Let \[ f(z)=2z^5+8z-1. \] Show that all five zeros of \(f\) lie inside the disk \(D(0,2)\) and that exactly one zero lies inside the disk \(D(0,1)\).
Solution
Let \[ p(z)=2z^5+8z-1. \] On \(|z|=2\), \[ |8z-1|\le 17<64=|2z^5|. \] By Rouché’s theorem, \(p\) and \(2z^5\) have the same number of zeros in \(D(0,2)\), counted with multiplicity, namely five. The strict inequality also shows that \(p\) has no zero on \(|z|=2\).
On \(|z|=1\), \[ |2z^5-1|\le 3<8=|8z|. \] Again by Rouché’s theorem, \(p\) and \(8z\) have the same number of zeros in \(D(0,1)\), counted with multiplicity, namely one. The strict inequality excludes zeros on \(|z|=1\).
Thus all five zeros lie in \(D(0,2)\) and exactly one lies in \(D(0,1)\).