1. **Problem Statement:** Find the modulus and argument of the complex number $$z = \frac{(\cos(\frac{\pi}{4}) - i \sin(\frac{\pi}{4}))^2 (\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}))^3}{(\cos(\frac{\pi}{24}) - i \sin(\frac{\pi}{24}))^4}$$ Then show that $z$ is a cube root of unity and simplify $(1 + 2z)(2 + z^2)$. 2. **Recall:** - Modulus of $z = r$ where $z = r(\cos \theta + i \sin \theta)$. - Argument of $z$ is $\theta$. - De Moivre's theorem: $(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$. 3. **Rewrite each term in polar form:** - $\cos(\frac{\pi}{4}) - i \sin(\frac{\pi}{4}) = \cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})$ (since $-\sin x = \sin(-x)$). - So first term: $(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}))^2 = \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2})$ by De Moivre. - Second term: $(\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}))^3 = \cos(\pi) + i \sin(\pi)$. - Denominator: $(\cos(\frac{\pi}{24}) - i \sin(\frac{\pi}{24}))^4 = (\cos(-\frac{\pi}{24}) + i \sin(-\frac{\pi}{24}))^4 = \cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})$. 4. **Combine numerator:** $$\cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) \times \cos(\pi) + i \sin(\pi) = \cos(-\frac{\pi}{2} + \pi) + i \sin(-\frac{\pi}{2} + \pi) = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})$$ 5. **Divide by denominator:** $$z = \frac{\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})}{\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})} = \cos\left(\frac{\pi}{2} - (-\frac{\pi}{6})\right) + i \sin\left(\frac{\pi}{2} - (-\frac{\pi}{6})\right) = \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right)$$ 6. **(a)(i) Modulus of $z$:** Each factor has modulus 1 (cos and sin on unit circle), so modulus of $z$ is $$|z| = 1$$ 7. **(a)(ii) Argument of $z$:** From step 5, $$\arg(z) = \frac{2\pi}{3}$$ 8. **(b) Show $z$ is a cube root of unity:** Using De Moivre's theorem, $$z^3 = \left(\cos\frac{2\pi}{3} + i \sin\frac{2\pi}{3}\right)^3 = \cos(2\pi) + i \sin(2\pi) = 1$$ Thus, $z$ is a cube root of 1. 9. **(c) Simplify $(1 + 2z)(2 + z^2)$:** - First, find $z^2$: $$z^2 = \cos\left(2 \times \frac{2\pi}{3}\right) + i \sin\left(2 \times \frac{2\pi}{3}\right) = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$$ - Write $z$ explicitly: $$z = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$$ - Compute $(1 + 2z)$: $$1 + 2\left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 1 -1 + i \sqrt{3} = i \sqrt{3}$$ - Compute $(2 + z^2)$: $$2 + \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \frac{3}{2} - i \frac{\sqrt{3}}{2}$$ - Multiply: $$(1 + 2z)(2 + z^2) = i \sqrt{3} \times \left(\frac{3}{2} - i \frac{\sqrt{3}}{2}\right) = i \sqrt{3} \times \frac{3}{2} - i \sqrt{3} \times i \frac{\sqrt{3}}{2}$$ - Simplify each term: $$= \frac{3 i \sqrt{3}}{2} - i^2 \frac{3}{2} = \frac{3 i \sqrt{3}}{2} + \frac{3}{2}$$ (since $i^2 = -1$) - Final form: $$\frac{3}{2} + i \frac{3 \sqrt{3}}{2}$$ **Answer:** - (a)(i) $|z| = 1$ - (a)(ii) $\arg(z) = \frac{2\pi}{3}$ - (b) $z^3 = 1$, so $z$ is a cube root of unity. - (c) $(1 + 2z)(2 + z^2) = \frac{3}{2} + i \frac{3 \sqrt{3}}{2}$