Show that if \(z_0\) is a zero of \(f'\) of order \(n-1\), then \(f\) is \(n\)-to-one in a neighborhood of \(z_0\).
Solution
Wlog, assume \(z_0 = 0\). We want to show that there exists discs \(U = \DD_r(0)\) and \(W = \DD_R(0)\) such that the fiber of \(f:U\to W\) has exactly \(n\) distinct points. Since \(0\) is a zero of order \(n\), expand \(f\) as \(\sum_{k\geq n} c_k z^k = z^n\sum_{k\geq 0} c_{k+n}z^k\). By dividing coefficients through, we may assume \(c_n = 1\), so \[ f(z) = z^n + \qty{ c_{n+1} z^{n+1} + c_{n+2}z^{n+2} + \cdots} = z^n + z^{n+1} \sum_{k\geq 0} c_{k+n+1}z^k \da z^n + g(z) .\]
By Rouché, \(f(z)\) and \(z^n\) have the same number of zeros in a small disc \(\DD_\rho(0)\).
(of claim)
Write \(m(z) = \sum_{k\geq 0}c_{k+n}z^k\) and \(M(z) = z^n\); then if \(\abs{m(z)} < \abs{M(z)}\) for any circle \(\abs{z} = \rho\) with \(\rho< 1\) then \(M\) and \(m+M = f\) will have the same number of zeros (\(n\) with multiplicity).
Bounding \(m\), the tail of the Laurent series of \(f\): by Cauchy’s integral formula, on a disc of radius \(R\), \[ c_k = {f^{(n)}(z_0) \over n!} = {1\over 2\pi i} \oint_{\abs{\xi} = R} { f(\xi) \over (\xi - z_0)^{n+1} } \dxi ,\] so \[ \abs{c_k} \leq \max_{\abs{\xi} = R}\abs{f(\xi)} R^{-k} \da {M_R \over R^{k}} .\]
We can now estimate \(g\): \[ \abs{g(z)} &= \abs{z^n \sum_{k\geq 0} c_{k+n+1} z^k} \\ &\leq \abs{z}^n \sum_{k\geq 0} \abs{ c_{k+n+1}} \abs{z}^k \\ &\leq \sum_{k\geq 0} {M_R \over R^{k+n+1}} \rho^k \\ &= \abs{z}^n {M_R \over R^{n+1}} \sum_{k\geq 0} \qty{\rho\over R}^k \\ &= \abs{z}^n {M_R \over R^{n+1}} \qty{1\over 1- {\rho \over R}} \\ &= \abs{z}^n {M_R \over R^{n+1}} {R\over R-\rho} \\ &= \abs{z}^n \qty{ {M_R\over R^n( R-\rho)} } \\ &\da \abs{z}^n C_{R, \rho} ,\] and \(R, \rho\) can be chosen such that \(C_{R, \rho} < 1\).
Thus on \(\abs{z} = \rho\), \[ \abs{m(z)} = \abs{g(z) } \leq C_{R, \rho} \abs{z} < \abs{z} = \abs{M(z)} .\]
So the fiber above \(z=0\) is of size \(n\), the claim is that this is also true in a neighborhood of zero. The above estimate also shows that for \(0 < \abs{z}\leq \rho\), \(\abs{g(z)} \leq \abs{z^n}\), and so \[ \abs{g(z)} = \abs{g(z) - z^n + z^n} \geq \abs{ \abs{g(z) - z^n} - \abs{z^n} } > 0 ,\] so \(g\) is nonzero on \(\DD_\rho(0)\smz\). For the zero-counting function \[ F(w) \da {1\over 2\pi i} \oint_{\abs{\xi} = \rho'} {f'(\xi) \over f(\xi) - w }\dxi .\] Taking \(\rho ' < \min_{\abs{\xi} = \rho} \abs{f(z)}\) makes this a holomorphic function of \(w\) on \(\DD_{\rho'}(0)\), and as a continuous \(\ZZ\dash\)valued function it is constant. Since \(F(0) = n\), this forces \(F(w) = n\) for all \(\abs{w} < \rho'\), so there are \(n\) solutions to \(f(z) = w\) in these discs. After shrinking these discs if necessary, \(f'\neq 0\) is nonvanishing on a punctured disc, so \(f\) is injective there and these solutions are distinct.