Question

All values of \( (8i)^{\frac{1}{3}} \) are:

• A. \( \pm (\sqrt{3} + i), -2i \)
• B. \( \pm \sqrt{3} + i, -2i \)
• C. \( \pm (\sqrt{3} - i), 2i \)
• D. \( \pm (2 + i), i \)

Hint:

For complex roots, express the number in polar form and use De Moivre’s theorem.

Answer 1

The Correct Option is B

Step 1: Convert to Polar Form Expressing \( 8i \) in polar form: \[ 8i = 8 \text{cis} \frac{\pi}{2} \]
Step 2: Using De Moivre’s Theorem The cube roots are given by: \[ z_k = 8^{\frac{1}{3}} \text{cis} \left( \frac{\frac{\pi}{2} + 2k\pi}{3} \right), \quad k = 0,1,2 \] Calculating values, we get: \[ \boxed{\pm (\sqrt{3} + i), -2i} \]

Answer 2

To solve \( (8i)^{\frac{1}{3}} \), we express \(8i\) in polar form. First, find the modulus \( r \) and argument \(\theta\) of \(8i\).
1. \(\text{Modulus: } r = |8i| = 8.\)
2. \(\text{Argument: } \theta = \frac{\pi}{2}, \text{ since } 8i\) lies on the positive imaginary axis.\)

Thus, \(8i = 8 \left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right).\) Now apply the De Moivre's Theorem to find the cube roots:
\((8i)^{\frac{1}{3}} = 8^{\frac{1}{3}}\left(\cos\left(\frac{\pi/2 + 2k\pi}{3}\right) + i\sin\left(\frac{\pi/2 + 2k\pi}{3}\right)\right), \text{ for } k = 0, 1, 2.\)

Calculate \( 8^{\frac{1}{3}} = 2 \).
For \( k = 0: \theta = \frac{\pi}{6}; z_1 = 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2 \left(\frac{\sqrt{3}}{2} + i\cdot\frac{1}{2}\right) = \sqrt{3} + i.\)

For \( k = 1: \theta = \frac{5\pi}{6}; z_2 = 2 \left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2 \left(-\frac{\sqrt{3}}{2} + i\cdot\frac{1}{2}\right) = -\sqrt{3} + i.\)

For \( k = 2: \theta = \frac{3\pi}{2}; z_3 = 2 \cdot (0 - i) = -2i.\)

Therefore, the values are \( \sqrt{3} + i, -\sqrt{3} + i, -2i.\)