Let \(0<r<1\). Show that polynomials \(P_n(z) = 1 + 2z + 3 z^2 + \cdots + n z^{n-1}\) have no zeros in \(|z|<r\) for all sufficiently large \(n\)’s.
Solution
Key observation: \[ P_n(z) = \sum_{1\leq k\leq n-1} kz^{k-1} = \dd{}{z}Q_n(z) \qquad Q(z) \da \sum_{0\leq k \leq n} z^k .\] Note that \(Q(z) \to \sum_{k\geq 0} z^k = {1\over 1-z}\) uniformly on \(\abs{z} \leq R < 1\) since this power series has radius of convergence 1. Similarly \(P_n(z)\) converges uniformly to \(\dd{}{z}{1\over 1-z} = {1\over (1-z)^2}\), so let \(P(z) \da {1\over (1-z)^2}\). Note that \(P\) is nonvanishing in \(\DD\).
Strategy:
- Small: \(m(z) = P(z) - P_n(z)\), look for an upper bound \(\abs{m(z)} < U\)
- Big: \(P(z)\), look for a lower bound \(L\) with \(\abs{P(z)} > L > U\).
Just by considering the geometry of circles of radius \(R < 1\) and \(1\) and measuring distances to the point \(1\), we can estimate \[ 0 < 1-R \leq \abs{1-z} < 1+R < 2 \implies \abs{P(z)} = {1\over \abs{1-z}^2} \geq {1\over 2^2} = {1\over 4} .\]
Now fix \(\eps < {1\over 4}\) and use uniform convergence of \(P_n\to P\) to produce an \(N\) such that \(n\geq N\) implies \(\norm{P-P_n}_\infty < \eps\) in \(\abs{z} \leq R\). Then on \(\abs{z} = R\), for \(n\geq N\), \[ \abs{m(z)} \da \abs{P(z) - P_n(z)} \leq \norm{P - P_n}_\infty < \eps < {1\over 4} \leq \abs{P(z)} = \abs{M(z)} ,\] so \(0 = Z_P = Z_{P_n}\) by Rouché.