For each \(n\in \ZZ^{\geq 1}\), let \[ P_n(z) = 1 + z + {1\over 2!} z^2 + \cdots + {1\over n!}z^n .\] Show that for sufficiently large \(n\), the polynomial \(P_n\) has no zeros in \(\abs{z} < 10\), while the polynomial \(P_n(z) - 1\) has precisely 3 zeros there.
Solution
Goal: With \(P_n(z) = \sum_{k=0}^n \frac{z^k}{k!}\), show that for all large \(n\), \(P_n\) has no zeros in \(\abs{z} < 10\) and \(P_n - 1\) has exactly 3 zeros there.
1. \(P_n \to e^z\) uniformly on the circle \(\abs{z} = 10\).
\(e^z - P_n(z) = \sum_{k=n+1}^\infty \frac{z^k}{k!}\), and \(\abs{e^z - P_n(z)} \leq \sum_{k=n+1}^\infty \frac{10^k}{k!} \to 0\) as \(n \to \infty\) (the tail of the convergent series \(e^{10} = \sum 10^k/k!\)).
2. On \(\abs{z} = 10\), \(\abs{e^z} \geq e^{-10} > 0\).
\(\abs{e^z} = e^{\Re z} \geq e^{-\abs{z}} = e^{-10}\).
3. For all sufficiently large \(n\), \(P_n\) has no zeros in \(\abs{z} < 10\).
Choose \(n\) so that \(\abs{e^z - P_n(z)} < e^{-10} \leq \abs{e^z}\) on \(\abs{z} = 10\) (possible by 1 and <1>2). Rouch'e’s theorem applied to \(e^z\) and \(P_n - e^z\) shows \(P_n\) and \(e^z\) have equally many zeros in \(\abs{z} < 10\); \(e^z\) has none, so neither does \(P_n\).
4. \(e^z - 1\) has exactly 3 zeros in \(\abs{z} < 10\), all simple.
The zeros of \(e^z - 1\) are \(z = 2\pi i k\), \(k \in \ZZ\); those with \(\abs{z} < 10\) are \(k = -1, 0, 1\), i.e. \(z = 0, \pm 2\pi i\), and each is simple since \(\ddd{z}(e^z - 1) = e^z \neq 0\) at each.
5. On \(\abs{z} = 10\), \(\abs{e^z - 1} \geq m\) for some \(m > 0\).
\(e^z - 1\) is continuous on the compact circle and has no zeros there (all zeros lie strictly inside by 4, since \(\abs{2\pi i} \approx 6.28 < 10\) and \(\abs{4\pi i} \approx 12.57 > 10\)), so its minimum \(m\) on the circle is positive.
6. For all sufficiently large \(n\), \(P_n - 1\) has exactly 3 zeros in \(\abs{z} < 10\).
Write \(P_n - 1 = (e^z - 1) - (e^z - P_n)\). Choose \(n\) so that \(\abs{e^z - P_n} < m\) on \(\abs{z} = 10\) (by <1>1); then Rouch'e with \(f = e^z - 1\) and \(g = -(e^z - P_n)\) gives that \(P_n - 1\) and \(e^z - 1\) have equally many zeros in \(\abs{z} < 10\), namely 3 by 4.
7. Q.E.D.
3 and 6 together establish both claims for all sufficiently large \(n\).