Show that the set U is unbounded

Question

I am working on a practice qualifier problem:

Let $f : \mathbb{C} → \mathbb{C}$ be an entire function with $f(z) \ne 0$ for all $z ∈ \mathbb{C}$. Define U = {z ∈ C : |f(z)| < 1}. Show that all connected component of U is unbounded.

I know that $f$ is holomorphic on all of $\mathbb{C}$

I'm assuming that I have to use consider $\frac{1}{f(x)}$ since $f(z) \ne 0$ anywhere. Any thoughts would be grealy appreciated.

Answer 1

does the maximum modulus principle come into play here? i.e. if a connected component were bounded, then $\frac1{|f(z)|}$ would have to have its maximum on the boundary, which cannot be the case

Answer 2

The maximum modulus principle is indeed the right idea here. Suppose that there is a bounded component $V$ of $U=\{z: |f(z)|<1\}$. Note that $|f|=1$ on $\partial V$.

Since $f$ does not vanish in $\mathbb{C}$, $1/f$ is entire. Moreover, $|1/f| \equiv 1$ on $\partial V$, so by the maximum modulus principle (which can be applied since $V$ is bounded), we get $|1/f| \leq 1$ on $V$. This contradicts the fact that $|f|<1$ on $V$...