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Let $f(z) = \sum_{k\geq 0}a_kz^k$ be an analytic function, where $a_k\in\mathbb{C}$ for $k\geq 0$. I am trying to get some conditions for $a_k$ that give us the general form of $f$ such that $(f(z))^n= f(z^n)$.

What I tried is to use the Multinomial Theorem, but I get stuck while trying to compute the coefficients of $(f(z))^n$.

I would like to obtain the general solution, but in fact I need only two cases: if either $a_0 = f(0) = 0$ or $a_1 = f'(0) = 0$.

Thank you!

dudas
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1 Answers1

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If $n \le 2$ the characterization is immediate, so assume otherwise.

Lemma. If $f(0) \ne 0$, then $f(z) \equiv \zeta$ where $\zeta$ is an $(n-1)$-th root of unity.

Proof. Iterate the equality, finding that for any $k \in \mathbb{N}$ we have

$$f(\lambda^{n^k}) = f(\lambda)^{n^k}$$

Applying this to $\lambda \in \mathbb{D}$ and letting $k \to \infty$, we find that $f(\lambda)^{n^k} \to f(0) \ne 0$ by assumption. Certainly this implies that $|f(\lambda)| = 1$, and thus $f$ is some unimodular constant. Applying the equation with $z = 1$ then shows that constant is an $(n-1)$-th root of unity.

So we can now consider the case that $f(0) = 0$. Let $m$ be the order of $0$ as a zero of $f$ (if defined) and look at

$$g(z) = \frac{f(z)}{z^m}$$

with the obvious definition at $z = 0$. This function $g$ satisfies the same functional equation that $f$ does and has non-zero value at $0$ (where we apply the identity theorem to $g(z^n) - g(z)^n$ to get equality at the origin). By the lemma, $f(z) = z^m \zeta$ for an $(n-1)$-th root of unity.

Thus, all solutions are of the form $f(z) = z^m \zeta$ with $\zeta^{n - 1} = 1$, and the zero function.

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    Thank you @user296602, I like it! Just a doubt. When you define $g(0) = a_m$ considering $f(z) = a_mz^m + a_{m+1}z^{m+1} + \dots$, a priori you do not know if $g$ satisfies the statement for $z=0$, but you assert it to use the Lemma. Is there a kind of Identity Theorem (https://en.wikipedia.org/wiki/Identity_theorem) that assure it (using that the functional equation is fulfilled for $z\neq 0$)? – dudas Mar 20 '18 at 11:50
  • Good point that I definitely didn't handle much in the answer. Yes, you can, because $f(z)^n - f(z^n)$ is holomorphic. By the way, evaluation at zero now immediately implies that $a_m^n = a_m$, which is what we wanted. I'll correct/expand the answer when I have time later today. –  Mar 20 '18 at 12:29
  • The lemma statement needs to say $(n-1)$-th root. And the lemma requires $n \geq 2$, but the special cases are easy: $n=0$ implies either $f$ is the zero function or $f(1)=1$, and $n=1$ allows any analytic function at all. – aschepler Mar 20 '18 at 12:44
  • Of course! Thank you for your contribution, @aschepler. – dudas Mar 20 '18 at 12:56
  • @aschleper Thanks! I think all the corrections have been made now. –  Mar 20 '18 at 14:13
  • Yes @user296602 , probably the only thing that is not written is how $|f(z)|=1$ for $z\in\mathbb{D}$ implies that $f$ is constant, which follows from the Open Mapping Theorem (because $f(\mathbb{D})\subset\mathbb{S}^1$). Maybe this is not necessary and it is an obviousness that I cannot see now. – dudas Mar 20 '18 at 16:02
  • @dudas It can also follow from the Cauchy-Riemann equations. –  Mar 20 '18 at 16:44
  • Perfect! P.s. https://math.stackexchange.com/questions/147834/why-a-holomorphic-function-with-constant-magnitude-must-be-constant for future readers. – dudas Mar 20 '18 at 16:50