Boundedness of an entire function

Question

Question. Let $f$ be an entire function which has a certain asymptote.

Answer 1

$f$ is entire, so it is represented everywhere by its Maclaurin series. An application of the Cauchy estimates gives

$|f^n(0)|\le \frac{n!M_R}{R^n}$ where $M_R= \max \{|f(z)|:|z|=R\}$,

and since by hypothesis $M_R\le R$ if $R$ is large enough, we have

$|f^n(0)|\le \frac{n!M_R}{R^n}\le \frac{n!}{R^{n-1}}\to 0$, as $R\to \infty$ so $f^n(0)=0$ for $n=2,3,\cdots.$

Therefore, $f(z)=az+b$ for some $a,b\in \mathbb C$. But $a=0$ since $\frac{f(z)}{z}\to 0$ as $z\to \infty$ and so $f$ is constant.

Answer 2

Let $g(z)=\frac {f(z)-f(0)} z$ for $z \neq 0$ and $g(0)=f'(0)$. This is an entire function. One way to see this is to use power series. This function tends to $0$ as $|z| \to \infty$. Apply Louiville's Theorem to this function and conclude that $f=cz+f(0)$ for some constant $c$. But the hypothesis implies that $c=0$.