prove that if $|f(z)|\geq |z|+|\sin(z)|$ then it cannot be an entire function

Question

Problem: Prove that if $\forall z \in \mathbb{C}.|f(z)|\geq |z|+|\sin(z)|$ then it cannot be an entire function.

I thought about claiming that $f$ must be a polynomial because it has a pole in infinity, but I stuck why it polynomial cannot satisfy this property.

Answer 1

Assume $f$ entire:

$|f(z)|\geq |z|+|\sin(z)| \ge |z|$ implies $f(z) \to \infty, z \to \infty$ hence $f$ is a polynomial of degree $n \ge 1$

But then if $f=\sum_{k=0}^n a_kz^k, M=\max |a_k|$ we have that $| f(iR)| \le M(n+1)R^n$ for $R >1$

On the other hand $2|\sin (iR)|=|e^{-R}-e^{R}| \ge e^R-1$ so one gets the inequality:

$e^R-1 \le 2M(n+1)R^n$ for all $R>1$ where $M,n$ are fixed and that is plainly impossible since $e^R/R^n \to \infty, R \to \infty$ so we get a contradiction!

Answer 2

Assume $f$ is entire and satisfies the given inequality. Then $|f(z)|\to \infty$ as $|z|\to \infty.$ Yes, you're right: This implies $f$ is a polynomial. (Make sure you know why.)

But notice $|f(iy)| \ge |\sin(iy)|,$ and $ |\sin(iy)|$ grows like $e^y$ as $y\to \infty.$ That is a growth rate no polynomial can match, and we have a contradiction.