Spring 2012 - Problem 11

construction

Let $P\p{z}$ be a polynomial. Show that there is an integer $n$ and a second polynomial $Q\p{z}$ so that

P\p{z}Q\p{z} = z^n \abs{P\p{z}}^2 \quad\text{whenever } \abs{z} = 1.

Write $P\p{z} = a_0 + \cdots + a_{n-1}z^{n-1} + a_nz^n$ . Then

\begin{aligned} \conj{P\p{z}} &= \conj{a_0} + \cdots + \conj{a_{n-1}} \p{\conj{z}}^{n-1} + \conj{a_n} \p{\conj{z}}^n \\ \implies z^n\conj{P\p{z}} &= \conj{a_0}z^n + \cdots + \conj{a_{n-1}}\abs{z}^{n-1}z + \conj{a_n} \abs{z}^n \end{aligned}

If $\abs{z} = 1$ , then

z^n\conj{P\p{z}} = \conj{a_0}z^n + \cdots + \conj{a_{n-1}}z + \conj{a_n}.

Set $Q\p{z} = \conj{a_n} + \cdots + \conj{a_0} z^n$ so that $Q\p{z} = z^n \conj{P\p{z}}$ on $\abs{z} = 1$ . Hence, for these $z$ ,

P\p{z}Q\p{z} = z^nP\p{z} \conj{P\p{z}} = z^n\abs{P\p{z}}^2,

which completes the proof.