Determine the number of zeroes of the polynomial
\[2z^5 - 6z^2 + z + 1 = 0\]
in the annulus \(1 \leq |z| \leq 2\).
Solution
There are exactly \(\boxed{3}\) zeros in the closed annulus. They are distinct. Put \(p(z)=2z^5-6z^2+z+1\).
1. There are two zeros inside the unit circle and none on it.
On \(|z|=1\), \[ |2z^5+z+1|\leq2+1+1=4<6=|-6z^2|. \] Rouche’s theorem says that \(p\) and \(-6z^2\) have the same number of zeros in \(|z|<1\), namely two counted with multiplicity [@SS03]. The strict inequality also gives \(|p(z)|\geq6-4>0\) on the circle, excluding boundary zeros.
2. There are five zeros inside the radius-two circle and none on it.
On \(|z|=2\), \[ |-6z^2+z+1|\leq24+2+1=27<64=|2z^5|. \] Rouche’s theorem gives five zeros in \(|z|<2\), counted with multiplicity, and the same strict inequality excludes zeros on \(|z|=2\) [@SS03]. Subtracting the two zeros in \(|z|<1\) leaves \(5-2=3\) in \(1<|z|<2\). Neither boundary has a zero, so this is also the count in the requested closed annulus.
3. All zeros of \(p\) are simple.
The derivative is \(p'(z)=10z^4-12z+1\). Reducing these two integer polynomials modulo three gives \[ \overline p=2z^5+z+1,\qquad \overline{p'}=z^4+1. \] Their Euclidean remainders satisfy \[ \overline p-2z\,\overline{p'}=1-z, \qquad \overline{p'}(1)=2\ne0 \quad\text{in }\mathbb F_3. \] Hence their gcd in \(\mathbb F_3[z]\) is one.
If \(p\) and \(p'\) had a nonconstant common factor over \(\mathbb Q\), Gauss’s lemma would give a primitive common factor \(h\in\mathbb Z[z]\) [@DF04]. Since \(h\) divides \(p\) in \(\mathbb Z[z]\), its leading coefficient divides \(2\), and so is nonzero modulo three. Its reduction would therefore be a nonconstant common factor of \(\overline p\) and \(\overline{p'}\), a contradiction. Thus \(p\) and \(p'\) are coprime over \(\mathbb Q\), hence over \(\mathbb C\) by their Bezout identity. A repeated zero would be common to both, so none exists. The three zeros counted in step 2 are consequently distinct.