Prove that every entire function satisfies this particular property

Question

This question was asked in an assignment given to me and I could not solve it .

Question: If $f$ is an entire function such that $|f(z)|\to \infty$ as $|z| \to \infty$ prove that $|f(z)|\geq c|z|$ for some positive number $c$ for all $z$ with $|z|$ sufficiently large.

I assumed that on the contrary that no such number $c$ exists. Then $\frac{|f(z)|}{|z|}=0$ if $|z|$ is sufficiently large. But I am unable to see any contradiction due to this.

Can anyone please tell is there any contradiction which can be deduced here or can anyone please tell me the another approach to prove it .

Thanks!!

Answer 1

Since $\lim_{|z|\to\infty}|f(z)|=\infty$, there is some $M>0$ such that $|z|>M\implies f(z)\ne0$. If $|z|<\frac1M$, let$$g(z)=\begin{cases}\frac1{f(1/z)}&\text{ if }z\ne0\\0&\text{ if }z=0.\end{cases}$$Then $g$ is analytic on $D\left(0,\frac1M\right)\setminus\{0\}$ and $\lim_{z\to0}g(z)=0$. Therefore, by Riemann's theorem, $g$ is analytic. Since $g(0)=0$, its Taylor series centered at $0$ is of the type$$g(z)=a_kz^k+a_{k+1}z^{k+1}+\cdots,$$for some $k\in\Bbb N$, with $a_k\ne0$. So, if $|z|>M$,\begin{align}f(z)&=\frac1{g(1/z)}\\&=\frac1{\frac{a_k}{z^k}+\frac{a_{k+1}}{z^{k+1}}+\cdots}\\&=\frac{z^k}{a_k+\frac{a_{k+1}}z+\cdots}\end{align}If $|z|\gg0$,$$\left|a_k+\frac{a_{k+1}}z+\cdots\right|<2|a_k|$$and therefore$$|f(z)|\geqslant\frac1{2|a_k|}|z|^k\geqslant\frac1{2|a_k|}|z|,$$if $|z|\geqslant1$.

Answer 2

Set $$ g(z)=\frac{1}{f\left(\frac{1}{z}\right)}. $$ Then $g$ is analytic with no roots in a neighborhood of $0$ minus $z=0$, say $D(0,r)\setminus\{0\}$, and $\lim_{z\to 0}g(z)=0$. Thus $g$ extends analytically to a neighborhood of $0$, and vanishes at $z=0$. Therefore $$ g(z)=zh(z) $$ where $h$ is also analytic in $D(0,r)$ with no roots in $D(0,r)\setminus\{0\}$. In particular, $$ |h(z)|\le c, \quad\text{for $|z|\le r'$} $$ whenever $0<r'<r$.

Hence $$ f(z)=\frac{1}{g(\frac{1}{z})}=\frac{z}{h(\frac{1}{z})} $$ and thus $$ |f(z)|=|z|\frac{1}{|h(\frac{1}{z})|}\ge \frac{|z|}{|c|} $$ whenever $|z|\ge \frac{1}{r'}$.