Let $z = 2 + 3i$ and $w = 1 + i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2001 - Paper 1

To find the reciprocal of $w$, we apply the formula:

$\frac{1}{w} = \frac{1}{1 + i} = \frac{1(1 - i)}{(1 + i)(1 - i)} = \frac{1 - i}{1 + 1} = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}i.$

First, we find the modulus:

$r = |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2.$

Next, we find the argument:

$\theta = \arg(1 + \sqrt{3}i) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.$

So, the modulus-argument form is:

$1 + \sqrt{3}i = 2 \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right).$

Using De Moivre's Theorem:

$(1 + \sqrt{3}i)^{10} = \left(2\right)^{10} \left(\cos\left(10 \cdot \frac{\pi}{3}\right) + i \sin\left(10 \cdot \frac{\pi}{3}\right)\right) = 1024 \left(\cos\left(\frac{10\pi}{3}\right) + i \sin\left(\frac{10\pi}{3}\right)\right).$

Now simplifying:

$\frac{10\pi}{3} = 3\pi + \frac{1\pi}{3} \Rightarrow\cos\left(\frac{10\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \sin\left(\frac{10\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.$

So,

$\implies (1 + \sqrt{3}i)^{10} = 1024 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 512 + 512\sqrt{3}i.$

To sketch the region defined by the inequalities:

1. The inequality $|z + 1 - 2i| \leq 3$ describes a circle centered at $(-1, 2)$ with a radius of 3.

2. The inequality $-\frac{\pi}{3} \leq \arg z \leq \frac{\pi}{4}$ describes a sector of the complex plane between these two angles.

Combining both, we look for the area inside the circle that also falls within the angles of the sector.

The equation can be rewritten as:

$z^{4} = e^{i\pi}$

By taking the fourth root:

$z = e^{i(\pi + 2k\pi)/4}, k = 0, 1, 2, 3.$

Calculating for each $k$ gives:

• For $k=0$: $z = e^{i\frac{\pi}{4}} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$.
• For $k=1$: $z = e^{i\frac{3\pi}{4}} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$.
• For $k=2$: $z = e^{i\frac{5\pi}{4}} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$.
• For $k=3$: $z = e^{i\frac{7\pi}{4}} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$.

So, the solutions in modulus-argument form are:

$z = 1 \text{ at } \arg \frac{\pi}{4}, \text{ and similar for the others.}$

Since triangle $ABC$ is isosceles and right-angled at $B$, the lengths of $AB$ and $BC$ are equal. Thus:

$|z_1 - z_2| = |z_3 - z_2|$

as both represent the lengths of the sides meeting at $B$.

Given that $ABCD$ is a square,

We can express $D$ as the point obtained by rotating $C$ around $B$ by 90 degrees:

$D = z_2 + (z_3 - z_2)i = z_2 + (z_3 - z_2)(i)$

This gives us the position of point $D$ expressed in terms of $z_1$, $z_2$, and $z_3$.