Prove that there is only one solution in the unit disc \(\{z : |z| < 1\}\) and there are three solutions on the annulus \(\{z : 1 < |z| < 2\}\) (counting multiplicities) for the equation \(z^4 - 6z + 3 = 0\).
Solution
Put \(p(z)=z^4-6z+3\).
1. There is exactly one zero in \(|z|<1\), counted with multiplicity.
On \(|z|=1\), \[ |z^4+3|\leq1+3=4<6=|-6z|. \] Both \(p\) and \(-6z\) are holomorphic on a neighborhood of the closed disk. Rouché’s theorem therefore gives the same number of interior zeros, counted with multiplicity, for these two functions [@SS03]. The polynomial \(-6z\) has exactly one simple zero. Thus the count for \(p\) is one. Moreover, \(|p(z)|\geq6-4>0\) on \(|z|=1\), so this circle contains no zero of \(p\).
2. There are exactly three zeros in \(1<|z|<2\), counted with multiplicity.
On \(|z|=2\), \[ |-6z+3|\leq12+3=15<16=|z^4|. \] Rouché’s theorem, now comparing \(p\) with \(z^4\), gives four zeros in \(|z|<2\), counted with multiplicity [@SS03]. The strict inequality also excludes zeros on \(|z|=2\). Step 1 accounts for one of these four zeros in the unit disk and excludes zeros on the intervening unit circle. Consequently the open annulus contains exactly \(4-1=3\) zeros, counted with multiplicity, as required.