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The image shows a problem asking to express the given complex number \( z \) in polar form \( z = r e^{i\theta} \), where:

\[
z = \frac{(1+i)^8 (2-2i)^6}{(1 - \sqrt{3} i)^{10}}
\]

### Steps to solve:

1. **Expressing the complex numbers in polar form:**

   - **\( 1 + i \):**
     - In polar form, this is \( \sqrt{2} e^{i\frac{\pi}{4}} \) because the modulus is \( \sqrt{1^2 + 1^2} = \sqrt{2} \) and the argument is \( \frac{\pi}{4} \) (45 degrees).
     - Therefore, \( (1+i)^8 = (\sqrt{2})^8 e^{i \cdot 8 \cdot \frac{\pi}{4}} = 16 e^{i 2\pi} \).
   
   - **\( 2 - 2i \):**
     - In polar form, this is \( 2\sqrt{2} e^{-i \frac{\pi}{4}} \), since the modulus is \( \sqrt{2^2 + (-2)^2} = 2\sqrt{2} \) and the argument is \( -\frac{\pi}{4} \).
     - Therefore, \( (2 - 2i)^6 = (2\sqrt{2})^6 e^{-i 6 \cdot \frac{\pi}{4}} = 64 \cdot 8 e^{-i \frac{3\pi}{2}} \).
   
   - **\( 1 - \sqrt{3} i \):**
     - In polar form, this is \( 2 e^{-i \frac{\pi}{3}} \) since the modulus is \( \sqrt{1^2 + (\sqrt{3})^2} = 2 \) and the argument is \( -\frac{\pi}{3} \).
     - Therefore, \( (1 - \sqrt{3} i)^{10} = 2^{10} e^{-i 10 \cdot \frac{\pi}{3}} = 1024 e^{-i \frac{10\pi}{3}} \).

2. **Simplifying \( z \):**
   - Combine the modulus terms:
     \[
     |z| = \frac{16 \cdot 512}{1024} = 8
     \]
   - Combine the argument terms (angles):
     \[
     \arg(z) = 2\pi - \frac{3\pi}{2} + \frac{10\pi}{3} = 2\pi - \frac{3\pi}{2} + \frac{10\pi}{3}
     \]
     Simplifying the argument:
     \[
     2\pi - \frac{3\pi}{2} + \frac{10\pi}{3} = \frac{12\pi}{6} - \frac{9\pi}{6} + \frac{20\pi}{6} = \frac{23\pi}{6}
     \]

3. **Final form:**
   \[
   z = 8 e^{i \frac{23\pi}{6}}
   \]

   The polar form of \( z \) is:
   \[
   r = 8, \quad \theta = \frac{23\pi}{6}
   \]

Would you like a more detailed explanation of any step?

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Here are five related questions you might explore:

1. How can you convert a complex number from Cartesian to polar form?
2. What is the general formula for multiplying and dividing complex numbers in polar form?
3. What is the significance of the modulus and argument in the polar representation of complex numbers?
4. How can you simplify complex numbers when their argument exceeds \( 2\pi \)?
5. How does De Moivre's Theorem apply to powers of complex numbers?

**Tip:** When working with complex numbers, always ensure to express the modulus and argument clearly for accurate results.

Écrivez le nombre complexe suivant: z = \frac{(1+i)^8 (2-2i)^6}{(1 - \sqrt{3} i)^{10}} sous la forme z = r e^{i\theta}. Donnez la valeur r et de \theta.

Solve the complex number expression (1+i)^8 (2-2i)^6 / (1-sqrt(3)i)^10 and express it in polar form as z = r e^{i\theta}. Learn to calculate the modulus and argument, applying De Moivre's Theorem and Euler's Formula. Suitable for advanced high school students.

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