Entire function such that $|f(z)| \geq 11|z|$ on $\mathbb {C}-closure({\mathbb {D}(0, 10))}$

Question

I'm learning complex analysis and encountered this question:

Let $f$ be an entire function such that $|f(z)| \geq 11|z|$ on $\mathbb {C}-closure({\mathbb {D}(0, 10))}$. Prove that there exists $z_{0}\in \Bbb{C}$ such that $f(z_{0})=z_{0}$.

I think I would know what to do if the inequality was on all of $\Bbb{C}$, but as it isn't, I have know idea. I'm still thinking in the direction on Casorati-Weierstrass somehow.

Thank you

Answer 1

You can do it by absurd. If $f$ has no fix points, then the function $$z \mapsto f(z)-z$$ is never $0$ and you can consider the entire function $g$ defined by $$g(z) = \frac{1}{f(z)-z}.$$ Now you have $$|g(z)|=\frac{1}{|z|}\frac{1}{\left|\frac{f(z)}{z}-1\right|} \leq \frac{1}{10} \frac{1}{\left|\frac{f(z)}{z}-1\right|} \leq \frac{1}{10} \frac{1}{\left|\left|\frac{f(z)}{z}\right|-1\right|}\leq\frac{1}{100},$$ if $z\in \mathbb{C}\backslash D(0,10)$. For the last inequality, I have used your hypothesis. Now by Liouville's theorem, you know that $g$ must be a constant $c\neq 0$. Hence $$f(z) = z+\frac{1}{c}, \quad \forall z\in \mathbb{C}.$$ This leads to a contradiction in view of your hypothesis.