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To determine the algebraic form of the complex numbers \( z_1 = (2 + 2i)^6 \) and \( z_2 = \left( \frac{1 + i \sqrt{3}}{1 - i} \right)^{20} \), we can approach each problem by using polar form and De Moivre's theorem.

### Solution for \( z_1 = (2 + 2i)^6 \):

1. **Convert \( 2 + 2i \) to polar form**:
   - Find the modulus \( r \) of \( 2 + 2i \):
     \[
     r = |2 + 2i| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}
     \]
   - Find the argument \( \theta \):
     \[
     \theta = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}
     \]
   - Therefore, in polar form:
     \[
     2 + 2i = 2\sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)
     \]

2. **Apply De Moivre's theorem**:
   - Since \( z_1 = (2 + 2i)^6 \), we raise both the modulus and the argument to the power of 6:
     \[
     z_1 = (2\sqrt{2})^6 \left(\cos\left(6 \cdot \frac{\pi}{4}\right) + i \sin\left(6 \cdot \frac{\pi}{4}\right)\right)
     \]
   - Calculate \( (2\sqrt{2})^6 \):
     \[
     (2\sqrt{2})^6 = 64
     \]
   - Simplify the argument \( 6 \cdot \frac{\pi}{4} = \frac{3\pi}{2} \):
     \[
     z_1 = 64 \left(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}\right)
     \]
   - Since \( \cos \frac{3\pi}{2} = 0 \) and \( \sin \frac{3\pi}{2} = -1 \):
     \[
     z_1 = 64 \cdot (0 - i) = -64i
     \]

Thus, \( z_1 = -64i \).

### Solution for \( z_2 = \left( \frac{1 + i \sqrt{3}}{1 - i} \right)^{20} \):

1. **Simplify \( \frac{1 + i \sqrt{3}}{1 - i} \)**:
   - To simplify, multiply the numerator and denominator by the conjugate of the denominator, \( 1 + i \):
     \[
     \frac{1 + i \sqrt{3}}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i \sqrt{3})(1 + i)}{(1 - i)(1 + i)}
     \]
   - Calculate the denominator:
     \[
     (1 - i)(1 + i) = 1^2 - (-1) = 2
     \]
   - Expand the numerator:
     \[
     (1 + i \sqrt{3})(1 + i) = 1 + i + i \sqrt{3} - \sqrt{3} = (1 - \sqrt{3}) + i (1 + \sqrt{3})
     \

Determine the algebraic form of the following complex numbers: z1 = (2 + 2i)^6 and z2 = ((1 + i√3) / (1 - i))^20.

This solution provides a detailed step-by-step approach to determine the algebraic form of the complex numbers (2 + 2i)^6 and ((1 + i√3) / (1 - i))^20 using polar form and De Moivre's Theorem. Key mathematical concepts include modulus, argument, and De Moivre’s theorem. Suitable for advanced high school or college-level students.

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