How many zeros does the complex polynomial
\[ 3z^9+8z^6+z^5+2z^3+1 \]
have in the annulus \(1<\abs{z}<2\)?
Solution
Let \[ p(z)=3z^9+8z^6+z^5+2z^3+1. \] All zero counts below are with multiplicity.
1. \(p\) has six zeros in \(\abs{z}<1\) and no zero on \(\abs{z}=1\).
On \(\abs{z}=1\), \[ \abs{8z^6}=8, \] while \[ \abs{3z^9+z^5+2z^3+1} \le3+1+2+1=7<8. \] By Rouché’s theorem, \(p\) and \(8z^6\) have the same number of zeros in \(\abs{z}<1\), namely six. The strict inequality also gives \(\abs{p(z)}\ge8-7>0\) on \(\abs{z}=1\).
2. \(p\) has nine zeros in \(\abs{z}<2\).
On \(\abs{z}=2\), \[ \abs{3z^9}=3\cdot2^9=1536, \] while \[ \begin{aligned} \abs{8z^6+z^5+2z^3+1} &\le 8\cdot2^6+2^5+2\cdot2^3+1\\ &=512+32+16+1\\ &=561<1536. \end{aligned} \] By Rouché’s theorem, \(p\) and \(3z^9\) have the same number of zeros in \(\abs{z}<2\), namely nine.
3. \(p\) has\[ \boxed{3} \]zeros in the annulus \(1<\abs{z}<2\).
By step 1, every zero in \(\abs{z}<2\) lies either in \(\abs{z}<1\) or in the annulus \(1<\abs{z}<2\). Steps 1 and 2 give \(9-6=3\) zeros in the annulus.
4. Q.E.D.
Step 3 gives the requested number.