1. **Problem 1:** Compute $$\frac{z_1^3 + z_2^3 + z_3^3}{z_1 z_2 z_3}$$ where $$z_1 = 1 + i\sqrt{3}, z_2 = \sqrt{3} - i, z_3 = -1 + i$$. 2. **Step 1:** Calculate each cube. - $$z_1^3 = (1 + i\sqrt{3})^3$$. Use binomial expansion or De Moivre's theorem. Note that $$1 + i\sqrt{3} = 2(\cos 60^\circ + i \sin 60^\circ)$$. So, $$z_1^3 = 2^3 (\cos 180^\circ + i \sin 180^\circ) = 8(-1 + 0i) = -8$$. 3. **Step 2:** Calculate $$z_2^3$$. Similarly, $$z_2 = \sqrt{3} - i = 2(\cos(-30^\circ) + i \sin(-30^\circ))$$. Then, $$z_2^3 = 2^3 (\cos(-90^\circ) + i \sin(-90^\circ)) = 8(0 - i) = -8i$$. 4. **Step 3:** Calculate $$z_3^3$$. $$z_3 = -1 + i = \sqrt{2}(\cos 135^\circ + i \sin 135^\circ)$$. Then, $$z_3^3 = (\sqrt{2})^3 (\cos 405^\circ + i \sin 405^\circ) = 2\sqrt{2} (\cos 45^\circ + i \sin 45^\circ) = 2\sqrt{2} \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = 2 + 2i$$. 5. **Step 4:** Sum the cubes. $$z_1^3 + z_2^3 + z_3^3 = -8 + (-8i) + (2 + 2i) = (-8 + 2) + (-8i + 2i) = -6 - 6i$$. 6. **Step 5:** Calculate the product $$z_1 z_2 z_3$$. Calculate stepwise: $$z_1 z_2 = (1 + i\sqrt{3})(\sqrt{3} - i) = \sqrt{3} + i3 - i - i^2 \sqrt{3} = \sqrt{3} + 3i - i + \sqrt{3} = 2\sqrt{3} + 2i$$. Then, $$(z_1 z_2) z_3 = (2\sqrt{3} + 2i)(-1 + i) = -2\sqrt{3} + 2\sqrt{3}i - 2i + 2i^2 = -2\sqrt{3} + 2\sqrt{3}i - 2i - 2 = (-2\sqrt{3} - 2) + (2\sqrt{3} - 2)i$$. 7. **Step 6:** Divide numerator by denominator: $$\frac{-6 - 6i}{(-2\sqrt{3} - 2) + (2\sqrt{3} - 2)i}$$. Multiply numerator and denominator by the conjugate of denominator: $$\frac{(-6 - 6i)((-2\sqrt{3} - 2) - (2\sqrt{3} - 2)i)}{((-2\sqrt{3} - 2) + (2\sqrt{3} - 2)i)((-2\sqrt{3} - 2) - (2\sqrt{3} - 2)i)}$$. 8. **Step 7:** Calculate denominator: $$a = -2\sqrt{3} - 2, b = 2\sqrt{3} - 2$$ $$a^2 + b^2 = (-2\sqrt{3} - 2)^2 + (2\sqrt{3} - 2)^2 = (4 \times 3 + 8\sqrt{3} + 4) + (4 \times 3 - 8\sqrt{3} + 4) = (12 + 8\sqrt{3} + 4) + (12 - 8\sqrt{3} + 4) = 16 + 16 + 0 = 32$$. 9. **Step 8:** Calculate numerator: Expand: $$(-6)(-2\sqrt{3} - 2) + (-6)(-(2\sqrt{3} - 2)i) + (-6i)(-2\sqrt{3} - 2) + (-6i)(-(2\sqrt{3} - 2)i)$$ Simplify each term: - $$(-6)(-2\sqrt{3} - 2) = 12\sqrt{3} + 12$$ - $$(-6)(-(2\sqrt{3} - 2)i) = 12\sqrt{3}i - 12i$$ - $$(-6i)(-2\sqrt{3} - 2) = 12\sqrt{3}i + 12i$$ - $$(-6i)(-(2\sqrt{3} - 2)i) = -6i \times - (2\sqrt{3} - 2)i = -6i \times -i (2\sqrt{3} - 2) = -6(-1)(2\sqrt{3} - 2) = 6(2\sqrt{3} - 2) = 12\sqrt{3} - 12$$ Sum real parts: $$12 + 12\sqrt{3} + 12\sqrt{3} - 12 = (12 - 12) + (12\sqrt{3} + 12\sqrt{3}) = 0 + 24\sqrt{3} = 24\sqrt{3}$$ Sum imaginary parts: $$(12\sqrt{3}i - 12i + 12\sqrt{3}i + 12i) = (12\sqrt{3}i + 12\sqrt{3}i) + (-12i + 12i) = 24\sqrt{3}i + 0 = 24\sqrt{3}i$$ 10. **Step 9:** Final division: $$\frac{24\sqrt{3} + 24\sqrt{3}i}{32} = \frac{24\sqrt{3}}{32} + \frac{24\sqrt{3}}{32}i = \frac{3\sqrt{3}}{4} + \frac{3\sqrt{3}}{4}i$$. --- **Final answer:** $$\boxed{\frac{3\sqrt{3}}{4} + \frac{3\sqrt{3}}{4}i}$$