Find the number of roots of \(p(z) = 4z^4 - 6z + 3\) in \(\abs{z} < 1\) and \(1 < \abs{z} < 2\) respectively.
Solution
Let \(x_0 = \qty{\tfrac38}^{1/3}\).
1. \(p\) has exactly \(4\) roots in \(\abs{z} < 2\), counted with multiplicity.
On \(\abs{z} = 2\), \(\abs{4z^4} = 64\) and \(\abs{-6z + 3} \le 12 + 3 = 15 < 64\). By Rouché’s theorem, \(p = 4z^4 + (-6z + 3)\) has as many zeros in \(\abs{z} < 2\) as \(4z^4\), namely \(4\).
2. On \(\RR\), \(p\) is decreasing on \((-\infty, x_0]\) and increasing on \([x_0, \infty)\).
\(p'(x) = 16x^3 - 6\) vanishes only at \(x_0\), is negative for \(x < x_0\), and is positive for \(x > x_0\).
3. \(p\) has exactly two real roots \(r_1<r_2\), with \(r_1 \in (\tfrac12, x_0)\) and \(r_2 \in (\tfrac34, 1)\).
Since \(\tfrac18 < \tfrac38 < \tfrac{27}{64}\) and \(\tfrac{8}{27}<\tfrac38\), one has \(\tfrac23 < x_0 < \tfrac34\). The values \[p\qty(\tfrac12) = \tfrac14 > 0, \qquad p(x_0) = \tfrac32x_0 - 6x_0 + 3 = 3 - \tfrac92x_0 < 0, \qquad p\qty(\tfrac34) = \tfrac{81}{64} - \tfrac92 + 3 = -\tfrac{15}{64} < 0, \qquad p(1) = 1 > 0\] use \(4x_0^4 = 4x_0\cdot\tfrac38\). By step 2 and the intermediate value theorem, \(p\) has exactly one root in \((\tfrac12, x_0)\), exactly one in \((\tfrac34, 1)\), and no other real root: \(p > p(\tfrac12) > 0\) on \((-\infty, \tfrac12]\) and \(p \ge p(1) > 0\) on \([1,\infty)\), while \(p<0\) on \([x_0,\tfrac34]\).
4. The two nonreal roots are a conjugate pair \(z,\bar z\) with \(1 < \abs{z} < \sqrt2\).
By step 1 and step 3, \(p\) has two further roots, which are nonreal; since \(p\) has real coefficients they are conjugate. The product of all four roots is \(\tfrac34\), so \(\abs{z}^2 r_1r_2 = \tfrac34\). Step 3 gives \(\tfrac38 < r_1r_2 < x_0 < \tfrac34\), hence \(\abs{z}^2 = \tfrac{3/4}{r_1r_2} \in (1, 2)\).
5. \(p\) has \(\boxed{2}\) roots in \(\abs{z} < 1\) and \(\boxed{2}\) roots in \(1 < \abs{z} < 2\).
4. These are all four roots.
6. Q.E.D.
Step 5 gives both counts.