Prove that \[ \max_{\abs z = 1} \abs{a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + z^n} \geq 1 .\]
Hint: the first part of the problem asks for a statement of Rouche’s theorem.
Solution
Rouché’s theorem: if \(f,g\) are holomorphic on a domain containing a simple closed contour \(\gamma\) and \(\abs{g(z)} < \abs{f(z)}\) for all \(z\in\gamma\), then \(f\) and \(f+g\) have the same number of zeros (counting multiplicity) inside \(\gamma\).
Write \(p(z) \da a_0 + \cdots + z^n\). Toward a contradiction, suppose not so that \(\abs{p(z)} < 1\) on \(\abs{z} = 1\). Then \[ \abs{f(z)} < 1 = \abs{z}^n \qquad \text{ on } \abs{z} = 1 .\] Taking \(m(z) \da f(z)\) and \(M(z) \da -z^n\), we have \[ n = \size Z_M = \size Z_{M+m} = \size Z_{f(z) - z^n} \leq n-1 ,\] since \(f(z) - z^n\) is degree at most \(n-1\), a contradiction.