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

First, we compute the modulus: $|1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2.$
Next, we find the argument: $\arg(1 + \sqrt{3}i) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.$
Therefore, the modulus-argument form is: $1 + \sqrt{3}i = 2 \left(\cos\frac{\pi}{3} + i \sin\frac{\pi}{3}\right).$

Using De Moivre's Theorem, we have: $(1 + \sqrt{3}i)^{10} = \left(2 \text{cis}\frac{\pi}{3}\right)^{10} = 2^{10} \text{cis}\left(\frac{10\pi}{3}\right) = 1024 \text{cis}\left(\frac{10\pi}{3} - 2\pi\right) = 1024 \text{cis}\left(\frac{4\pi}{3}\right).$
Calculating the Cartesian form: $x + iy = 1024 \left(-\frac{1}{2}\right) + i 1024 \left(-\frac{\sqrt{3}}{2}\right) = -512 - 512\sqrt{3}i.$

We need to break this into two inequalities:

1. For $|z + 1 - 2i| \leq 3$, the circle is centered at $(-1, 2)$ with a radius of 3.
2. The second inequality, $-\frac{\pi}{3} \leq \arg z \leq \frac{\pi}{4}$, restricts $z$ to the angles in the arguments, resulting in a wedge shape.

The intersection of these two describes the desired region in the complex plane.

Rewriting –1 in modulus-argument form gives: $-1 = 1\text{cis}\left(\pi + 2k\pi\right),$ for k = 0, 1, 2, 3. Taking the fourth root: $z_k = \sqrt[4]{1} \text{cis}\left(\frac{\pi + 2k\pi}{4}\right) = 1\text{cis}\left(\frac{\pi}{4} + \frac{k\pi}{2}\right), k = 0, 1, 2, 3.$
This results in:

• For k=0: $z_0=\text{cis}\frac{\pi}{4}$,
• For k=1: $z_1=\text{cis}\frac{3\pi}{4}$,
• For k=2: $z_2=\text{cis}\frac{5\pi}{4}$,
• For k=3: $z_3=\text{cis}\frac{7\pi}{4}$.
  Thus, the solutions are: $z_0, z_1, z_2, z_3$.

Since triangle ABC is isosceles and right-angled at B, we know:

1. $AB = AC$.
2. The lengths can be expressed in terms of the complex coordinates: $|z_1 - z_2| = |z_3 - z_2|.$
   Squaring both sides gives: $|z_1 - z_2|^2 = |z_3 - z_2|^2$
   This holds due to the properties of the isosceles triangle, making the respective sides equal.

For points A, B, and C defined by $z_1, z_2, and z_3$ respectively:

1. The coordinates of D can be constructed by taking the translation and rotation from point A.
   Specifically for square ABCD: $D = z_1 + (z_1 - z_2)i.$
   Thus, expressing $D$ in terms of $z_1, z_2, z_3$ reflects both the translation and resulting vertices.