Find the number of zeros in \(\abs{z} < 2\) of \[ h(z) \da z^5 + 3z + 1 .\]
Solution
Strategy: bound the difference.
- Big: \(F(z) \da z^5\) so \(\abs{F(z)} = 2^5 = 32\) on \(\abs{z} = 2\)
- Small: \(g(z) \da p(z) - F(z) = 3z+1\), so \(\abs{g(z)} \leq 3\abs{z}+ 1 = 7\) on \(\abs{z} = 2\).
Then \(\abs{g}\leq \abs{F}\) on \(\abs{z} = 2\), \(Z_{p} = Z_F = 5\).