# Show that entire function $f:\mathbb{C}\to \mathbb{C}$ so that $|f(\cos z)|\leq A |z|^n$ is constant [closed]

Let $f:\mathbb{C}\to \mathbb{C}$ a holomorphic function so that $|f(\cos z)|\leq A |z|^n$ for all $z$ with $|z|>1$ and positive numbers $A$ and $n$. Show that $f$ is constant.

I want to apply Liouville's theorem on $f$. So we have to show that $f$ is bounded. But from that inequality how we can say that $f$ is bounded ?

## closed as off-topic by Did, Daniel W. Farlow, colormegone, Michael Albanese, LeucippusAug 2 '16 at 2:51

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• Hint: Try invoking Liouville's Theorem... – LordVader007 Jul 27 '16 at 23:05

An entire function that grows like a polynomial is a polynomial. Let's take this as known. From the given inequality we then see $f(\cos z)$ is a polynomial; let's call it $P(z).$ It follows that $P(2\pi n) = f(\cos (2\pi n)) = f(1)$ for all $n\in \mathbb N.$ Because $P$ is a polynomial, we have $P\equiv f(1).$ So $f(\cos z) = f(1)$ everywhere, which implies $f\equiv f(1)$ as desired.
$$g(z) = \dfrac{f(\cos(z)) - f(1)}{z^n}$$ is analytic on $\mathbb C \backslash \{0\}$, bounded at $\infty$, and has at most a pole of order $n-1$ at $0$. Therefore its Laurent series can contain only terms in $z^j$, $1-n \le j \le 0$, which says $z^n g(z)$ is a polynomial. But this polynomial is $0$ at $2\pi n$, so...