Prove that \(z^5 + 3z^3 + 7\) has all its zeros in the disk \(D(0,2) = \{z \in \mathbb{C} : |z| < 2\}\).
Solution
1. The polynomial has five zeros in \(|z|<2\), counted with multiplicity.
On \(|z|=2\), \[ |3z^3+7|\leq3\cdot2^3+7=31<32=|z^5|. \] Both summands are entire. Rouché’s theorem therefore gives the same number of zeros in this circle for \(z^5\) and \(z^5+3z^3+7\), namely five with multiplicity [@SS03, Chapter 3, Theorem 4.3]. The strict estimate also gives \(|z^5+3z^3+7|\geq1\) on \(|z|=2\).
2. There are no further zeros on or outside the circle.
A nonzero degree-five polynomial has at most five zeros counted with multiplicity: each zero of multiplicity \(m\) contributes a factor \((z-a)^m\), and division by these factors cannot remove more than the total degree [@DF04]. Step 1 already accounts for five inside the disk. No additional zero can occur elsewhere, proving the assertion.