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: $1 - i\sqrt{3}$. $$Z = \frac{(4\sqrt{3} + 4i)(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})}$$ 3. Calculate denominator: $$(1)^2 - (i\sqrt{3})^2 = 1 - (-3) = 4$$ 4. Calculate numerator: $$(4\sqrt{3})(1) = 4\sqrt{3}$$ $$(4\sqrt{3})(-i\sqrt{3}) = -4i \times 3 = -12i$$ $$(4i)(1) = 4i$$ $$(4i)(-i\sqrt{3}) = -4i^2 \sqrt{3} = 4\sqrt{3}$$ Sum numerator real parts: $4\sqrt{3} + 4\sqrt{3} = 8\sqrt{3}$ Sum numerator imaginary parts: $-12i + 4i = -8i$ 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 $$r = |Z| = \sqrt{a^2 + b^2}$$ $$\theta = \tan^{-1}\left(\frac{b}{a}\right)$$ 2. Calculate magnitude: $$r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{4 \times 3 + 4} = \sqrt{12 + 4} = \sqrt{16} = 4$$ 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. For $n=4$: $$Z^4 = 4^4 \left( \cos \left(4 \times -\frac{\pi}{6}\right) + i \sin \left(4 \times -\frac{\pi}{6}\right) \right) = 256 \left( \cos \left(-\frac{2\pi}{3}\right) + i \sin \left(-\frac{2\pi}{3}\right) \right)$$ 3. Evaluate: $$\cos \left(-\frac{2\pi}{3}\right) = \cos \frac{2\pi}{3} = -\frac{1}{2}$$ $$\sin \left(-\frac{2\pi}{3}\right) = -\sin \frac{2\pi}{3} = -\frac{\sqrt{3}}{2}$$ 4. 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: $$Z_k = r^{1/3} \left( \cos \left( \frac{\theta + 2k\pi}{3} \right) + i \sin \left( \frac{\theta + 2k\pi}{3} \right) \right), \quad k=0,1,2$$ 2. Calculate $r^{1/3}$: $$r^{1/3} = 4^{1/3} = \sqrt[3]{4}$$ 3. Calculate angles for $k=0,1,2$: $$\theta = -\frac{\pi}{6}$$ $$\alpha_k = \frac{-\frac{\pi}{6} + 2k\pi}{3} = -\frac{\pi}{18} + \frac{2k\pi}{3}$$ 4. So the cube roots are: $$Z_0 = \sqrt[3]{4} \left( \cos \left(-\frac{\pi}{18}\right) + i \sin \left(-\frac{\pi}{18}\right) \right)$$ $$Z_1 = \sqrt[3]{4} \left( \cos \left(\frac{11\pi}{18}\right) + i \sin \left(\frac{11\pi}{18}\right) \right)$$ $$Z_2 = \sqrt[3]{4} \left( \cos \left(\frac{23\pi}{18}\right) + i \sin \left(\frac{23\pi}{18}\right) \right)$$ **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: $Z_k = \sqrt[3]{4} \left( \cos \left(-\frac{\pi}{18} + \frac{2k\pi}{3} \right) + i \sin \left(-\frac{\pi}{18} + \frac{2k\pi}{3} \right) \right)$ for $k=0,1,2$