Show that \(h(z) =z^5 + 3z + 1\) has 5 zeros in \(\abs z \leq 2\).
Solution
Goal: Show that \(h(z) = z^5 + 3z + 1\) has exactly 5 zeros in \(\abs{z} \le 2\).
1. Setup: write \(h = f + g\) with \(f(z) = z^5\) and \(g(z) = 3z + 1\).
\(h(z) = z^5 + 3z + 1\).
2. On \(\abs{z} = 2\): \(\abs{g(z)} < \abs{f(z)}\).
\(\abs{g(z)} \le 3\abs{z} + 1 = 7\) on \(\abs{z} = 2\), while \(\abs{f(z)} = \abs{z}^5 = 32\).
Since \(7 < 32\), the strict inequality holds everywhere on the circle.
3. \(h\) and \(f\) have the same number of zeros inside \(\abs{z} < 2\).
Rouch'e’s theorem with \(f(z) = z^5\), \(g(z) = 3z+1\), \(\gamma = \abs{z}=2\), using 2.
4. \(f(z) = z^5\) has exactly 5 zeros in \(\abs{z} < 2\) (counting multiplicity).
The only zero is \(z = 0\) with multiplicity 5, and \(0\) lies in the disk.
5. Q.E.D.
3 and 4 give that \(h\) has exactly \(5\) zeros in \(\abs{z} < 2\), hence in \(\abs{z} \le 2\).