How many roots does the following polynomial have in the open disc \(\abs{z} < 1\)? \[ f(z) = z^7 - 4z^3 - 1 .\]
Solution
Big: \(M(z) = -4z^3\). Small: \(m(z) = z^7 - 1\). Then on \(\abs{z} = 1\), \[ \abs{m(z)} = \abs{z^7-1} \leq \abs{z}^7 + 1 = 2 < 4 = \abs{-4z^3} ,\] so \(f\) and \(M\) have the same number of zeros: three.