- How many zeros does the polynomial
\[ z^9 + z^6 + 30 z^5 - 3z + 2 \]
have in the annulus \(\{ 1 \leq | z | \leq 3 \}\) . Justify your answer.
Solution
Let \(P(z)=z^9+z^6+30z^5-3z+2\), and count zeros with multiplicity.
1. \(P\) has \(9\) zeros in \(\abs z<3\) and none on \(\abs z=3\).
On \(\abs z=3\), \(\abs{z^9}=3^9=19683\), while \(\abs{z^6+30z^5-3z+2}\le 3^6+30\cdot3^5+3\cdot3+2=8030\). By Rouché’s theorem \(P\) has as many zeros in \(\abs z<3\) as \(z^9\), namely \(9\), and \(\abs P\ge19683-8030>0\) on the circle.
2. \(P\) has \(5\) zeros in \(\abs z<1\) and none on \(\abs z=1\).
On \(\abs z=1\), \(\abs{30z^5}=30\), while \(\abs{z^9+z^6-3z+2}\le1+1+3+2=7\). By Rouché’s theorem \(P\) has as many zeros in \(\abs z<1\) as \(30z^5\), namely \(5\), and \(\abs P\ge30-7>0\) on the circle.
3. Q.E.D.
By steps 1 and 2, the annulus \(1\le\abs z\le3\) contains \(9-5=\boxed{4}\) zeros.