MHB Proving an entire function is a polynomial under certain conditions

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An entire function f that satisfies the condition |f(z)| ≤ C(1 + |z|)^n for all z in the complex plane can be shown to be a polynomial of degree less than or equal to n. The function can be expressed as a power series, and the growth condition implies that the coefficients of the power series must vanish for terms of degree greater than n. By applying Cauchy's integral formula, a uniform bound for the derivatives f^(n)(w) can be established, indicating that they must also be bounded. This leads to the conclusion that f is indeed a polynomial of degree at most n. Thus, the conditions provided confirm that f is a polynomial.
Bingk1
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Hello,
This was an exam question which I wasn't sure how to solve:

Suppose f is entire and |f(z)| \leq C(1+ |z|)^n for all z \in \mathbb{C} and for some n \in \mathbb{N}.
Prove that f is a polynomial of degree less than or equal to n.

I know that f can be expressed as a power series, but I'm not sure how to show that the upper limit of the sum has to be less than or equal to n.

Thanks!
 
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Hints:

  • $f^{(n)}(z)$ is entire for all $n\in{\mathbb{N}}$.
  • $\frac{1}{2\pi\cdot i}\cdot \oint_\Gamma \frac{f(z)}{(z-w)^{n+1}}dz = \frac{f^{(n)}(w)}{n!}$ where $\Gamma$ is, say, a circle centered at $w$ of radius $R$.
  • What can you say, then, about $f^{(n)}(w)$ for some $n$ ? (Hint: try to find a uniform bound for $f^{(n)}(w)$ on the whole plane)
 

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