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

Now, calculate $zw$:

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

Next, compute $6/w$:

To find the modulus of $\beta$:

$|\beta| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.$

For the argument, we have:

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

Thus, in modulus-argument form: $\beta = 2\text{cis}\left(-\frac{\pi}{3}\right).$

Using the modulus and argument found: $\beta^3 = (2)^3\text{cis}\left(3 \cdot -\frac{\pi}{3}\right) = 8\text{cis}(-\pi) = -8.$

The form $x + iy$ derived from $\beta^3$ is simply: $\beta^3 = -8 + 0i.$

To sketch the inequalities $|z - z| < 2$ and $|z - 1| \ge 1$:

1. The first inequality represents a region within a circle of radius 2 centered at $z$.
2. The second inequality represents the area outside or on the boundary of the circle of radius 1 centered at $1$.

The intersection of these conditions should be marked distinctly on the Argand diagram.

When point $P$ is reflected across the line $l$, the angle from the positive real axis to $P$ plus the angle from the positive real axis to $Q$ equals twice the angle of the line $l$, which is $2\alpha$. This is due to the nature of angle reflection in geometry.