if f is integer and satisfies $| f (z) | ≤ | z |^n$ then f must be a polynomial

Question

Show that if f is integer and satisfies an inequality of the form $| f (z) | ≤ | z |^n$ for some $n ∈ \mathbb N$ and for all $| z |$ big enough, then f must be a polynomial.

I know that since f is integer it's equal to a power series centered at zero$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$ but I don't know how to proceed

Am I right that you mean $f$ being entire function? – richrow Dec 08 '21 at 06:43 — richrow, Dec 08 '21 at 06:43
Look at Cauchy's inequalities. – copper.hat Dec 08 '21 at 06:51 — copper.hat, Dec 08 '21 at 06:51

score 0 · Answer 1 · answered Dec 08 '21 at 06:46

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Hint: try to use the Cauchy integral formula for the $n$-th derivative $$ f^{(n)}(0)=\frac{n!}{2\pi i}\oint_{|\zeta|=R}\frac{f(\zeta)}{\zeta^{n+1}}d\zeta $$ for sufficiently large $R$.

answered Dec 08 '21 at 06:46

richrow

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the last thing I get is $ |f^{(n)}(0)| ≤ \frac{n! |z|^n}{r^n}$ @richrow – Joe Dec 08 '21 at 18:38

if f is integer and satisfies $| f (z) | ≤ | z |^n$ then f must be a polynomial

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