1. Problem: Simplify the given complex expressions and powers. 2. Recall the formula for powers of complex numbers in rectangular form: For $z = a + bi$, $$z^n = (a + bi)^n$$ Use binomial expansion or convert to polar form for simplification. 3. i) Simplify $$\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3$$ Note that $$-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos 120^\circ + i \sin 120^\circ = e^{i 2\pi/3}$$ So, $$\left(e^{i 2\pi/3}\right)^3 = e^{i 2\pi} = 1$$ 4. ii) Simplify $$\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)^3$$ Similarly, $$-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos 240^\circ + i \sin 240^\circ = e^{i 4\pi/3}$$ Then, $$\left(e^{i 4\pi/3}\right)^3 = e^{i 4\pi} = 1$$ 5. iii) Simplify $$\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^{-2} \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)$$ Using polar form, $$\left(e^{i 2\pi/3}\right)^{-2} e^{i 4\pi/3} = e^{-i 4\pi/3} e^{i 4\pi/3} = e^{0} = 1$$ 6. iv) Simplify $$(a + bi)^2$$ Using binomial expansion: $$(a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + 2ab i$$ 7. v) Simplify $$(a + bi)^{-2}$$ Recall, $$(a + bi)^{-1} = \frac{a - bi}{a^2 + b^2}$$ So, $$(a + bi)^{-2} = \left(\frac{a - bi}{a^2 + b^2}\right)^2 = \frac{(a - bi)^2}{(a^2 + b^2)^2} = \frac{a^2 - 2ab i + b^2 i^2}{(a^2 + b^2)^2} = \frac{a^2 - 2ab i - b^2}{(a^2 + b^2)^2} = \frac{a^2 - b^2 - 2ab i}{(a^2 + b^2)^2}$$ 8. vi) Simplify $$(a + bi)^3$$ Using binomial expansion: $$(a + bi)^3 = a^3 + 3a^2 bi + 3a (bi)^2 + (bi)^3 = a^3 + 3a^2 bi - 3ab^2 - b^3 i = (a^3 - 3ab^2) + (3a^2 b - b^3) i$$ 9. vii) Simplify $$(a - bi)^3$$ Similarly, $$(a - bi)^3 = a^3 - 3a^2 bi + 3a ( -bi)^2 - (bi)^3 = a^3 - 3a^2 bi - 3ab^2 + b^3 i = (a^3 - 3ab^2) + (-3a^2 b + b^3) i$$ 10. viii) Simplify $$\left(3 - \sqrt{-4}\right)^{-3}$$ Note that $$\sqrt{-4} = 2i$$ So, $$3 - \sqrt{-4} = 3 - 2i$$ Then, $$\left(3 - 2i\right)^{-3} = \frac{1}{(3 - 2i)^3}$$ Calculate $(3 - 2i)^3$: $$(3 - 2i)^3 = (3 - 2i)(3 - 2i)^2$$ First, $$(3 - 2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i$$ Then, $$(3 - 2i)^3 = (3 - 2i)(5 - 12i) = 15 - 36i - 10i + 24 i^2 = 15 - 46i - 24 = -9 - 46i$$ Therefore, $$\left(3 - 2i\right)^{-3} = \frac{1}{-9 - 46i} = \frac{-9 + 46i}{(-9)^2 + 46^2} = \frac{-9 + 46i}{81 + 2116} = \frac{-9 + 46i}{2197}$$ Final answers: 1) 1 2) 1 3) 1 4) $$(a^2 - b^2) + 2ab i$$ 5) $$\frac{a^2 - b^2 - 2ab i}{(a^2 + b^2)^2}$$ 6) $$(a^3 - 3ab^2) + (3a^2 b - b^3) i$$ 7) $$(a^3 - 3ab^2) + (-3a^2 b + b^3) i$$ 8) $$\frac{-9 + 46i}{2197}$$