- Let \(f\) be non-constant and holomorphic in an open set containing the closed unit disc.
Show that if \(|f(z)| = 1\) whenever \(|z| = 1\), then the image of \(f\) contains the unit disc. [Hint: One must show that \(f(z) = w_0\) has a root for every \(w_0 \in \mathbb{D}\). To do this, it suffices to show that \(f(z) = 0\) has a root (why?). Use the maximum modulus principle to conclude.]
If \(|f(z)| \geq 1\) whenever \(|z| = 1\) and there exists a point \(z_0 \in \mathbb{D}\) such that \(|f(z_0)| < 1\), then the image of \(f\) contains the unit disc.
Solution
- First, \(f\) must vanish somewhere in \(\mathbb D\). Otherwise \(1/f\) would be holomorphic on a neighborhood of the closed disc and satisfy \(|1/f|=1\) on \(|z|=1\). The maximum modulus principle applied to both \(f\) and \(1/f\) would give \(|f|\le1\) and \(|f|\ge1\) in \(\mathbb D\), so \(|f|\equiv1\) there; then \(f\) would be constant, contrary to hypothesis.
Now fix \(w_0\in\mathbb D\). On \(|z|=1\), \[ |w_0|<1=|f(z)|. \] By Rouché’s theorem, \(f\) and \(f-w_0\) have the same number of zeros in \(\mathbb D\). Since \(f\) has at least one, so does \(f-w_0\). Thus \(w_0\in f(\mathbb D)\), and the image contains the unit disc.
- Choose \(z_0\in\mathbb D\) with \(|f(z_0)|<1\). If \(f\) had no zero in \(\mathbb D\), then \(1/f\) would be holomorphic there and continuous on the closed disc. Since \(|f|\ge1\) on the boundary, \(|1/f|\le1\) there; the maximum modulus principle would give \(|1/f|\le1\) throughout the disc, contradicting \(|f(z_0)|<1\). Hence \(f\) has a zero in \(\mathbb D\).
For any \(w_0\in\mathbb D\), on \(|z|=1\) we have \[ |w_0|<1\le|f(z)|. \] Rouché again shows that \(f-w_0\) has as many zeros as \(f\), hence at least one. Therefore \(f(\mathbb D)\) contains \(\mathbb D\).