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**slevvio** Let f be an entire function satisfying $\displaystyle |f(z)| \le K|z^2|$ for $\displaystyle |z| \ge 1$, where K is a positive constant. Show using Cauchy's inequality, that $\displaystyle f^{(n)}(0) = 0$ for$\displaystyle n > 2 $ and deduce that f is a quadratic polynomial.

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$\displaystyle |f^{(n)}(0)| = \left| \frac{n!}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz \right| \le \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R}\left| \frac{f(z)}{z^{n+1}} \right|$ by the estimation lemma. I understand it all up until this point.

However , now the solution I have says that

$\displaystyle \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le Rn!\frac{KR^2}{R^{n+1}}$

which I think tells me that

(*) $\displaystyle \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le \frac{KR^2}{R^{n+1}}$ but I have no idea where this comes from! Does anyone have an idea how to help?