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

The modulus of a complex number $a + bi$ is given by $|a + bi| = \sqrt{a^2 + b^2}$.
For $w = 2 - 3i$:
[ |w| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}. ]

To express $\frac{w}{z}$ in the form $a + bi$, we calculate:
[ \frac{w}{z} = \frac{2 - 3i}{3 + 4i} ]
Multiply 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{-6 - 17i}{25}. ]
Thus,
[ \frac{w}{z} = -\frac{6}{25} - \frac{17}{25} i. ]

Since $0$, $1 + i\sqrt{3}$, $\sqrt{3} + i$, and $z$ form a rhombus, we can find $z$ given the points.
The diagonals bisect each other at right angles.
The coordinates of $z$ can be calculated as:
[ z = 3 + 4i. ]

The interior angles of a rhombus are equal and can be calculated using trigonometry.
With the vertices known, we can find $\theta$ as follows:
In the triangle formed, use the tangent function:
[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{3}}{1}. ]
Thus,
[ \theta = 60^\circ \text{ or } \frac{\pi}{3}. ]

Express 8 in polar form:
[ 8 = 8\text{cis}(0) ] (where $\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)$)
Therefore,
The modulus is 2 and the arguments are:
[ \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{6\pi}{3} ]
Thus,
[ z = 2\text{cis}\left( k\frac{2\pi}{3} \right), k = 0, 1, 2. ]

Using the binomial theorem, we can expand:
[ (\cos \theta + i \sin \theta)^3 = \sum_{k=0}^3 {3 \choose k} (\cos \theta)^{3-k} (i \sin \theta)^k ]
Which simplifies to:
[ \cos^3 \theta + 3\cos^2 \theta(i \sin \theta) + 3\cos \theta(i \sin \theta)^2 + (i \sin \theta)^3. ]

We apply de Moivre's theorem:
[ \cos(3\theta) + i\sin(3\theta) = \left( \cos \theta + i \sin \theta \right)^3 = \cos^3 \theta - 3\cos \theta \sin^2 \theta. ]
From the previous part:
[ \cos^3 \theta = \frac{1}{4} \cos 3\theta + \frac{3}{4}. ]

To solve the equation $4\cos^3 \theta - 3\cos \theta = 0$:
Factor out $\cos \theta$:
[ \cos \theta(4\cos^2 \theta - 3) = 0. ]
Thus, $\cos \theta = 0$ or $4\cos^2 \theta - 3 = 0$.
Solving gives $\cos^2 \theta = \frac{3}{4}$, leading to:
[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ, 150^\circ. ]
The smallest positive solution is $30^\circ$.