}$ such that $f(z^n)=f(z)^n$. We consider first the case where $f=g$ where $g$ is a function such that $g(0)\neq 0$.

We want to differentiate the expression $g(z^n)$, $k$ times. For $k\geq 1$, $$\frac{d^k}{dz^k}g(z^n)=\sum_{j=1}^kg^{(j)}(z^n)h_j(z)$$ for some entire functions $h_j$ where $h_k(0)=0$. We proceed by induction. For $k=1$ we have $$\frac{d}{dz}g(z^n)=g'(z^n)nz^{n-1}.$$ We thus have $h_0(z)=0$ and $h_1(z)=nz^{n-1}$. Note that $h_1(0)=0$. Assume now that it holds for $k=l$. This gives us $$\frac{d^l}{dz^l}g(z^n)=\sum_{j=1}^lg^{(j)}(z^n)h_j(z)$$ and thus $$\frac{d^{l+1}}{dz^{l+1}}g(z^n)=\frac d{dz}\sum_{j=1}^lg^{(j)}(z^n)h_j(z)=\sum_{j=1}^l\frac d{dz}g^{(j)}(z^n)h_j(z)=$$$$\sum_{j=1}^lg^{(j+1)}(z^n)nz^{n-1}h_j(z)+g^{(j)}(z^n)h_j'(z)=\sum_{j=1}^{l+1}g^{(j)}(z^n)\tilde{h}_j(z).$$ Note again that $\tilde{h}_{l+1}(0)=n\cdot0^{n-1}h_{l}(0)=0$ which proves the statement above.

We now try to differentiate $g(z)^n$, $k$ times. For $k\geq1$, $$\frac{d^k}{dz^k}g(z)^n=\sum_{j=1}^kg^{(j)}(z)h_j(z)$$ for some entire functions $h_j$ where $h_k(0)\neq0$. For $k=1$ we have $$\frac{d}{dz}g(z)^n=g'(z)ng(z)^{n-1}.$$ Here $h_1(0)=ng(0)^{n-1}\neq0$ by the assumption on $g$. Assume now that it holds for $k=l$, we then have $$\frac{d^l}{dz^l}g(z)^n=\sum_{j=1}^lg^{(j)}(z)h_j(z)$$ where $h_l(0)\neq0$ and it follows that $$\frac{d^{l+1}}{dz^{l+1}}g(z)^n=\frac d{dz}\sum_{j=1}^lg^{(j)}(z)h_j(z)=\sum_{j=1}^l\frac d{dz}g^{(j)}(z)h_j(z)=$$$$\sum_{j=1}^lg^{(j+1)}(z)h_j(z)+g^{(j)}(z)h_j'(z)=\sum_{j=1}^{l+1}g^{(j)}(z)\tilde{h}_j(z)$$ where $\tilde{h}_{l+1}(0)=h_l(0)\neq0$. This proves the second induction.

Now once again by induction we prove that $g^{(k)}(0)=0$ for all $k\geq1$. $g(z^n)=g(z)^n$ implies $g'(z^n)nz^{n-1}=ng(z)^{n-1}g'(z)$ and $0=0g'(0)=g(0)^{n-1}g'(0)$ where $g'(0)=0$ because $g(0)\neq0$. The statement thus holds for $k=1$. Assume now that $g^{(k)}(0)=0$ for $k\leq l$. By the two induction arguments above we get $$\sum_{j=1}^{l+1}g^{(j)}(z^n)h_j(z)=\frac{d^{l+1}}{dz^{l+1}}g(z^n)=\frac{d^{l+1}}{dz^{l+1}}g(z)^n=\sum_{j=1}^{l+1}g^{(j)}(z)\tilde{h}_j(z).$$ This gives $$\sum_{j=1}^{l+1}g^{(j)}(0)h_j(0)=\sum_{j=1}^{l+1}g^{(j)}(0)\tilde{h}_j(0)$$ which is the same as $$g^{({l+1})}(0)h_{l+1}(0)=g^{({l+1})}(0)\tilde{h}_{l+1}(0).$$ Note that $h_{l+1}(0)=0$ but $\tilde{h}_{l+1}(0)\neq 0$. We must thus have $g^{(l+1)}(0)=0$. We finally arrive with the conclusion that $g(z)=g(0)$ by a taylor expansion of the entire function $g$ and $g$ is thus constant. In particular $g(0)=g(0)^n$ and $g(0)=\zeta$ must be an $n-1$'th root of unity.

Suppose now instead that $f(0)=0$. Either $f(z)=0$ for all $z$ or $f$ has a root of some order $k$ at $0$, in this case $$g(z)=\frac{f(z)}{z^k}$$ is an entire function such that $g(0)=0$. It is not hard to show that $g$ satisfies the same functional equation and by the reasoning above $g(z)=\zeta$. We thus have $f(z)=\zeta z^k$ for an $n-1$'th root of unity $\zeta$. In conclusion we have shown that the only solutions are $f(z)=0$ or $f(z)=\zeta z^k$ for some $k\geq 0$.

  • $\begingroup$ Thank you very much for the proof, I wasn't able to go that far in my first tries at the problem! $\endgroup$ Oct 22, 2023 at 21:23

Perhaps there is a more elegant and self-contained way of doing this, but I think we can use normal numbers to our advantage. I'll focus on the specific case of $n=2$, but similar arguments should work for other natural values of $n$. This is not a complete proof, but I think it's a good starting point.

Let $\omega$ be a normal number in base $2$, and let $f$ be an entire function satisfying $f(z^2) = f(z)^2$ for all $z\in\mathbb C$. Because $\omega$ is normal, we have that the numbers taking the form $$a_n = \exp\Big(2\pi i \omega \cdot 2^n\Big)$$ are dense in the unit circle. Further, we have that $$f(a_n) = f(a_0)^{2^n}$$ By the identity principle, since the $a_n$ have an accumulation point in $\mathbb C$ (because they are dense in $\mathbb S^1$) this means that $f$ is uniquely determined by its image of $a_0$.

Note that if $|f(a_0)| > 1$, then $f$ is unbounded on the unit circle, which is impossible for an entire function, so this cannot be. On the other hand, if $|f(a_0)| < 1$, then $f$ gets arbitrarily close to zero in arbitrary neighborhoods of any point on the unit circle, which is not possible except for the zero function $f=0$. So aside from the zero function (a trivial case which you did not mention in your question), we only need to consider cases when $|f(a_0)|=1$.

From here, we just need to consider the case when $f(a_0)=e^{2\pi i\lambda}$. When $\lambda$ is an integer multiple of $\omega$, then we have one of the power functions you mentioned (and when it is a nonnegative integer multiple, we have an entire power function).

Maybe someone else can take it from here? Some casework on the ratio $\lambda/\omega$ might allow us to narrow down the values that $\lambda$ can take, and thereby classify the possible functions satisfying this equation.

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    $\begingroup$ Thanks a lot for your contribution. I have never worked with normal numbers, but they seem usefull indeed to reduce the problem! This is marked as an "easy" question for third year undergrads, so I think the teacher made a mistake writing the exercise. Thanks a lot though, now I know about normal numbers! $\endgroup$ Oct 22, 2023 at 19:53

As all the other answers state, $f$ is either $=0$ or a monomial $f(z) = az^k$ with $a = \zeta \in\{e^{\frac{2\pi i r}{n-1}}: r=0,1,..., n-2\}$, i.e. an ($n-1$)-th root of unity.

Proof: By $f(0) = (f(0))^n$, we have that either:

  • $f(0)=\zeta$, an ($n-1$)-th root of unity. Then by iterating the assumption, and continuity of $f$, for all $\lvert z \rvert <1$, $\lim_{r\to \infty}|f(z)|^{n^r} = |f(0)|=1$ which is possible only if $|f(z)| =1$. That is, $f$ maps the open unit disk into the unit circle. By the open mapping theorem, $f$ has to be constant, i.e. $f(z) = \zeta$ for all $z$.


  • $f(0)=0$. Then if $f$ is not constantly $0$, say $0$ is a zero of order $k$, i.e. the continuation of $g(z) = \frac{f(z)}{z^k}$ to $z=0$ is still an entire function but with $g(0) \neq 0$. But also, $g(z^n) = (g(z))^n$ for all $z$, i.e. $g$ is a function like in the first case! So $g(z) = \zeta$, thus $f(z) = \zeta \cdot z^k$.

Note that the only "complex-analytic" ingredients here are the open mapping theorem, and the fact that a zero of a certain finite order can be "removed" as in the second case (which is trivial if one knows analytic functions as power series). In fact, this short proof can be easily adapted to any $f$ which is analytic in a neighbourhood of $0$, and satisfies $f(z^n) = (f(z))^n$ in that neighbourhood.


Focus on $\overline B(0,1)$ and its boundary $S^1$. If there was a point $z\in S^1$ such that $\vert f(z)\vert \gt 1$, then there is a $\vert z'\vert \lt 1$ where $\vert f(z')\vert \gt 1$ implying $f$ is unbounded in $\overline B(0,1)$, violating compactness.

If there is a point $z\in S^1$ such that $\vert f(z)\vert \lt 1$, then there is a $\delta$ neighborhood of $z$ where the image under $f$ still has modulus $\lt 1$ and for all primes $p$ large enough (also insisting $p\gt n$) there is a $p$th root of unity $\omega \neq 1$ within that $\delta$ neighborhood of $z$, i.e. $\vert f(\omega)\vert \lt 1$. We may denote the finite group of $p$th roots of unity as $G$ and the map $\phi(z)=z^n$ acts as an automorphism (e.g. by the Counting Formula). By finiteness of the group, there is some $g' \in G-\big\{1\big\}$ which has a minimum modulus $M\lt 1$ under the image of $f$. Then $\phi(g') \in G-\big\{1\big\}$ so $0\leq M \leq \big\vert f\big(\phi(g')\big)\big\vert = \big\vert\phi\big( f(g')\big)\big\vert =\phi\big(\vert f(g')\vert\big) = M^n \leq M \lt 1\implies M =0$; that is, $f$ maps a $p$th root of unity $\neq 1$ to $0$ for all $p$ large enough $\implies f =0$ since there are countably infinite number of primes and $S^1$ is compact.

[Addendum: to handle the stricter problem where we only know $f$ is analytic on $B(0,1)$ and continuous on its boundary, we can iterate on the preceding argument. Suppose $f\Big(G-\big\{1\big\}\Big)$ has cardinality $\geq 2$ and let $\big\vert f(g^*)\big\vert =M^* \in (0,1]$ for some $g^* \in G-\big\{1\big\}$ be the second smallest modulus, where the smallest modulus is $=0$ per the above. Then as above $\phi\big(g^*\big) \in G-\big\{1\big\}$ so $M^*\leq \big\vert f\big(\phi(g^*)\big)\big\vert=\phi\big(\big\vert f(g^*)\big\vert\big)$, giving the chain $0\lt M^*\leq \phi\big(\big\vert f(g^*)\big\vert\big) = (M^*)^n\leq M^*\implies M^*=1$. But every $z \in S^1$ is the limit of a sequence of pth roots of unity $\omega_p^{(k)}\neq 1 $ (by selecting increasingly large $p$) $\implies \big\vert \phi\big(S^1\big)\big \vert =\big\{0,1\big\}$ which contradicts $S^1$ being connected. Conclude $M^*$ doesn't exist and $\big\vert \phi\big(S^1\big)\big \vert =\big\{0\big\}$ and $f=0$ by maximum modulus theorem.]

It remains to consider the case of $f\big(S^1\big)\subseteq S^1$. $f$ must be non-zero on the punctured ball $B^*\big(0,1\big)$, otherwise it would have countably infinite zeros in $\overline B(0,1)$ and be identically zero. Let $f(0) = 0$ with multiplicity $k$ (a non-negative integer) and $F(z):=\frac{f(z)}{z^k}$ with $F(0)$ being the value implied by continuity and observe $F\big(S^1\big)\subseteq S^1$. Since $F$ has no zeros in $B(0,1)$ the max and min modulus theorems tell us it attains a max and min modulus on the boundary, i.e. $1\leq \big \vert F\big(B(0,1)\big)\big\vert \leq 1$ hence $F$ is constant by Max Modulus Theorem, and $F(1) = \lambda =F(1^n)=F(1)^n =\lambda^n$ i.e. $\lambda$ is an $n-1$th root of unity $\implies f(z) = \lambda \cdot z^k$.


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