- How many zeros does the polynomial
\[ z ^ { 7 } - 4 z ^ { 3 } + z - { \textstyle { \frac { 1 } { 2 } } } \]
have in the unit disk \(\{|z|<1\}\)? How many zeros does it have in the disk \(\{|z|<2\}\) of radius two? Justify your answers.
Solution
Write \(p(z)=z^7-4z^3+z-1/2\). The requested zero counts, with multiplicities, are \(\boxed{3}\) in \(|z|<1\) and \(\boxed{7}\) in \(|z|<2\).
1. The cubic term determines the unit-disk count.
For \(|z|=1\), \[ |p(z)-(-4z^3)|=|z^7+z-1/2| \leq1+1+\frac12=\frac52<4=|-4z^3|. \] Rouché’s theorem applies because both polynomials are holomorphic on a neighborhood of the closed disk [@SS03]. It gives the same number of zeros for \(p\) and \(-4z^3\) in that disk. The latter has a zero of multiplicity three at zero and no other zero. The strict inequality also gives \(|p|\geq4-5/2>0\) on the boundary.
2. The leading term determines the radius-two count.
For \(|z|=2\), \[ |p(z)-z^7|=|-4z^3+z-1/2| \leq32+2+\frac12=\frac{69}{2}<128=|z^7|. \] Another application of Rouché’s theorem gives seven zeros in \(|z|<2\), counted with multiplicity, the same count as for \(z^7\) [@SS03]. The strict comparison excludes zeros on \(|z|=2\). Thus the two counts refer to the open disks specified in the question, with no boundary ambiguity.