Let $w = 2 - 3i$ and $z = 3 + 4i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2011 - Paper 1

The modulus of a complex number $w = a + bi$ is given by the formula:

$|w| = \sqrt{a^2 + b^2}.$
For $w = 2 - 3i$, we have $a = 2$ and $b = -3$. Thus,

$|w| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}.$

To express $\frac{w}{z}$ in the form $a + ib$, we start by dividing the complex numbers:

$\frac{w}{z} = \frac{2 - 3i}{3 + 4i}.$
Multiply the numerator and denominator by the conjugate of the denominator:

$\frac{(2 - 3i)(3 - 4i)}{(3 + 4i)(3 - 4i)}.$
Calculating the denominator:

$(3 + 4i)(3 - 4i) = 9 + 16 = 25.$
Now for the numerator:

$(2\cdot3 + 2\cdot(-4i) - 3i\cdot3 - 3i\cdot(-4i)) = (6 + 12) + (-8 - 9)i = 18 - 17i.$
Thus,

$\frac{w}{z} = \frac{18 - 17i}{25} = \frac{18}{25} - \frac{17}{25}i.$

Given the triangle formed by the points $0$, $1 + i\sqrt{3}$, and $\sqrt{3} + i$, we need to find $z$.
Using the properties of a rhombus, the diagonals bisect each other perpendicularly.
We find the midpoints and use vector addition to find that:

z = 1 + rac{3}{2}i.
Thus, $z = 1 + i\frac{3}{2}.$

In a rhombus, opposite angles are equal. The interior angle $\theta$ can be derived using trigonometric identities.
A right triangle in the rhombus gives:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1}.$
Thus,

$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.$

Using the modulus-argument form:
The modulus is $|8| = 8$, and the argument is $\arg(8) = 0$. The general solution is given by:

$z = 2 \text{cis} \left(\frac{2k\pi}{3}\right), \; k = 0, 1, 2.$
Calculating yields:

• For $k = 0$: $z_0 = 2$.
• For $k = 1$: $z_1 = -1 + i\sqrt{3}$.
• For $k = 2$: $z_2 = -1 - i\sqrt{3}$.

Using Binomial theorem, we write:

$(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k.$
For our expression:

$\left(\cos \theta + i \sin \theta\right)^3 = \sum_{k=0}^{3} {3 \choose k} (\cos \theta)^{3-k} (i \sin \theta)^k.$
Calculating gives:

$= \cos^3 \theta + 3\cos^2 \theta (i \sin \theta) - 3\cos \theta (\sin^2 \theta) - i\sin^3 \theta.$

Using de Moivre's theorem, we know:

$\left(\cos \theta + i\sin \theta\right)^3 = \cos(3\theta) + i\sin(3\theta).$
Separating real and imaginary parts gives:

$\cos^3 \theta - 3\cos \theta \sin^2 \theta = \cos(3\theta).$
Rewriting:

$\cos^3 \theta = \frac{1}{4}\cos(3\theta) + \frac{3}{4}.$

We rearrange to:

$4\cos^3 \theta - 3\cos \theta = 0.$
Factoring gives:

$\cos \theta (4\cos^2 \theta - 3) = 0.$
Thus, solutions are: $\cos \theta = 0$ or $\cos^2 \theta = \frac{3}{4} \Rightarrow \cos \theta = \pm \frac{\sqrt{3}}{2}.$ The smallest positive solution occurs at:

$\theta = \frac{\pi}{6}.$