Function satisfying inequality has no root

Question

Let $f$ be an entire function such that, for all $z \in \mathbb{C}$ with $|z| > 1$, $$ |f'(z)| < \frac{|f(z)|}{|z|^2} < 1 $$ Show that there is no $a \in \mathbb{C}$ such that $f(a) = 0$.

Assume towards a contradiction that such a $a \in \mathbb{C}$ exists. If $|a| > 1$, then $$ |f'(a)| < \frac{|f(a)|}{|a|^2} = 0 $$ (note: $|a| \neq 0$), a contradiction.

But what about the case where $|a| \leq 1$?

So far I only had to prove statements of the form "prove $f$ does have a zero" (because then you can just look at $1/f$, which is then analytic, and use Liouville's Theorem or the Maximal Modulus Principle).

In the two follow-up questions, I have to show that $f'/f$ is constant, and then that $f$ is constant (so I can't first prove that and then prove the original question).

Any help is greatly appreciated. I know quite a lot of theorems from Complex Analysis.

Answer 1

Using the inequality show that $$ \lim_{R\to\infty}\frac{1}{2\,\pi\,i}\int_{|z|=R}\frac{f'(z)}{f(z)}\,dz=0. $$ Now apply the Argument Principle.

Answer 2

$\textbf{Hint}$: Suppose $f$ is an entire function. If $|f(z)|<|z^n|$ on $\mathbb{C}\setminus\mathbb{\overline D}$, $n\geq0$, then $f$ is a polynomial of degree at most $n$. If $n\leq0$, then $f$ is constant.

First try proving this using the Cauchy Estimates, then use this repeatedly.