Let \(P(x)\) be a monic polynomial with real coefficients. Suppose \[ P(0)=-1 \] and \(P\) has no complex zeros in the open unit disk. Prove that \[ P(1)=0. \]
Solution
Let \(d=\deg P\), and let \[ z_1,\ldots,z_d \] be the complex roots of \(P\), counted with multiplicity.
1. Every root of \(P\) has modulus \(1\).
Since \(P\) is monic, \[ P(0)=(-1)^d\prod_{j=1}^d z_j=-1. \] Taking absolute values gives \[ \prod_{j=1}^d\abs{z_j}=1. \] By hypothesis no root lies in the open unit disk, so \(\abs{z_j}\ge1\) for every \(j\). A finite product of numbers at least \(1\) can equal \(1\) only if every factor equals \(1\). Thus \[ \abs{z_j}=1 \qquad(1\le j\le d). \]
2. At least one root of \(P\) equals \(1\).
Because \(P\) has real coefficients, every nonreal root occurs with its complex conjugate, with the same multiplicity. By step 1, each nonreal conjugate pair has product \[ z\overline z=\abs z^2=1. \] Every real root on the unit circle is either \(1\) or \(-1\).
Suppose, for contradiction, that \(1\) is not a root. Then every real root is \(-1\). Let \(r\) be the number of real roots counted with multiplicity. The remaining \(d-r\) roots occur in conjugate pairs, so \[ d\equiv r\pmod2. \] The product of all roots is then \((-1)^r\). On the other hand, the constant-term identity gives \[ \prod_{j=1}^d z_j=(-1)^{d+1}. \] Since \(d\) and \(r\) have the same parity, these two values are opposites, a contradiction.
3. \(P(1)=0\).
By step 2, \(1\) is a root of \(P\).
4. Q.E.D.
Step 3 proves the assertion.