Find the number of zeros in \(\abs{z} \in (1, 2)\) of \[ f(z) \da z^4 + 5z + 3 .\] Note the strict inequality.
Solution
On \(\abs{z} = 1\):
- Big: \(M(z) = 5z\)
- Small: \(m(z) = z^4 + 3\)
- The estimate: \[ \abs{m(z)} \leq \abs{z}^4 + 3 = 4 < 5 = \abs{M(z)} \implies 1 = \size Z_M = \size Z_f .\]
On \(\abs{z} = 2\):
- Big: \(M(z) = z^4\)
- Small: \(m(z) = 5z + 3\)
- The estimate: \[ \abs{m(z)} \leq 5\abs{z} + 3 = 13 < 16 = 2^4 = \abs{M(z)} \implies 4 = \size Z_M = \size Z_f .\]
So \(f\) has \(4-1=3\) zeros on \(\abs{z} \in (0, 2)\sm (0, 1) = [1, 2)\). To get the strict inequality, it suffices to show \(f\) has no zeros on \(\abs{z} = 1\). Estimate a lower bound: \[ \abs{z^4 + 5z + 3} \geq \abs{\abs{5z+3} - \abs{z}^4 } = \abs{\abs{5z+3} - 1 } \geq \abs{ 2-1} > 0 .\] The nontrivial bound at the end comes from the fact that \(5z+3 = 3 + 5e^{it}\) whose modulus ranges between \(3-5 = -2\) and \(3+5 = 8\).