Question

Two values of $ (-8 - 8\sqrt{3}i)^{1/4} $ are

• A. \( \sqrt{3} - i,\ -1 - \sqrt{3}i \)
• B. \( \sqrt{3} + i,\ 1 + \sqrt{3}i \)
• C. \( -\sqrt{3} + i,\ \sqrt{3} + i \)
• D. \( 1 - \sqrt{3}i,\ \sqrt{3} + i \)

Hint:

To extract roots of complex numbers, always convert to polar form and use De Moivre's Theorem.

Answer 1

The Correct Option is A

Let \( z = -8 - 8\sqrt{3}i \)
Convert to polar form: \[ r = \sqrt{(-8)^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16 \] \[ \arg(z) = \tan^{-1}\left( \frac{-8\sqrt{3}}{-8} \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \text{ but in third quadrant, so } \pi + \frac{\pi}{3} = \frac{4\pi}{3} \] Now: \[ z^{1/4} = \sqrt[4]{16} \text{cis} \left( \frac{4\pi}{12} + \frac{2k\pi}{4} \right),\ k = 0,1,2,3 \] Calculate roots for \( k = 0,1 \), which yield the required complex roots.