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To find the cube roots of \( z = 4 \left( \cos \left( \frac{2\pi}{3} \right) + i \sin \left( \frac{2\pi}{3} \right) \right) \), we will use De Moivre's Theorem.

Given \( z = r (\cos \theta + i \sin \theta) \), the \( n \)-th roots are given by:
\[ z^{1/n} = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \]
for \( k = 0, 1, \ldots, n-1 \).

For this problem:
- \( r = 4 \)
- \( \theta = \frac{2\pi}{3} \)
- \( n = 3 \)

Thus, the cube roots of \( z \) are given by:
\[ z^{1/3} = 4^{1/3} \left( \cos \left( \frac{\frac{2\pi}{3} + 2k\pi}{3} \right) + i \sin \left( \frac{\frac{2\pi}{3} + 2k\pi}{3} \right) \right) \]

Let's calculate the cube roots for \( k = 0, 1, 2 \).

### When \( k = 0 \):
\[ z^{1/3} = 4^{1/3} \left( \cos \left( \frac{\frac{2\pi}{3} + 0}{3} \right) + i \sin \left( \frac{\frac{2\pi}{3} + 0}{3} \right) \right) \]
\[ = 4^{1/3} \left( \cos \left( \frac{2\pi}{9} \right) + i \sin \left( \frac{2\pi}{9} \right) \right) \]

### When \( k = 1 \):
\[ z^{1/3} = 4^{1/3} \left( \cos \left( \frac{\frac{2\pi}{3} + 2\pi}{3} \right) + i \sin \left( \frac{\frac{2\pi}{3} + 2\pi}{3} \right) \right) \]
\[ = 4^{1/3} \left( \cos \left( \frac{8\pi}{9} \right) + i \sin \left( \frac{8\pi}{9} \right) \right) \]

### When \( k = 2 \):
\[ z^{1/3} = 4^{1/3} \left( \cos \left( \frac{\frac{2\pi}{3} + 4\pi}{3} \right) + i \sin \left( \frac{\frac{2\pi}{3} + 4\pi}{3} \right) \right) \]
\[ = 4^{1/3} \left( \cos \left( \frac{14\pi}{9} \right) + i \sin \left( \frac{14\pi}{9} \right) \right) \]

Now, let's simplify the expressions.

### Simplified Results:
1. For \( k = 0 \):
\[ z^{1/3} = \sqrt[3]{4} \left( \cos \left( \frac{2\pi}{9} \right) + i \sin \left( \frac{2\pi}{9} \right) \right) \]

2. For \( k = 1 \):
\[ z^{1/3} = \sqrt[3]{4} \left( \cos \left( \frac{8\pi}{9} \right) + i \sin \left( \frac{8\pi}{9} \right) \right) \]

3. For \( k = 2 \):
\[ z^{1/3} = \sqrt[3]{4} \left( \cos \left( \frac{14\pi}{9} \right) + i \sin \left( \frac{14\pi}{9} \right) \right) \]

If you have any questions or want more details, please let me know!

Here are 8 related questions for further practice:
1. What is the general form of the nth root of a complex number?
2. How do you convert a complex number from rectangular form to polar form?
3. What is De Moivre's Theorem and how is it used?
4. How do you find the cube roots of a real number using complex numbers?
5. What are the geometric interpretations of complex roots on the complex plane?
6. How do you verify the correctness of the roots of a complex number?
7. What is the principal root of a complex number?
8. How do you find the fourth roots of a complex number?

**Tip:** When dealing with complex roots, always express the number in polar form for easier computation and apply De Moivre's Theorem for roots and powers.

Explore how to find the cube roots of a complex number using De Moivre's Theorem. This problem involves calculating cube roots in polar form, applying trigonometric identities, and understanding complex number operations. Suitable for advanced high school and college students.

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