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}) \