Find all the cube roots of \(-2 - 2i\).
Solution
1. \(-2 - 2i = 2\sqrt{2}\, e^{-3\pi i/4}\).
\(|-2-2i| = \sqrt{4+4} = 2\sqrt{2}\), and \(\arg(-2-2i) = -3\pi/4\).
2. The cube roots have modulus \((2\sqrt{2})^{1/3} = \sqrt{2}\) and arguments \(\frac{-3\pi/4 + 2\pi k}{3}\) for \(k = 0, 1, 2\).
the cube roots of \(re^{i\theta}\) are \(r^{1/3} e^{i(\theta + 2\pi k)/3}\).
3. The three cube roots are:\[1 - i,\qquad \frac{\sqrt3 - 1}{2} + \frac{\sqrt3 + 1}{2}i,\qquad -\frac{\sqrt3 + 1}{2} + \frac{1 - \sqrt3}{2}i.\]
evaluate 2 for \(k = 0, 1, 2\):
- \(k=0\): \(\sqrt2 e^{-i\pi/4} = \sqrt2(\frac{\sqrt2}{2} - i\frac{\sqrt2}{2}) = 1 - i\);
- \(k=1\): \(\sqrt2 e^{5\pi i/12} = \frac{\sqrt3 - 1}{2} + \frac{\sqrt3 + 1}{2}i\);
- \(k=2\): \(\sqrt2 e^{13\pi i/12} = -\frac{\sqrt3 + 1}{2} + \frac{1 - \sqrt3}{2}i\).
4. Q.E.D.
3.