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

The modulus of $w$ is given by:

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

We calculate:

$\frac{w}{z} = \frac{2 - 3i}{3 + 4i}.$

To express this in the form $a + bi$, we multiply the numerator and denominator by the conjugate of the denominator:

$= \frac{(2 - 3i)(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{(6 - 8i - 9i + 12)}{9 + 16} = \frac{(18 - 17i)}{25} = \frac{18}{25} - \frac{17}{25}i.$

Thus, we have:

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

By observing the rhombus, we see that the points $0$, $1 + i\sqrt{3}$, and $\sqrt{3} + i$ together form two sides of the rhombus. The diagonal from $0$ to $\sqrt{3} + i$ can be computed as:

$z = \sqrt{3} + i = \sqrt{3} + 1i.$

In a rhombus, opposite angles are equal. The angle $\theta$ can be derived from the slopes of the sides.

Using trigonometry, we can find the interior angle:

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

Thus,

$\theta = 60^\circ = \frac{\pi}{3}.$

This can be expressed in modulus-argument form as:

$z = 2\text{cis}(\frac{2k\pi}{3}) \text{ for } k = 0, 1, 2,$

where $\text{cis}\theta = \cos\theta + i\sin\theta.$ Therefore, the solutions are:

$z_0 = 2\text{cis}(0), \quad z_1 = 2\text{cis}(\frac{2\pi}{3}), \quad z_2 = 2\text{cis}(\frac{4\pi}{3}).$

Using the binomial theorem, we expand:

$(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3.$

If we let $x = \cos\theta$ and $y = \sin\theta$, we have:

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

According to De Moivre's theorem, we express:

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

From our expansion, we identify that:

$\sin^2\theta = 1 - \cos^2\theta,$

therefore:

$\cos(3\theta) = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta).$

Rearranging yields:

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

This is equivalent to:

$4\cos^3\theta - 3\cos\theta - 1 = 0.$

Using the previous relationship found, we can solve this polynomial equation for $\cos\theta$, obtaining:

Let:

$x = \cos\theta,$

Leading to:

$4x^3 - 3x - 1 = 0.$

Using numerical methods or the rational root theorem, we find that the smallest positive solution is at another specified angle. The solution yields:

$\theta \approx 0.52.$