How many zeros, counted with multiplicity, does \[ 2z^5-6z^3+z+1 \] have in the annular region \[ 1\le |z|\le2? \]
Solution
Let \[ P(z)=2z^5-6z^3+z+1. \]
1. The polynomial \(P\) has exactly three zeros in \(\abs z<1\), counted with multiplicity.
On \(\abs z=1\), \[ \abs{-6z^3}=6, \] while \[ \abs{2z^5+z+1} \leq 2+1+1 = 4. \] Thus \[ \abs{2z^5+z+1} < \abs{-6z^3} \] on the unit circle. By Rouché’s theorem, \(P\) and \(-6z^3\) have the same number of zeros in \(\abs z<1\), namely three.
2. The polynomial \(P\) has exactly five zeros in \(\abs z<2\), counted with multiplicity.
On \(\abs z=2\), \[ \abs{2z^5}=2\cdot2^5=64, \] while \[ \begin{aligned} \abs{-6z^3+z+1} &\leq 6\cdot2^3+2+1\\ &= 51. \end{aligned} \] Hence \[ \abs{-6z^3+z+1} < \abs{2z^5} \] on \(\abs z=2\). Rouché’s theorem gives the same number of zeros as \(2z^5\), namely five.
3. The polynomial \(P\) has no zero on either boundary circle \(\abs z=1\) or \(\abs z=2\).
Each strict Rouché inequality in steps 1 and 2 implies in particular that the corresponding dominant term cannot be cancelled by the remaining terms. Thus \(P(z)\neq0\) on both circles.
4. The number of zeros of \(P\) in\[ 1\leq\abs z\leq2 \]is\[ \boxed{2}. \]
By step 2 there are five zeros in \(\abs z<2\), and by step 1 three of them lie in \(\abs z<1\). Step 3 shows that neither boundary circle contains zeros. Hence the annulus contains \[ 5-3=2 \] zeros, counted with multiplicity.
5. Q.E.D.
Step 4 is the required count.