- Let \(\begin{array} { r } { P ( z ) = z ^ { 7 } + z ^ { 3 } + \frac { 1 } { 1 6 } } \end{array}\)
(5 points) Show that P has no multiple zeros.
(15 points) Determine the number of zeros of P that lie in the closed disc \(| z | \leq { \frac { 1 } { 2 } }\)
Solution
1. Part (a): \(P\) and \(P'\) have no common zero.
Let \[ P(z)=z^7+z^3+\frac1{16}. \] Then \[ P'(z)=7z^6+3z^2=z^2(7z^4+3). \] A multiple zero of \(P\) would be a common zero of \(P\) and \(P'\). The point \(z=0\) is not a zero of \(P\). Hence any common zero would satisfy \[ z^4=-\frac37. \] Substituting this into \(P(z)=0\) gives \[ z^3\left(z^4+1\right)+\frac1{16}=0, \] so \[ \frac47 z^3=-\frac1{16}, \qquad z^3=-\frac7{64}. \] But then \[ z^{12}=(z^4)^3=-\frac{27}{343}<0, \] while simultaneously \[ z^{12}=(z^3)^4=\left(\frac7{64}\right)^4>0, \] a contradiction. Thus \(P\) has no multiple zeros.
2. Part (b): Rouché’s theorem gives exactly three zeros in the closed disk.
On \(|z|=1/2\), \[ |z^7|=\frac1{128}, \] whereas \[ \left|z^3+\frac1{16}\right| \ge |z|^3-\frac1{16} =\frac18-\frac1{16} =\frac1{16}. \] Hence \[ |z^7|<\left|z^3+\frac1{16}\right| \] on the entire boundary circle. Rouché’s theorem shows that \(P\) and \(z^3+1/16\) have the same number of zeros in \(|z|<1/2\), counted with multiplicity.
The three zeros of \(z^3+1/16\) satisfy \[ |z|=\left(\frac1{16}\right)^{1/3}=2^{-4/3}<\frac12, \] so all three lie inside the circle. Therefore \(P\) has exactly three zeros in \(|z|<1/2\). The strict Rouché inequality also implies that \(P\) has no zero on \(|z|=1/2\). Thus the closed disk \(|z|\le1/2\) contains exactly \[ \boxed{3} \] zeros.