Let \[ p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n, \qquad R=\sqrt{|a_0|^2+\cdots+|a_{n-1}|^2+1}. \]
If \(R=1\), show that the zero set of \(p\) is \(\{0\}\).
If \(R>1\) and \(|z|=R\), show, using Cauchy-Schwarz, that \[ |z^n-p(z)|<|z^n|. \]
Conclude that every zero of \(p\) lies in the open disk \(|z|<R\).