Prove that \[ \max_{\abs{z}=1} \abs{a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + z^n} \ge 1. \]
Hint: The first part of the problem asks for a statement of Rouché’s Theorem.
Solution
Set \[ p(z) = a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n. \]
1. Suppose, for contradiction, that\[ \max_{\abs{z}=1}\abs{p(z)}<1. \]Then on the unit circle,\[ \abs{p(z)}<\abs{z^n}. \]
If \(\abs{z}=1\), then \[ \abs{z^n}=1. \] The assumed strict upper bound on \(\abs{p(z)}\) therefore gives the displayed inequality.
2. Under the assumption of step 1, the polynomial\[ q(z) = z^n-p(z) = -a_0-a_1z-\cdots-a_{n-1}z^{n-1} \]has exactly \(n\) zeros in the open unit disk, counting multiplicity.
1. Rouché’s theorem implies that \(q\) and \(z^n\) have the samenumber of zeros in the unit disk. The polynomial \(z^n\) has exactly \(n\),counting multiplicity.
3. The conclusion of step 2 is impossible.
The polynomial \(q\) has degree at most \(n-1\). It is not identically zero: if \(q\equiv0\), then \(p(z)=z^n\), which has \[ \abs{p(z)}=1 \] on the unit circle, contradicting step 1.
A nonzero polynomial of degree at most \(n-1\) has at most \(n-1\) zeros in \(\CC\), counting multiplicity. It therefore cannot have the \(n\) zeros in the unit disk asserted by step 2.
4. Therefore\[ \boxed{ \max_{\abs{z}=1} \abs{ a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n } \geq1. } \]
Steps <1>1–<1>3 show that the strict inequality \[ \max_{\abs{z}=1}\abs{p(z)}<1 \] leads to a contradiction. Hence the maximum is at least \(1\).
5. Q.E.D.
Step 4 is the required inequality.