Let $ z = 1 - ext{i}oldsymbol{ oot{3}} $ and $ w = 1 + ext{i} $ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2017 - Paper 1

To find the argument of $\frac{z}{w}$, we compute:

$\text{arg}\left(\frac{z}{w}\right) = \text{arg}(z) - \text{arg}(w).$

From part (i), we already found ( \text{arg}(z) = -\frac{\pi}{3} ). Now, we calculate ( \text{arg}(w) ):

$w = 1 + \text{i} \implies \text{arg}(w) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}.$

Now substitute the values:

$\text{arg}\left(\frac{z}{w}\right) = -\frac{\pi}{3} - \frac{\pi}{4} = -\frac{4\pi + 3\pi}{12} = -\frac{7\pi}{12}.$

For the hyperbola given by ( \frac{x^2}{12} - \frac{y^2}{4} = 1 ), the asymptotes are given by:

$\frac{x^2}{12} - \frac{y^2}{4} = 0 \implies \frac{y^2}{4} = \frac{x^2}{12} \implies y = \pm \frac{2}{\sqrt{3}}x.$

The slope of the asymptote is given by:

$m = \frac{2}{\sqrt{3}}.$

To find the angle ( \alpha ) with the positive x-axis, we use:

$\tan(\alpha) = m = \frac{2}{\sqrt{3}}.$

Thus, we have:

$\alpha = \tan^{-1}(\frac{2}{\sqrt{3}}) = \frac{\pi}{6}.$

To sketch the region defined by the inequalities:

1. $-\frac{\pi}{4} \leq \text{arg} z \leq 0$ creates a sector in the Argand plane covering the angles from $-\frac{\pi}{4}$ to $0$. This is the right side of the negative x-axis.
2. The inequality $|z - (-1 + \text{i})| \leq 1$ represents a circle centered at $(-1, 1)$ with a radius of 1.

We combine these two regions: the sector from $-\frac{\pi}{4}$ to $0$ intersected with the area inside the circle leads to a shaded region in the Argand diagram.