Show that if \(z_0\) is a zero of \(f'\) of order \(n\), then \(f\) is \((n+1)\)-to-one in a neighborhood of \(z_0\).
Solution
Goal: Show that if \(z_0\) is a zero of \(f'\) of order \(n\), then \(f\) is \((n+1)\)-to-one in a neighborhood of \(z_0\) (for \(z\) near \(z_0\), the equation \(f(z) = w\) has exactly \(n+1\) solutions in a punctured neighborhood of \(z_0\), for \(w\) near \(f(z_0)\), \(w \neq f(z_0)\)).
1. Setup: expand \(f\) about \(z_0\).
Since \(z_0\) is a zero of \(f'\) of order \(n\), we have \(f'(z) = (z - z_0)^n h(z)\) with \(h\) analytic near \(z_0\), \(h(z_0) \neq 0\).
Integrating, \(f(z) - f(z_0) = (z - z_0)^{n+1} \phi(z)\) where \(\phi\) is analytic near \(z_0\) and \(\phi(z_0) = h(z_0)/(n+1) \neq 0\).
2. \(\phi\) is nonzero on a neighborhood of \(z_0\), so \(\phi\) has an \((n+1)\)-st root \(\psi\) with \(\psi^{n+1} = \phi\), \(\psi\) analytic, \(\psi(z_0) \neq 0\).
\(\phi(z_0) \neq 0\) by 1; by continuity \(\phi \neq 0\) in a disk around \(z_0\), where a holomorphic branch of the \((n+1)\)-st root exists (the disk is simply connected and \(\phi\) avoids \(0\)).
3. Define \(\zeta(z) = (z - z_0)\psi(z)\); then \(\zeta\) is a biholomorphism from a neighborhood of \(z_0\) onto a neighborhood of \(0\), with \(\zeta(z_0) = 0\) and \(\zeta'(z_0) = \psi(z_0) \neq 0\).
\(\zeta'(z_0) = \psi(z_0) \neq 0\), so the inverse function theorem gives the local biholomorphism.
4. In the \(\zeta\) coordinate, \(f(z) = f(z_0) + \zeta(z)^{n+1}\).
1 and 2: \(f(z) - f(z_0) = (z - z_0)^{n+1}\phi(z) = ((z - z_0)\psi(z))^{n+1} = \zeta(z)^{n+1}\).
5. The map \(w \mapsto f(z_0) + w^{n+1}\) is \((n+1)\)-to-one on a punctured neighborhood of \(0\).
For \(w' \neq 0\) near \(0\), the equation \(w^{n+1} = w'\) has exactly \(n+1\) distinct solutions (the \(n+1\) roots of \(w'\)), all lying in a small disk around \(0\).
6. Q.E.D.
3 shows the change of variables \(z \leftrightarrow \zeta\) is one-to-one, and 5 shows \(f\) in these coordinates is \((n+1)\)-to-one; composing, \(f\) is \((n+1)\)-to-one near \(z_0\).