1. **State the problem:** We need to find $Z^6$ where $Z=\frac{4i}{1 - i\sqrt{3}}$. 2. **Simplify the denominator:** Multiply numerator and denominator by the conjugate of the denominator: $$Z = \frac{4i}{1 - i\sqrt{3}} \times \frac{1 + i\sqrt{3}}{1 + i\sqrt{3}} = \frac{4i(1 + i\sqrt{3})}{(1)^2 - (i\sqrt{3})^2}$$ 3. **Calculate denominator:** $$1 - (i\sqrt{3})^2 = 1 - (i^2)(3) = 1 - (-1)(3) = 1 + 3 = 4$$ 4. **Calculate numerator:** $$4i(1 + i\sqrt{3}) = 4i + 4i^2\sqrt{3} = 4i - 4\sqrt{3} \quad \text{(since } i^2 = -1)$$ 5. **So,** $$Z = \frac{4i - 4\sqrt{3}}{4} = i - \sqrt{3}$$ 6. **Rewrite $Z$ in standard form:** $$Z = -\sqrt{3} + i$$ 7. **Convert $Z$ to polar form:** - Magnitude: $$r = |Z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$$ - Argument $\theta$: $$\theta = \arctan\left(\frac{1}{-\sqrt{3}}\right)$$ Since real part is negative and imaginary part positive, $Z$ is in the second quadrant. $$\arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$ So, $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ 8. **Express $Z$ in polar form:** $$Z = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$$ 9. **Use De Moivre's theorem to find $Z^6$:** $$Z^6 = r^6 \left(\cos(6\theta) + i\sin(6\theta)\right) = 2^6 \left(\cos\left(6 \times \frac{5\pi}{6}\right) + i\sin\left(6 \times \frac{5\pi}{6}\right)\right)$$ 10. **Calculate:** $$2^6 = 64$$ $$6 \times \frac{5\pi}{6} = 5\pi$$ 11. **Evaluate trigonometric functions:** $$\cos(5\pi) = \cos(\pi) = -1$$ $$\sin(5\pi) = \sin(\pi) = 0$$ 12. **Final result:** $$Z^6 = 64(-1 + 0i) = -64$$ **Answer:** $Z^6 = -64$