Condition to be polynomial.

Question

Let $f$ be entire function on complex plane.

If for any sequence $z_n$ in $\mathbb{C}$ satisfying $z_n \to \infty, f(z_n) \to \infty$ , then $f$ is nonconstant polynomial.

I think so $f \to \infty $ as $z \to \infty$. so $f$ has all solutions in some disk centered at 0. What shall I do? Thank you in advance.

Answer 1

By Casorati-Weierstrass, this implies that $\infty$ is a pole of $f$. The Laurent series has only finitely many nonzero coefficients, and therefore is a polynomial.

Answer 2

Suppose $f$ is not a non-constant polynomial with given condition.

Because of given condition $f$ could not be a constant function.

Then, the set of zeros of $f$ (say $E$) is an infinite set. (Am I correct?)

Now $E$ must be discrete unbounded set. (If it is bounded then it has a limit point so $f \equiv 0$.)

So we can find a sequence $z_n \in E$ such that $z_n \to \infty$.

But $f(z_n) =0 , \forall n$. Which contradicts the hypothesis.

Hence $f$ must be a non-constant polynomial.