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

To convert to the form $x + iy$, multiply the numerator and denominator by the conjugate of the denominator:

$\frac{(1 + i\sqrt{3})(1 - i)}{(1 + i)(1 - i)}$

Calculating the denominator: $(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2$

For the numerator:

distributing gives: $(1 + i\sqrt{3} - i - i^2\sqrt{3}) = (1 + \sqrt{3}) + (\sqrt{3} - 1)i$

Thus:

$\frac{(1 + \sqrt{3}) + (\sqrt{3} - 1)i}{2} = \frac{1 + \sqrt{3}}{2} + i\frac{\sqrt{3} - 1}{2}$

So, $x = \frac{1 + \sqrt{3}}{2}$ and $y = \frac{\sqrt{3} - 1}{2}$.

First, we find the modulus and argument of $1 + i\sqrt{3}$:

• Modulus: $|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$
• Argument: $\tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$

Thus, in modulus-argument form, it is: $2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$

Now for $1 + i$:

• Modulus: $|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$
• Argument: $\tan^{-1}(1) = \frac{\pi}{4}$

So, in modulus-argument form, it is: $\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

Thus, $\frac{1 + i\sqrt{3}}{1 + i} = \frac{2}{\sqrt{2}} e^{i(\frac{\pi}{3} - \frac{\pi}{4})} = \sqrt{2} e^{i(\frac{\pi}{12})}$.

From the modulus-argument expression, we established: $\frac{1 + i\sqrt{3}}{1 + i} = \sqrt{2} e^{i(\frac{\pi}{12})}$

Taking the real part: $\cos\frac{\pi}{12} = \frac{1 + \sqrt{3}}{\sqrt{2}} = \sqrt{2}$

Thus, the surd form of $\cos\frac{\pi}{12}$ is: $$\frac{\sqrt{6} + \sqrt{2}}{4}.$

Since $\frac{1 + i}{1 + i} = 1$, the square root is simply:

$\sqrt{1} = 1.$

Starting with the equation:

$z^{2} + z = S$

Substituting $z = x + iy$ gives: $(x + iy)^{2} + (x + iy) = S$

Expanding: $x^{2} - y^{2} + 2xyi + x + iy = S$

Separating real and imaginary parts results in:

• Real: $x^{2} - y^{2} + x = Re(S)$
• Imaginary: $2xy + y = Im(S)$

The resulting equations describe a quadratic curve, specifically a parabola.

Knowing that $M$ is the midpoint of $QR$, we can express it as: $M = \frac{Q + R}{2} = \frac{\omega + \overline{\omega}}{2}$

Since $Q$ and $R$ are conjugates, simplifies to:

double sum $M = \frac{2\cos{\frac{2\pi}{3}}}{2} = \cos{\frac{2\pi}{3}} = z.$

In a parallelogram, we know that: $z + S = P + Q \Rightarrow S = P + Q - z.$

Substituting values: $S = z + \omega - z = \omega.$

Thus, the complex number represented by $S$ is: $$S = \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}}.$