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If $f$ is an entire function such that $\lim_{z\to \infty}\left|\frac{f(z)}{z}\right|=0$ then find the function $f$.

Replacing $z$ by $\frac{1}{z}$, we get $$\lim_{z\to 0}|zf(1/z)|=0$$This shows that $f(1/z)$ has removable singularity at $z=0$ , so $f(z)$ has removable singularity at $z=\infty$. As $f$ is entire so , $f$ must be constant.

Is it correct?

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Yes correct. Here I present another argument that uses Liouville's theorem : Let $$ g(z):=\frac{f(z)-f(0)}{z} $$ Clearly $g$ is also entire and bounded and thus constant! Since $$ \lim_{z \to \infty} g(z) = 0 $$ then $f(z)=f(0)$.

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    I was about to write just this before I rad it , Alonso. I guess great minds think alike :) . – Gary. Jul 31 '15 at 03:49