- How many zeros does the polynomial
\[ z ^ { 9 } + z ^ { 5 } - 8 z ^ { 3 } - z + 2 \]
have between the circles \(\{ | z | = 1 \}\) and \(\{ | z | = 2 \}\) . Justify your answer.
Solution
There are \(\boxed{6}\) zeros between the circles, counted with multiplicity. Let \(p(z)=z^9+z^5-8z^3-z+2\).
1. The unit disk contains three zeros.
On \(|z|=1\), \[ |z^9+z^5-z+2|\leq1+1+1+2=5<8=|-8z^3|. \] Rouché’s theorem gives the same number of zeros for \(p\) and \(-8z^3\) in this disk, namely three counted with multiplicity [@SS03]. The strict inequality also gives \(|p(z)|\geq3\) on the circle, so there are no boundary zeros.
2. The radius-two disk contains nine zeros, leaving six in the annulus.
On \(|z|=2\), \[ |z^5-8z^3-z+2|\leq32+64+2+2=100<512=|z^9|. \] Rouché’s theorem now compares \(p\) with \(z^9\), giving nine zeros in \(|z|<2\), with multiplicity [@SS03]. Again the strict inequality excludes zeros on the boundary. Subtracting the three inside \(|z|<1\) gives \(9-3=6\) in \(1<|z|<2\). Since neither circle contains a zero, including the boundary circles would not change this answer.