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

The modulus of the complex number is calculated as follows:
$r = |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.$
For the argument:
$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.$
Thus, in modulus-argument form, we have:
$1 + \sqrt{3}i = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right).$

Using De Moivre's theorem:
$(1 + \sqrt{3}i)^{10} = \left(2\right)^{10}\left(\cos\left(10 \cdot \frac{\pi}{3}\right) + i\sin\left(10 \cdot \frac{\pi}{3}\right)\right).$
Calculating the modulus:
$2^{10} = 1024.$
Calculating the argument:
$10 \cdot \frac{\pi}{3} = \frac{10\pi}{3} = \frac{4\pi}{3} + 2\pi.$
This can be reduced to (\frac{4\pi}{3} \.
Now substituting back:
$\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \ \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}.$
Therefore,
$(1 + \sqrt{3}i)^{10} = 1024\left(-\frac{1}{2} + i\left(-\frac{\sqrt{3}}{2}\right)\right) = -512 - 512\sqrt{3}i.$

The inequality $|z + 1 - 2i| \leq 3$ describes a circle with center at $(-1, 2)$ and radius $3$. The second part indicates that the argument of $z$ lies between two angles. In the complex plane, this results in a slice of the plane bounded between the angles corresponding to $-\frac{\pi}{3}$ and $\frac{\pi}{4}$ from the origin, while the entire area must also be within the circle described. Thus, the sketch will depict a circular region bounded within these angular limits.

To find the solutions to the equation: $z^4 = -1$
we can express $-1$ in modulus-argument form: $-1 = 1\left(\cos(\pi) + i\sin(\pi)\right).$
Applying De Moivre's theorem, we find the fourth roots: $z_k = \sqrt[4]{1}\left(\cos\left(\frac{\pi + 2k\pi}{4}\right) + i\sin\left(\frac{\pi + 2k\pi}{4}\right)\right), k = 0, 1, 2, 3.$
Calculating for each $k$:

• For $k=0$: $z_0 = e^{i\frac{\pi}{4}}$
• For $k=1$: $z_1 = e^{i\frac{3\pi}{4}}$
• For $k=2$: $z_2 = e^{i\frac{5\pi}{4}}$
• For $k=3$: $z_3 = e^{i\frac{7\pi}{4}}$
  The solutions are:
  $z_0 = 1\left(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})\right), z_1 = 1\left(\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})\right), \ \text{etc.}$

Given that triangle $ABC$ is isosceles and right-angled at $B$, we have:
$|z_1 - z_2| = |z_3 - z_2|$
which implies:
$|z_1 - z_2|^2 = (z_1 - z_2)(\overline{z_1 - z_2}) = |z_3 - z_2|^2 = (z_3 - z_2)(\overline{z_3 - z_2}).$
Therefore, we establish that
$|z_1 - z_2|^2 = (z_3 - z_2)^2$
is a true statement.

The point $D$ can be defined geometrically as the vertex of the square $ABCD$, which can be derived using:
$D = z_1 + (z_2 - z_1)i.$
Expressing in terms of $z_1, z_2$ and $z_3$ would yield:
$D = z_2 + (z_3 - z_2)i.$
This directly relates the vertices together in the square formation.