Find the number of zeros in \(\abs{z} < 1\) of \[ p(z) \da z^6 + 9z^4 + z^3 + 2z + 4 .\]
Solution
Strategy: bound the difference. Find the big and small term:
- Big: \(F(z) = 9z^4\), so \(\abs{F(z)} = 9\) on the boundary
- Small: \(g(z) = p(z) - F(z) = z^6 + z^3 + 2z + 4\), so \(\abs{g(z)}\leq 1+1+2+4=8\) on the boundary.
So \(\abs{p-F} \leq \abs{F}\) on \(\abs{z} = 2\), meaning \(Z_{p} = Z_F = 4\).