Show that \(P(z) \da z^4 + 6z + 3\) has 3 zeros in \(\ts{1\leq \abs{z} \leq 2}\).
Solution
Take \(P(z) = z^4 + 6z + 3\).
On \(\abs{z} < 2\):
Set \(f(z) = z^4\) and \(g(z) = 6z + 3\), then \(\abs{g(z)} \leq 6\abs{z} + 3 = 15 < 16= \abs{f(z)}\).
So \(P\) has 4 zeros here.
On \(\abs{z} < 1\):
Set \(f(z) = 6z\) and \(g(z) = z^4 + 3\).
Check \(\abs{g(z)} \leq \abs{z}^4 + 3 = 4 < 6 = \abs{f(z)}\).
So \(P\) has 1 zero here.