Let $z = 3 + i$ and $w = 1 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2005 - Paper 1

We multiply $z$ and $w$:

$zw = (3 + i)(1 - i) = 3 - 3i + i + 1 = 4 - 2i$

The conjugate of $w$ is $\overline{w} = 1 + i$.

Thus, $6 \overline{w} = 6(1 + i) = 6 + 6i$

The complex number $\beta = 1 - i\sqrt{3}$ can be expressed in modulus-argument form as:

$r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$

The argument $\theta$ is given by:

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

Thus, we have:

$\beta = 2 \left( \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) \right)$

Using De Moivre's theorem:

$\beta^3 = r^3 \left(\cos(3\theta) + i\sin(3\theta)\right) = 2^3 \left(\cos(-\pi) + i\sin(-\pi)\right)$

This simplifies to:

$\beta^3 = 8(-1 + 0i) = -8$

From part (ii), we have:

$\beta^3 = -8 + 0i$

To sketch the region for $| z - z_0 | < 2$ and $| z - 1 | \geq 1$:

1. The inequality $|z - z_0| < 2$ represents a circle of radius 2 centered at $z_0$.
2. The inequality $|z - 1| \geq 1$ represents the area outside (and including) a circle of radius 1 centered at 1.

The intersection of these regions is the part of the area outside the second circle and inside the first circle.

The reflection of a point across a line creates an angle that is equal to the angle the point makes with the line. Since $P$ is positioned such that its argument is $\arg(z_1)$ and $Q$ is its reflection, the angles satisfy:

$\arg(z_1) + \arg(z_2) = \alpha + \alpha = 2\alpha$