Given a polynomial equation
\[ x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 = 0 \]
with real or complex coefficients. Show that if \(\abs{a_{n-1}} + \cdots + \abs{a_1} + \abs{a_0} < 1\), then all the roots of the equation lie interior to the unit ball \(B^2\). [Hint: Let \(g(x) = 1 + a_{n-1}x + \cdots + a_1 x^{n-1} + a_0 x^n\), and show that \(g(x) \neq 0\) for \(x \in B^2\).]
Solution
Let \[ p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0 \] and assume \[ |a_{n-1}|+\cdots+|a_0|<1. \] Suppose \(p(z)=0\) and \(|z|\ge1\). Since \(z\ne0\), divide by \(z^n\): \[ 1=-\left(\frac{a_{n-1}}z+\frac{a_{n-2}}{z^2}+\cdots+\frac{a_0}{z^n}\right). \] Taking absolute values and using \(|z|\ge1\), \[ 1\le \frac{|a_{n-1}|}{|z|}+\frac{|a_{n-2}|}{|z|^2}+\cdots+\frac{|a_0|}{|z|^n} \le |a_{n-1}|+\cdots+|a_0|<1, \] a contradiction. Hence every root satisfies \(|z|<1\), so all roots lie in the interior of the unit disk.
Equivalently, with \(w=1/z\), the polynomial from the hint \[ g(w)=1+a_{n-1}w+\cdots+a_0w^n \] has no zero with \(|w|\le1\), since the nonconstant part has absolute value strictly less than \(1\).