Let \(f\) be non-constant and holomorphic in an open set containing the closed unit disc.
- Show that if \(\abs{f(z)} = 1\) whenever \(\abs{z} = 1\), then the image of \(f\) contains the unit disc.
Hint: Show that \(f(z) = w_0\) has a root for every \(w_0 \in \DD\), for which it suffices to show that \(f(z) = 0\) has a root. Conclude using the maximum modulus principle.
- If \(\abs{f(z)} \geq 1\) whenever \(\abs{z} = 1\) and there exists a \(z_0\in \DD\) such that \(\abs{f(z_0)} < 1\), then the image of \(f\) contains the unit disc.
Solution
For (a), first show that \(f\) has a zero in \(\mathbb D\). If it did not, then \(1/f\) would be holomorphic in a neighborhood of \(\overline{\mathbb D}\) and \(|1/f|=1\) on \(\partial\mathbb D\). The maximum-modulus principle applied to \(f\) and \(1/f\) would give both \(|f|\le1\) and \(|f|\ge1\) in \(\mathbb D\); hence \(|f|\equiv1\), forcing \(f\) to be constant, contrary to hypothesis.
Now fix \(w\in\mathbb D\). On \(|z|=1\), \[ |w|<1=|f(z)|. \] Rouché’s theorem shows that \(f-w\) and \(f\) have the same number of zeros in \(\mathbb D\). Since \(f\) has at least one zero, so does \(f-w\). Thus every \(w\in\mathbb D\) lies in \(f(\mathbb D)\).
For (b), if \(f\) had no zero in \(\mathbb D\), then \(1/f\) would be holomorphic there and satisfy \(|1/f|\le1\) on the boundary. The maximum principle would give \(|f|\ge1\) throughout \(\mathbb D\), contradicting the given point \(z_0\). Hence \(f\) has a zero. For any \(|w|<1\) we have on the boundary \[ |w|<1\le|f(z)|, \] so Rouché again implies that \(f-w\) has a zero. Therefore \[ \boxed{\mathbb D\subset f(\mathbb D)} \] in both cases.