How to prove that $f(z)$ is bounded if we know that $\lim\limits_{|z|\to \infty}\frac{f(z)}{z}= 0$?

Question

Knowing that $f(z)$ is analytic on the entire complex plane and that $\lim\limits_{|z| \to \infty} \frac{f(z)}{z} = 0$.
How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$

Answer 1

If $n\in\mathbb N$, then$$f^{(n)}(0)=\frac1{2\pi i}\oint_{\partial D(0,R)}\frac{f(z)}{z^{n+1}}\,\mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,b\in\mathbb C$ such that $(\forall z\in\mathbb{C}):f(z)=az+b$. But, since $\lim_{z\to\infty}\frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.

Answer 2

Hint If $f(z)$ is entire, so is $$g(z)=\frac{f(z)-f(0)}{z}$$

Next, it is easy to see that $$\lim_{z \to \infty}g(z)=0$$ which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C \forall x$.

Finally, $$0= \lim_{z \to \infty} g(z)= C$$