1. Given the complex numbers $z_1 = 4\sqrt{3} + 4i$, $z_2 = 1 + i\sqrt{3}$, and $Z = \frac{z_1}{z_2}$. **i. Write $Z$ in the form $a + ib$** 1. Write $Z = \frac{4\sqrt{3} + 4i}{1 + i\sqrt{3}}$. 2. Multiply numerator and denominator by the conjugate of the denominator: $1 - i\sqrt{3}$. 3. Calculate numerator: $$ (4\sqrt{3} + 4i)(1 - i\sqrt{3}) = 4\sqrt{3} \cdot 1 - 4\sqrt{3} \cdot i\sqrt{3} + 4i \cdot 1 - 4i \cdot i\sqrt{3} $$ $$ = 4\sqrt{3} - 4i \cdot 3 + 4i - 4i^2 \sqrt{3} $$ $$ = 4\sqrt{3} - 12i + 4i - 4(-1)\sqrt{3} $$ $$ = 4\sqrt{3} - 8i + 4\sqrt{3} = 8\sqrt{3} - 8i $$ 4. Calculate denominator: $$ (1 + i\sqrt{3})(1 - i\sqrt{3}) = 1^2 - (i\sqrt{3})^2 = 1 - (-3) = 4 $$ 5. So, $$ Z = \frac{8\sqrt{3} - 8i}{4} = 2\sqrt{3} - 2i $$ **Answer:** $Z = 2\sqrt{3} - 2i$ **ii. Find the Polar form of $Z$** 1. Recall polar form: $Z = r(\cos \theta + i \sin \theta)$ where $$ r = |Z| = \sqrt{a^2 + b^2}, \quad \theta = \tan^{-1}\left(\frac{b}{a}\right) $$ 2. Calculate magnitude: $$ r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{4 \cdot 3 + 4} = \sqrt{12 + 4} = \sqrt{16} = 4 $$ 3. Calculate argument: $$ \theta = \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} $$ 4. Polar form: $$ Z = 4 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) $$ **iii. Find $Z^4$** 1. Use De Moivre's theorem: $$ Z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right) $$ 2. Here $n=4$, $r=4$, $\theta = -\frac{\pi}{6}$ 3. Calculate: $$ Z^4 = 4^4 \left( \cos\left(4 \times -\frac{\pi}{6}\right) + i \sin\left(4 \times -\frac{\pi}{6}\right) \right) = 256 \left( \cos\left(-\frac{2\pi}{3}\right) + i \sin\left(-\frac{2\pi}{3}\right) \right) $$ 4. Evaluate trigonometric functions: $$ \cos\left(-\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, \quad \sin\left(-\frac{2\pi}{3}\right) = -\sin\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} $$ 5. So, $$ Z^4 = 256 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -128 - 128 i \sqrt{3} $$ **iv. Find the cube root of $Z$** 1. Cube roots of $Z$ are given by: $$ Z^{1/3} = r^{1/3} \left( \cos\left(\frac{\theta + 2k\pi}{3}\right) + i \sin\left(\frac{\theta + 2k\pi}{3}\right) \right), \quad k=0,1,2 $$ 2. Calculate $r^{1/3}$: $$ r^{1/3} = 4^{1/3} = \sqrt[3]{4} $$ 3. Calculate angles for $k=0,1,2$: $$ \theta_0 = \frac{-\frac{\pi}{6} + 0}{3} = -\frac{\pi}{18} $$ $$ \theta_1 = \frac{-\frac{\pi}{6} + 2\pi}{3} = \frac{11\pi}{18} $$ $$ \theta_2 = \frac{-\frac{\pi}{6} + 4\pi}{3} = \frac{23\pi}{18} $$ 4. So the cube roots are: $$ Z_k = \sqrt[3]{4} \left( \cos \theta_k + i \sin \theta_k \right), \quad k=0,1,2 $$ **Final answers:** - $Z = 2\sqrt{3} - 2i$ - Polar form: $Z = 4 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right)$ - $Z^4 = -128 - 128 i \sqrt{3}$ - Cube roots: $$ Z_k = \sqrt[3]{4} \left( \cos \left(\frac{-\frac{\pi}{6} + 2k\pi}{3}\right) + i \sin \left(\frac{-\frac{\pi}{6} + 2k\pi}{3}\right) \right), k=0,1,2 $$