Prove that the following polynomial has its roots outside of the unit circle: \[ p(z) = z^3 + 2z + 4 .\]
Hint: What is the maximum value of the modulus of the first two terms if \(\abs{z} \leq 1\)?
Solution
Goal: Prove that all roots of \(p(z) = z^3 + 2z + 4\) lie outside the unit circle.
1. On \(\abs{z} = 1\), the modulus of the first two terms is at most \(3\).
\(\abs{z^3 + 2z} \leq \abs{z}^3 + 2\abs{z} = 1 + 2 = 3\) by the triangle inequality.
2. On \(\abs{z} = 1\), \(\abs{z^3 + 2z} < \abs{4}\).
By 1, \(\abs{z^3 + 2z} \leq 3 < 4 = \abs{4}\).
3. \(p(z) = z^3 + 2z + 4\) has the same number of zeros in \(\abs{z} < 1\) as the constant \(4\).
Apply Rouch'e’s theorem to \(f(z) = 4\) and \(g(z) = z^3 + 2z\) on the circle \(\abs{z} = 1\); the strict inequality \(\abs{g(z)} < \abs{f(z)}\) from 2 holds on the whole circle, so \(f\) and \(f + g = p\) have equally many zeros inside.
4. \(p\) has no zeros in \(\abs{z} < 1\) and none on \(\abs{z} = 1\).
The constant \(4\) has no zeros, so by 3 neither does \(p\) inside the circle; and 2 rules out zeros on the circle itself (there \(\abs{p(z)} \geq \abs{4} - \abs{z^3 + 2z} \geq 4 - 3 = 1 > 0\)).
5. Q.E.D.
4 shows every root of \(p\) satisfies \(\abs{z} > 1\), i.e. lies strictly outside the unit circle.