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If we have a function $f(z)$ be an entire function which satisfies $\underset{% z\rightarrow \infty }{\lim}f(z)/z^{2}=0$; how can we go about proving that $f(z)$ is a linear function of $z$. What if this entire function satisfies $\underset{% z\rightarrow \infty }{\lim}f(z)/z^{n}=0$, where n is a positive integer?

Any suggestions are welcome, thank you.

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    $\begingroup$ The Cauchy inequalities yield that immediately. $\endgroup$ – Daniel Fischer Feb 17 '14 at 21:16
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In the general, assuming $$ \lim_{z\to\infty}\frac{f(z)}{z^n}=0, $$ implies that, there exists an $M>0$, such that $$ \sup_{|z|\ge M}\left|\frac{f(z)}{z^n}\right|\le 1. $$

According to Cauchy Integral Formula: For every $m>n$, and $r>M$, $$ f^{(m)}(0)=\frac{m!}{2\pi i}\int_{|z|=r}\frac{f(w)\,dw}{w^{m+1}}, $$ and hence $$ \left\lvert f^{(m)}(0)\right\rvert\le \frac{m!}{r^{m-n}}\sup_{|z|=r}\, \left|\,\frac{f(w)}{w^{n}}\right|\le \frac{m!}{r^{m-n}}\to 0, $$ as $r\to\infty$. Hence $f^{(m)}(0)=0$, for all $m>n$. In the case of $m=n$: $$ f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)\,dw}{w^{n+1}}, $$ and hence $$ \left\lvert f^{(n)}(0)\right\rvert\le n!\sup_{|z|=r}\, \left|\,\frac{f(w)}{w^{n}}\right|\to 0, $$ as $r\to\infty$.

So $f^{(m)}(0)=0$, for all $m\ge n$, which means that $f$ is a polynomial of degree at most $n-1$.

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Your hypothesis means $f(z) \leq A z^{n-1}+B$, you write the cauchy formula, takes the derivative at the n-order, and you find zero when the radius of your disk goes to infinity (here is the hypothesis you got an entire function). By integrating, you find that $f$ is a polynomial of degree smaller or equal than n-1.

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