Use De Moivre's Theorem to find the cube root of $$ z=8 \angle \pi $$. Choose the correct answer for the question: a. $$ \quad 1+\sqrt{3} i, \quad-2, \quad 1-\sqrt{3} i $$ b. $$ 1+\sqrt{3} i, \quad-2-\sqrt{2} i, \quad-1-\sqrt{3} i $$ c. $$ 2 \angle \frac{\pi}{4}, \backslash i \backslash 2 \angle \frac{3 \pi}{4}, \backslash \backslash i \backslash \frac{5 \pi}{4} $$ d. $$ \quad 1-\sqrt{3} i $$, $$ -2+\sqrt{2} i, \quad \mid-1+\sqrt{3} i $$ e. None of the options

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To find the cube roots of the complex number $$ z = 8 \angle \pi $$ using De Moivre's Theorem, we first express $$ z $$ in polar form:

$$
z = 8 (\cos \pi + i \sin \pi)
$$

The modulus $$ r $$ is 8, and the argument $$ \theta $$ is $$ \pi $$.

According to De Moivre's Theorem, the $$ n $$-th roots of a complex number can be found using the formula:

$$
z_k = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right)
$$

where $$ k = 0, 1, \ldots, n-1 $$.

In this case, we want the cube roots ($$ n = 3 $$), so we have:

1. Calculate $$ r^{1/3} $$:
   $$
   r^{1/3} = 8^{1/3} = 2
   $$

2. Calculate the arguments for $$ k = 0, 1, 2 $$:
   - For $$ k = 0 $$:
     $$
     z_0 = 2 \left( \cos \left( \frac{\pi + 0}{3} \right) + i \sin \left( \frac{\pi + 0}{3} \right) \right) = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 1 + \sqrt{3} i
     $$

- For $$ k = 1 $$:
     $$
     z_1 = 2 \left( \cos \left( \frac{\pi + 2\pi}{3} \right) + i \sin \left( \frac{\pi + 2\pi}{3} \right) \right) = 2 \left( \cos \pi + i \sin \pi \right) = 2(-1 + 0i) = -2
     $$

- For $$ k = 2 $$:
     $$
     z_2 = 2 \left( \cos \left( \frac{\pi + 4\pi}{3} \right) + i \sin \left( \frac{\pi + 4\pi}{3} \right) \right) = 2 \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = 2 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 1 - \sqrt{3} i
     $$

Thus, the three cube roots of $$ z = 8 \angle \pi $$ are:

$$
1 + \sqrt{3} i, \quad -2, \quad 1 - \sqrt{3} i
$$

Now, we can compare these results with the provided options:

a. $$ 1+\sqrt{3} i, \quad -2, \quad 1-\sqrt{3} i $$

This matches our calculated roots.

Therefore, the correct answer is:

**a. $$ 1+\sqrt{3} i, \quad -2, \quad 1-\sqrt{3} i $$**
The correct answer is a. $$ 1+\sqrt{3} i, \quad -2, \quad 1-\sqrt{3} i $$

Use De Moivre’s Theorem to find the cube root of
.
Choose the correct answer for the question:
a.
b.
c.
d. ,
e. None of the options

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