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
More is true: this will hold for any disc of arbitrary radius \(R\), with \(n\) depending on \(R\). Fix \(R\), then use that \(P_n(z) \convergesto{n\to\infty} e^z\) uniformly on the compact disc \(\abs{z} \leq R\). Consequently, setting \(g_n(z) \da {P_n(z)\over e^z}\), we have \(g_n(z) \to 1\) uniformly on this disc, for any \(\eps> 0\) this can be used to produce an \(n\gg 1\) such that \(\abs{ g_n(z) - 1 } < \eps\) for all \(\abs{z} \leq R\).
So take \(\eps \da 1\) and define \(h(z) \da 1\), then for \(\abs{z} = R\) \[ \abs{g_n(z) - 1} < 1 = \abs{h(z)} ,\] so by Rouché, \[ 0 = \size Z_{h} = \size Z_{h + (g_n - 1)} = \size Z_{g_n} ,\] since \(h\) has no zeros at all. Take \(R=10\) to get the stated result.
For \(P_n(z) - 1\), note that \(e^z-1=0\) has three solutions in \(\abs{z} < 10\), namely \(z=0, \pm 2\pi i\). We similarly have \(P_n(z)-1\to e^z-1\) uniformly, so on a disc of radius \(R\) choose \(n\) large enough so that \[ \abs{{P_n(z) -1 \over e^z - 1} - 1} &< 1 \\ \implies \abs{ (P_n(z) - 1) - (e^z-1) \over e^z-1} &< 1 \\ \implies \abs{ (P_n(z) - 1) - (e^z-1)} &< \abs{e^z-1} \\ \da \abs{m(z)} &< \abs{M(z)} ,\] so \[ 3 = \size Z_M = \size Z_{M+m} = \size Z_{P_n - 1} .\]