Let $ z = 1 - ext{j} imes ext{sqrt}(3) $ and $ w = 1 + ext{j} $ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2017 - Paper 1

To find the argument of the quotient of two complex numbers, we use: $\text{arg}\left(\frac{z}{w}\right) = \text{arg}(z) - \text{arg}(w)$

From the previous calculation, we have: $\text{arg}(z) = -\frac{\pi}{3}.$

Now, for $w = 1 + ext{j}$, we find: $\text{arg}(w) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}.$

Thus, we compute: $\text{arg}\left(\frac{z}{w}\right) = -\frac{\pi}{3} - \frac{\pi}{4}.$

To combine these, we convert to a common denominator (12): $-\frac{\pi}{3} = -\frac{4\pi}{12}, \quad -\frac{\pi}{4} = -\frac{3\pi}{12}.$

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

To find the angle $\alpha$ of the asymptote for the hyperbola $\frac{x^2}{12} - \frac{y^2}{4} = 1$, we first recognize it can be expressed in the form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a^2 = 12$ and $b^2 = 4$, thus $a = 2\sqrt{3}$ and $b = 2$. The slopes of the asymptotes are given by: $\tan(\alpha) = \frac{b}{a} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}.$

Therefore, $\alpha = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.$

To sketch the region defined by $-\frac{\pi}{4} \leq \text{arg} z \leq 0$ and $|z - (1 + j)| \leq 1$, we first consider:

• The argument condition means that $z$ lies in a sector between the angles $-\frac{\pi}{4}$ and $0$ in the Argand plane.
• The second condition describes a circle centered at the point $(1, 1)$ with radius 1.

The resulting area is the intersection of this sector and the circular region, which can be sketched accordingly with proper labeling.