Let $w = 2 - 3i$ and $z = 3 + 4i.$ (i) Find $ar{w} + z.$ (ii) Find $|w|.$ (iii) Express \( \frac{w}{z} \) in the form $a + ib$, where $a$ and $b$ are real numbers - HSC - SSCE Mathematics Extension 2 - Question 2 - 2011 - Paper 1

The modulus of ( w ) is calculated using the formula ( |w| = \sqrt{a^2 + b^2} ), where ( w = a + bi ). Thus: [ |w| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}. ]

To express ( \frac{w}{z} ) in the form ( a + ib ), we calculate: [ \frac{w}{z} = \frac{2 - 3i}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} = \frac{(2)(3) + (3)(4) + i(-(3)(4) + (2)(-4))}{(3^2 + 4^2)} = \frac{6 + 12 + i(-12 - 8)}{9 + 16} = \frac{18 - 20i}{25} = \frac{18}{25} - \frac{20}{25}i. ] Thus, ( a = \frac{18}{25} ) and ( b = -\frac{20}{25}. )

The coordinates of the points suggest the diagonal relationship in the rhombus. Thus, we need to find the point diagonally opposite to ( 1 + i\sqrt{3} ) and ( \sqrt{3} + i ): By analyzing the midpoints, we find that the diagonal forms yield: [ z = (2 + i), \text{ where } a = 2, b = 1. ]

Using the properties of a rhombus, we deduce that the opposite angles are equal, while each interior angle is bisected. Knowing the coordinates: [ \cos\theta = \frac{1}{2}, \text{ allowing us to conclude that } \theta = \frac{\pi}{3} ext{ or } 60^\circ. ]

We express ( 8 ) in polar form: [ 8 = 8(\cos 0 + i\sin 0), ] thus: To find ( z = r(\cos \theta + i\sin \theta) ), we have: [ r = \sqrt[3]{8} = 2, ] and the angles give: [ \theta = \frac{0 + 2k\pi}{3} \text{ for } k = 0, 1, 2. ] This yields the solutions:\n- ( z_0 = 2(\cos 0 + i\sin 0) )

• ( z_1 = 2(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}) )
• ( z_2 = 2(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}) )

Using the binomial theorem: [ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k ] We get: [ (\cos \theta + i \sin \theta)^3 = \cos^3 \theta + 3\cos^2 \theta (i\sin \theta) + 3\cos \theta (i\sin \theta)^2 + (i\sin \theta)^3 ] This simplifies to: [ \cos^3 \theta - 3\cos \theta \sin^2 \theta + 3i\sin \theta \cos^2 \theta - i\sin^3 \theta ]

Applying de Moivre's theorem: [ \cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta, ] Incorporating ( \sin^2 \theta = 1 - \cos^2 \theta ), we can rewrite: [ \cos 3\theta = \cos^3 \theta - 3\cos \theta (1 - \cos^2 \theta) = \cos^3 \theta - 3\cos \theta + 3\cos^3 \theta, ] which leads to: [ 4\cos^3 \theta = \cos 3\theta + 3, ] therefore, rearranging gives: [ \cos^3 \theta = \frac{1}{4}\cos 3\theta + \frac{3}{4}. ]

Rearranging the equation yields: [ 4\cos^3 \theta - 3\cos \theta = 0, ] Factoring out ( \cos \theta ): [ \cos \theta (4\cos^2 \theta - 3) = 0, ] This gives two cases:

1. ( \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} + n\pi )
2. ( 4\cos^2 \theta - 3 = 0 \Rightarrow \cos^2 \theta = \frac{3}{4} \Rightarrow \cos \theta = \pm\frac{\sqrt{3}}{2} ] Thus, the smallest positive solution is: [ \theta = \frac{\pi}{6} \text{ or } \frac{5\pi}{6} \text{ giving the final answer of } \frac{\pi}{6}. ]