Counting zeros of a polynomial on and between two circles
Let p(z)=z^5+3z²+1. A control engineer needs the number of unstable modes, which are the zeros of p in the annulus 1<|z|<2, and must also confirm that no zero lies exactly on the unit circle, the stability boundary. Direct root finding is unreliable at this scale, so Rouché's theorem is to be used. The engineer first considers |z|=1, where the term 3z² dominates the remaining terms, and then |z|=2, where z^5 dominates. A colleague warns that Rouché requires a strict inequality on the whole contour and that equality at even one point invalidates the count; the engineer must check this carefully. The final report must separate zeros inside |z|<1, zeros inside |z|<2, and zeros in the annulus, and must justify the absence of boundary zeros independently of Rouché.
1. Use Rouché's theorem to count the zeros of p inside |z|<1 and inside |z|<2.
Read the worked answer
On |z|=1, |3z²|=3 while |z^5+1|≤|z|^5+1=2<3. With g(z)=3z² and h(z)=z^5+1, Rouché gives that p=g+h has the same number of zeros inside |z|<1 as g, namely 2, counted with multiplicity. On |z|=2, |z^5|=32 while |3z²+1|≤3·4+1=13<32. Taking g(z)=z^5 and h(z)=3z²+1, p has the same number of zeros inside |z|<2 as z^5, namely 5. Therefore 5-2=3 zeros lie in the annulus 1<|z|<2.
2. Show that p has no zeros on |z|=1, and explain why this is needed in addition to Rouché.
Read the worked answer
Suppose p(z)=0 with |z|=1. Then 3z²=-(z^5+1), so 3=|3z²|=|z^5+1|≤|z|^5+1=2, a contradiction. Hence there are no boundary zeros on the unit circle. Rouché counts zeros strictly inside a contour; it says nothing about zeros on the contour itself. Without this separate argument, the count of 2 inside and 3 in the annulus would not exclude a zero sitting exactly on |z|=1. Strict inequalities on both circles also ensure Rouché applies cleanly.
3. For p_ε(z)=z^5+3z²+1+εz³, find a range of ε for which the number of zeros inside |z|<1 remains 2.
Read the worked answer
On |z|=1, compare 3z² with the perturbation: |p_ε(z)-3z²|=|z^5+1+εz³|≤|z|^5+1+|ε||z|³=2+|ε|. If |ε|<1, then 2+|ε|<3=|3z²|, so Rouché gives exactly 2 zeros of p_ε inside |z|<1. The zeros inside cannot cross the unit circle under this inequality. Any additional zeros created by the perturbation must lie outside or on the boundary, but the strict inequality also excludes the boundary for |ε|<1. Thus |ε|<1 is a safe sufficient range.