Prove that \[ p(z)=z^{47}-z^{23}+2z^{11}-z^5+4z^2+1 \] has at least one root in the disk \(|z|<1\).
Solution
Write \[ p(z)=\sum_{k=0}^{47}a_kz^k, \] and let \(\alpha_1,\ldots,\alpha_{47}\) be the roots of \(p\), counted with multiplicity.
1. If \(p\) has no root in the disk \(\abs{z}<1\), then\[ \abs{\alpha_j}\geq 1 \]for every \(j\).
This is exactly the negation of the existence of a root with modulus strictly less than \(1\).
2. Under the assumption in step 1, every root of \(p\) lies on theunit circle.
The polynomial is monic of odd degree \(47\) and has constant term \(1\). Vieta’s formula therefore gives \[ 1 = (-1)^{47}\prod_{j=1}^{47}\alpha_j = -\prod_{j=1}^{47}\alpha_j. \] Hence \[ \prod_{j=1}^{47}\abs{\alpha_j}=1. \] By step 1, every factor in this product is at least \(1\), so every factor must equal \(1\).
3. Define the reciprocal-conjugate polynomial\[ p^*(z) \coloneqq \sum_{k=0}^{47}\overline{a_k}z^{47-k}. \]For \(z\neq0\),\[ p^*(z)=z^{47}\overline{p\left(1/\overline z\right)}. \]Under the assumption in step 1, one has\[ p^*(z)=p(z). \]
The leading coefficient of \(p^*\) is \(\overline{a_0}=1\), so \(p^*\) is monic of degree \(47\). Its roots, counted with multiplicity, are \[ \frac{1}{\overline{\alpha_1}},\ldots, \frac{1}{\overline{\alpha_{47}}}. \] By step 2, \(\abs{\alpha_j}=1\), hence \[ \frac{1}{\overline{\alpha_j}}=\alpha_j \] for every \(j\). Thus \(p\) and \(p^*\) are monic polynomials of the same degree with the same roots counted with multiplicity, so they are equal.
4. The equality in step 3 is impossible.
Comparing the coefficient of \(z^{23}\) in \[ p^*(z)=\sum_{k=0}^{47}\overline{a_k}z^{47-k} \] with the coefficient of \(z^{23}\) in \(p\) gives \[ a_{23}=\overline{a_{24}}. \] For the given polynomial, \[ a_{23}=-1 \qquad\text{and}\qquad a_{24}=0, \] which is a contradiction.
5. The polynomial \(p\) has at least one root with\[ \abs{z}<1. \]
If no such root existed, step 1 would hold, and step 4 would give a contradiction.
6. Q.E.D.
Step 5 is the required conclusion.