How many zeros, counted with multiplicity, does \[ p(z)=z^5+z^3+5z^2+2 \] have in the annulus \[ 1<|z|<2? \]
Solution
1. The polynomial \(p\) has exactly five zeros in\[ \abs{z}<2, \]counted with multiplicity.
On the circle \(\abs{z}=2\), \[ \abs{z^5}=32, \] while \[ \abs{z^3+5z^2+2} \le 8+20+2 =30. \] Thus \[ \abs{z^3+5z^2+2}<\abs{z^5} \] on the whole circle. By Rouché’s theorem, \(p\) and \(z^5\) have the same number of zeros in \(\abs{z}<2\). Hence \(p\) has five there.
2. The polynomial \(p\) has exactly two zeros in\[ \abs{z}<1, \]counted with multiplicity.
On the circle \(\abs{z}=1\), \[ \abs{5z^2}=5, \] while \[ \abs{z^5+z^3+2} \le 1+1+2 =4. \] Therefore \[ \abs{z^5+z^3+2}<\abs{5z^2} \] on the whole circle. Rouché’s theorem shows that \(p\) and \(5z^2\) have the same number of zeros in \(\abs{z}<1\), namely two.
3. The annulus\[ 1<\abs{z}<2 \]contains exactly\[ \boxed{3} \]zeros of \(p\), counted with multiplicity.
The strict inequalities in steps 1 and 2 also show that \(p\) has no zeros on either boundary circle. Thus the zeros in \(\abs{z}<2\) split into those in \(\abs{z}<1\) and those in the annulus. Their multiplicities give \[ 5-2=3. \]
4. Q.E.D.
Step 3 is the requested count.