Find real numbers $a$ and $b$ such that $(1 + 2i)(1 - 3i) = a + ib.$ (b) (i) Write $\frac{1 + i\sqrt{3}}{1 + i}$ in the form $x + iy$, where $x$ and $y$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2008 - Paper 1

We simplify:

$\frac{1 + i\sqrt{3}}{1 + i} = \frac{(1 + i\sqrt{3})(1 - i)}{(1 + i)(1 - i)} = \frac{(1 - i) + i\sqrt{3} - \sqrt{3}}{1 + 1} = \frac{(1 - \sqrt{3}) + (\sqrt{3} - 1)i}{2}.$
Thus, $x = \frac{1 - \sqrt{3}}{2}, \quad y = \frac{\sqrt{3} - 1}{2}.$

First, we compute:

• For $\frac{1 + \sqrt{3}}{i}$, we have a magnitude of $\sqrt{(1 + \sqrt{3})^2 + 1^2} = 2$ and an argument of $\frac{\pi}{6}$.
• For $1 + i$, the magnitude is $\sqrt{2}$ and the argument is $\frac{\pi}{4}$.

Putting it all together: $\frac{1 + \sqrt{3}}{i + 1} = \frac{(1 + \sqrt{3}) e^{i\frac{\pi}{6}}}{\sqrt{2} e^{i\frac{\pi}{4}}} = \frac{2(1 + \sqrt{3})}{\sqrt{2}} e^{i(\frac{\pi}{6} - \frac{\pi}{4})}.$

Using the half-angle identity, we find $\cos\frac{\pi}{12} = \sqrt{\frac{1 + \cos{\frac{\pi}{6}}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}.$

Using modulus-argument form from part (ii): $\left(\frac{1+i\sqrt{3}}{1+i}\right)^{\frac{1}{2}} = \left(\sqrt{2} e^{i(\frac{\pi}{6} - \frac{\pi}{4})}\right)^{\frac{1}{2}} = \sqrt{\sqrt{2}} e^{i\frac{1}{2}(\frac{\pi}{6} - \frac{\pi}{4})} = \sqrt[4]{2} e^{i\frac{\pi}{12}}.$

To solve the equation:

$z^2 + z = 0 \\ \Rightarrow z(z + 1) = 0.$ This implies $z = 0$ or $z = -1$, representing two points on the Argand diagram. Since this is an equation with solutions that are discrete points, the locus consists of these two points. The type of curve represented here is a pair of points.

The midpoint $M$ of points $Q$ and $R$ can be expressed as: $M = \frac{\omega z + \overline{\omega} z}{2} = \frac{z(\omega + \overline{\omega})}{2} = z \cdot \frac{\cos\frac{2\pi}{3} + \cos\frac{2\pi}{3}}{2} = z \cdot \frac{-1}{2}.$

Since $PQSR$ is a parallelogram, we can state that: $\overline{PQ} + \overline{RS} = 0.$ From this, we have: $S = z + (\overline{\omega} - \omega) z = z(1 + \overline{\omega} - \omega).$
Thus, the complex number represented by $S$ is: $S = z (1 + \overline{\omega} - \omega).$