If f is holomorphic in all $\Bbb{C}$ and $f(z)=f(1-\cos(z))$ $\forall z$, is f necessarily a constant?

Question

Today I had Theory of Complex Variable Functions exam and I really want to know how to solve this exercise:

If f is holomorphic in all $\Bbb{C}$ and $f(z)=f(1-\cos(z))$ $\forall z$, is f necessarily a constant?

Answer 1

Yes. Write $f(z) = f(0) + \sum_{k=1}^{\infty} a_k z^k$. If $f$ is non-constant, there exists $n \geq 1$ such that $a_n \neq 0$. Then with $g(z) = \sum_{k=0}^{\infty} a_{n+k}z^k$, we have

$$ f(z) = f(0) + z^n g(z). $$

Plugging this to $f(z) = f(1-\cos z$), we obtain

$$ z^n g(z) = (1-\cos z)^n g(1-\cos z). $$

By noting that $1-\cos z = \mathcal{O}(z^2)$ as $z\to 0$, however,

$$ a_n = g(0) = \lim_{z\to0} \frac{(1-\cos z)^n g(1-\cos z)}{z^n} = 0, $$

which is a contradiction.

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To grasp the logic of this argument, you may consider the following simpler question first: If $f : \mathbb{C} \to \mathbb{C}$ is entire and satisfies $f(z) = f(z^2)$, does this imply that $f$ is constant?

Answer 2

Note that $1 - \cos(0) = 0$. If $f$ is not constant, there exist some positive integer $n$ and $c \in \mathbb C \backslash \{0\}$ such that $f(z) = f(0) + c z^n + O(z^{n+1})$ as $z \to 0$. But this is also $$f(1-\cos(z)) = f(0) + c (1-\cos(z))^n + O((1-\cos(z))^n) = f(0) + z^{2n}/2^n + O(z^{2n+2})$$