Find real numbers a and b such that (1 + 2i)(1 - 3i) = a + ib - HSC - SSCE Mathematics Extension 2 - Question 2 - 2008 - Paper 1

To write (\frac{1 + i\sqrt{3}}{1 + i}) in the form (x + iy), we multiply the numerator and denominator by the conjugate of the denominator:

[ \frac{(1 + i\sqrt{3})(1 - i)}{(1 + i)(1 - i)} = \frac{(1 + i\sqrt{3} - i - \sqrt{3})}{1^2 + 1^2} = \frac{(\sqrt{3} - 1) + (\sqrt{3} - 1)i}{2}. ]

This simplifies to (x + iy), where (x = \frac{\sqrt{3} - 1}{2}) and (y = \frac{\sqrt{3} - 1}{2}).

First, we find the modulus and argument for both complex numbers.

For ( 1 + i ):

• Modulus: ( r_1 = \sqrt{1^2 + 1^2} = \sqrt{2} )
• Argument: ( \theta_1 = \frac{\pi}{4} )

For ( 1 + i\sqrt{3} ):

• Modulus: ( r_2 = \sqrt{1^2 + (\sqrt{3})^2} = 2 )
• Argument: ( \theta_2 = \frac{\pi}{3} )

Thus,

[\frac{1 + i\sqrt{3}}{1 + i} = \frac{r_2}{r_1} e^{i(\theta_2 - \theta_1)} = \frac{2}{\sqrt{2}} e^{i(\frac{\pi}{3} - \frac{\pi}{4})}.]

This results in: ( e^{i\frac{\pi}{12}} \cdot \sqrt{2} ).

Using the half-angle identity:

[\cos \frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}.]

We have from part (ii):

[\frac{1 + i\sqrt{3}}{1 + i} = \sqrt{2} e^{i\frac{\pi}{12}} \quad \text{and thus} \quad \left( \sqrt{2} e^{i\frac{\pi}{12}} \right)^{\frac{1}{2}} = \sqrt{\sqrt{2}} e^{i\frac{\pi}{24}} = \sqrt[4]{2} e^{i\frac{\pi}{24}}.]

To find the equation of the locus of P, we rewrite the equation:

[ z^2 + z - 2 = 0.]

Applying the quadratic formula:

[ z = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}.]

This gives us the solutions: ( z = 1 ) and ( z = -2 ), indicating a linear relationship which describes a line. The locus is a straight line.

The midpoint M of segment QR can be found as follows:

Given (\omega = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}) and (\overline{\omega} = \omega^*), the midpoint M is given by:

[ M = \frac{\omega + \overline{\omega}}{2} = \frac{\left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}{2} = -\frac{1}{2}.]

Thus, we can express M in terms of z as:

( M = -\frac{1}{2} z. )

To find S in the parallelogram PQRS, we use the vector equation for the sides:

[ \text{Vector } S = P + Q - R. ]

Given that we know the relations of the points, we can derive:

[ S = z + \omega - \overline{\omega}. ]

Substituting the expressions provides the required representation of S.