Find the cube roots of the complex number -8i, giving the answers in the form a + bi, where a and b are real numbers.

Steps to Solve

1. Convert to Polar Form

To find the cube roots of the complex number $-8i$, we first express it in polar form. A complex number $z = re^{i\theta}$, where ( r ) is the modulus and ( \theta ) is the argument.

• The modulus ( r ) is calculated as: $$ r = \sqrt{(-8)^2 + 0^2} = \sqrt{64} = 8 $$

• The argument ( \theta ) is found using: $$ \theta = \text{atan2}(y, x) = \text{atan2}(-8, 0) = -\frac{\pi}{2} $$

Thus, we have: $$ z = 8e^{-i\frac{\pi}{2}} $$

2. Apply De Moivre's Theorem

To find the cube roots, we use De Moivre's theorem, which states that if ( z = re^{i\theta} ), then the $n$-th roots are given by: $$ z_k = r^{1/n} \cdot e^{i(\theta + 2k\pi)/n} $$

For the cube roots (where ( n = 3 )):

• The modulus for the cube root: $$ r^{1/3} = 8^{1/3} = 2 $$

3. Find the Angles for Each Root

We will calculate the argument for each of the three cube roots:

• For ( k = 0 ): $$ \theta_0 = \frac{-\frac{\pi}{2} + 2(0)\pi}{3} = -\frac{\pi}{6} $$

• For ( k = 1 ): $$ \theta_1 = \frac{-\frac{\pi}{2} + 2(1)\pi}{3} = \frac{\pi}{2} $$

• For ( k = 2 ): $$ \theta_2 = \frac{-\frac{\pi}{2} + 2(2)\pi}{3} = \frac{5\pi}{6} $$

4. Convert Each Root Back to Rectangular Form

Now we convert each root from polar to rectangular form ( z_k = r\left(\cos(\theta_k) + i\sin(\theta_k)\right) ).

• For ( k = 0 ): $$ z_0 = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right) = 2\left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = \sqrt{3} - i $$

• For ( k = 1 ): $$ z_1 = 2\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = 2(0 + i) = 2i $$

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

5. Compile the Results

The cube roots of $-8i$ are: $$ z_1 = \sqrt{3} - i, \quad z_2 = 2i, \quad z_3 = -\sqrt{3} + i $$