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Here is my guess.

Let $f$ be a entire function From $\mathbb C$ to $\mathbb C$

(1) If for $f$, there exist a sequence $z_n$ such that $|z_{n}|\rightarrow \infty$, with $|f(z_{n})|\rightarrow\infty$, Then $f$ must be polynomial.

(2) If for all $z_n$ such that $|z_{n}|\rightarrow \infty$, $|f(z_{n})|\rightarrow z<\infty$, then $f$ is bounded, so that $f$ is constant by Liouville's theorem.

Is this true?

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  • $\begingroup$ Am I misunderstanding (2)? I think that, as you state it, the hypothesis is satisfied by all entire functions. $|f(z_n)|<\infty$ is true for every $z_n\in \mathbb C$ if $f$ is entire. "Bounded" doesn't mean "$|f(z)|<\infty$". It means there is an $R$ with "$|f(z)|<R$" for for all $z$ in the domain of $f$. $\endgroup$ – MPW Dec 31 '15 at 5:34
  • $\begingroup$ @MPW you are right, sorry to confuse you. I'm not familar with '$' things. My first try. I'll fix it. $\endgroup$ – nicksohn Dec 31 '15 at 5:58
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(1) is not correct. Let $f(z)=e^z$. Consider sequence $1,2,3,\cdots$. It satisfies the condition. But $f$ is not polynomial. [For $f$ to be polynomial, the condition should be stronger; it should hold for all the possible sequences].

(2) Seems to be correct with your reason.

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    $\begingroup$ (1) is somewhat worse than you make it out to be here; for any non-constant function, such a sequence $z_n$ exists. $\endgroup$ – Milo Brandt Dec 31 '15 at 4:51
  • $\begingroup$ yes; just negation of his (2); provided $f$ is entire. [Are you saying that for any non-constant, even non-entire function, such a sequence exists]? $\endgroup$ – p Groups Dec 31 '15 at 4:52
  • $\begingroup$ Thanks a lot! so, is it impossible to find all entire function? $\endgroup$ – nicksohn Dec 31 '15 at 4:56
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    $\begingroup$ @nicksohn What does "find all entire functions" mean? What would you consider a successful classification? $\endgroup$ – Noah Schweber Dec 31 '15 at 6:29

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