Complex Expression C3Ad05

1. **State the problem:** Calculate $$T = \frac{(2z_1 \times 3z_2)^3}{(2z_2)^2}$$ where $$z_1 = 3 + 3i$$ and $$z_2 = 1 + \sqrt{3}i$$. 2. **Substitute the values:** $$T = \frac{(2(3+3i) \times 3(1+\sqrt{3}i))^3}{(2(1+\sqrt{3}i))^2}$$ 3. **Simplify inside the parentheses:** Calculate $$2z_1 = 2(3+3i) = 6 + 6i$$ Calculate $$3z_2 = 3(1+\sqrt{3}i) = 3 + 3\sqrt{3}i$$ 4. **Multiply $$2z_1$$ and $$3z_2$$:** $$(6 + 6i)(3 + 3\sqrt{3}i) = 6 \times 3 + 6 \times 3\sqrt{3}i + 6i \times 3 + 6i \times 3\sqrt{3}i$$ $$= 18 + 18\sqrt{3}i + 18i + 18\sqrt{3}i^2$$ Since $$i^2 = -1$$, $$= 18 + 18\sqrt{3}i + 18i - 18\sqrt{3}$$ Group real and imaginary parts: $$= (18 - 18\sqrt{3}) + (18\sqrt{3} + 18)i$$ 5. **Simplify the expression:** $$= 18(1 - \sqrt{3}) + 18(\sqrt{3} + 1)i$$ 6. **Raise to the power 3:** Let $$A = 18(1 - \sqrt{3}) + 18(\sqrt{3} + 1)i$$, then $$A^3$$ is calculated using binomial expansion or De Moivre's theorem. 7. **Calculate denominator:** $$(2z_2)^2 = (2(1 + \sqrt{3}i))^2 = (2 + 2\sqrt{3}i)^2$$ Expand: $$(2)^2 + 2 \times 2 \times 2\sqrt{3}i + (2\sqrt{3}i)^2 = 4 + 8\sqrt{3}i + 4 \times 3 i^2$$ Since $$i^2 = -1$$, $$= 4 + 8\sqrt{3}i - 12 = -8 + 8\sqrt{3}i$$ 8. **Final expression:** $$T = \frac{A^3}{-8 + 8\sqrt{3}i}$$ 9. **Calculate $$A^3$$:** First find magnitude and argument of $$A$$: $$|A| = 18 \sqrt{(1 - \sqrt{3})^2 + (\sqrt{3} + 1)^2}$$ Calculate inside: $$(1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}$$ $$(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$$ Sum: $$4 - 2\sqrt{3} + 4 + 2\sqrt{3} = 8$$ So, $$|A| = 18 \sqrt{8} = 18 \times 2\sqrt{2} = 36\sqrt{2}$$ 10. **Argument of $$A$$:** $$\theta = \arctan\left(\frac{18(\sqrt{3} + 1)}{18(1 - \sqrt{3})}\right) = \arctan\left(\frac{\sqrt{3} + 1}{1 - \sqrt{3}}\right)$$ Calculate numerator and denominator: $$\frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1 - 3} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{-2}$$ Calculate numerator: $$(\sqrt{3} + 1)(1 + \sqrt{3}) = \sqrt{3} + 3 + 1 + \sqrt{3} = 4 + 2\sqrt{3}$$ So, $$\frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$$ Thus, $$\theta = \arctan(-2 - \sqrt{3})$$ 11. **Use De Moivre's theorem:** $$A^3 = |A|^3 (\cos(3\theta) + i \sin(3\theta))$$ Calculate magnitude: $$|A|^3 = (36\sqrt{2})^3 = 36^3 \times (\sqrt{2})^3 = 46656 \times 2\sqrt{2} = 93312 \sqrt{2}$$ 12. **Calculate $$T$$ by dividing:** $$T = \frac{A^3}{-8 + 8\sqrt{3}i}$$ Multiply numerator and denominator by conjugate: $$\frac{A^3}{-8 + 8\sqrt{3}i} \times \frac{-8 - 8\sqrt{3}i}{-8 - 8\sqrt{3}i} = \frac{A^3(-8 - 8\sqrt{3}i)}{(-8)^2 - (8\sqrt{3})^2}$$ Calculate denominator: $$64 - 64 \times 3 = 64 - 192 = -128$$ 13. **Final simplified form:** $$T = \frac{A^3(-8 - 8\sqrt{3}i)}{-128} = \frac{-8A^3 - 8\sqrt{3}i A^3}{-128} = \frac{8A^3 + 8\sqrt{3}i A^3}{128} = \frac{A^3(8 + 8\sqrt{3}i)}{128} = \frac{A^3(1 + \sqrt{3}i)}{16}$$ 14. **Answer:** $$T = \frac{(18(1 - \sqrt{3}) + 18(\sqrt{3} + 1)i)^3 (1 + \sqrt{3}i)}{16}$$ This is the exact simplified form of $$T$$.