Question: Let $f\colon \mathbb{C} \to \mathbb{C}$ be an entire function. Suppose that there are real constants $r, R > 0$ so that $|f(z)| > R$ for all $z \in \mathbb{C}$ with $|z| > r$. Show that either $f$ is a constant or that $f$ has a zero in the disk $\{z \colon |z| < r\}$.
My attempt: Suppose $f$ has no zeros in the disk $\{z \colon |z| < r\}$ and if for a contradiction that $f$ is not a constant. Then, the set of zeros of $f$ lies in the circle $\{z\colon |z| = r\}$, if $f$ attains no zeros in $\mathbb{C}$ then $g\colon \mathbb{C} \to \mathbb{C}$ defined by $g(z) = 1/f(z)$ is entire and bounded so by Liouville's theorem $f$ is a constant, contradicting the assumption.
Given any $z_0 \in \{z\colon |z| = r\}$, let $(z_n)_n \subset \{|z| > r\}$ so that $z_n \to z_0$. By continuity, we have $$|f(z_0)| = \lim_{n\to \infty} |f(z_n)| \geq R$$ Since $z_0\in \{z\colon |z| = r\}$ is arbitrary, it is impossible for $f$ to attain any zeros in $\{z\colon |z| = r\}$.