Let \(\Delta=\{z\in\mathbb C:|z|<1\}\), and let \(f : \Delta \to \Delta\) be a holomorphic function with \(f(0) = 0\) and \(|f'(0)| = M\). If \(0 \neq w \in \Delta\) is any other zero of \(f(z)\), show that:
\[\frac{M}{1 + M} \leq |w|.\]
Solution
In fact the stronger estimate \[ \boxed{M\le |w|} \] holds.
1. Dividing by the fixed zero produces a holomorphic disk map.
Because \(f(0)=0\), define \[ g(z)=\begin{cases}f(z)/z,&z\ne0,\\ f'(0),&z=0.\end{cases} \] The Taylor expansion of \(f\) at zero shows that \(g\) is holomorphic on \(\Delta\). Schwarz’s lemma [@SS03] applied to \(f:\Delta\to\Delta\) gives \(|f(z)|\le |z|\), so \(|g(z)|\le1\) throughout the disk. Since \(w\ne0\) and \(f(w)=0\), one has \(g(w)=0\). Thus \(g\) is not a constant of modulus one. The maximum modulus principle therefore gives \(|g(z)|<1\) for every \(z\in\Delta\); hence \(g:\Delta\to\Delta\) is a holomorphic self-map.
2. Schwarz-Pick gives a stronger inequality than required.
Apply the Schwarz-Pick inequality [@SS03] to \(g\) at the two points \(0\) and \(w\): \[ \left|\frac{g(0)-g(w)}{1-\overline{g(w)}g(0)}\right| \le \left|\frac{0-w}{1-\overline w\,0}\right|. \] Since \(g(0)=f'(0)\) and \(g(w)=0\), this reduces to \[ M=|f'(0)|\le |w|. \] Finally \(M\ge0\), so \[ \frac{M}{1+M}\le M\le |w|. \] This is the claimed estimate.