Let $z = 4 + i$ and $w = ar{z}$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2007 - Paper 1

Subtracting $z$ from $w$, we get:

$w - z = (4 - i) - (4 + i) = -2i.$

Now we compute:

$\frac{z}{w} = \frac{4 + i}{4 - i}.$

To simplify, multiply by the conjugate of the denominator:

$\frac{(4 + i)(4 + i)}{(4 - i)(4 + i)} = \frac{16 + 8i - 1}{16 + 1} = \frac{15 + 8i}{17} = \frac{15}{17} + \frac{8}{17}i.$

To convert $1 + i$ into polar form:

• Calculate the modulus:

  $r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}.$

• Calculate the argument:

  $\theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}.$

Thus, we have:

$1 + i = \sqrt{2}(cos \frac{\pi}{4} + i sin \frac{\pi}{4}).$

Using De Moivre's theorem:

$(1 + i)^{17} = (\sqrt{2})^{17}\left( cos\left( 17 \frac{\pi}{4} \right) + i sin\left( 17 \frac{\pi}{4} \right) \right).$

Calculating the modulus:

d = $(\sqrt{2})^{17} = 2^{17/2} = 128\sqrt{2}.$

Finding the angle:

$17 \frac{\pi}{4} = 4\pi + \frac{\pi}{4} \text{ (cycle back)}$

Therefore:

$cos\left( 17 \frac{\pi}{4} \right) = cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2},$

$sin\left( 17 \frac{\pi}{4} \right) = sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.$

Thus, the polar form becomes:

$128\sqrt{2} \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 128 + 128i.$

From the equation:

$\frac{1}{z} + \frac{1}{\bar{z}} = 1,$

we can rearrange as:

$\frac{z + \bar{z}}{z\bar{z}} = 1$

which represents the real parts of $z$ where the locus describes the unit circle in the Argand diagram.

The complex number $\omega$ represents a rotation and scaling on the Argand plane. Since $P$, $Q$, and $R$ are on a unit circle, multiplying $z_1$ by $\omega$ gives $z_2$. This is due to the angle of 60 degrees corresponding to $\frac{\pi}{3}$.

Using the property of complex multiplication:

$z_1 z_2 = z_1 (\omega z_1) = \omega |z_1|^2 = \omega (1) = \omega.$

Therefore:

$z_1 z_2 = \omega^2$.

Given $z_1$ and $z_2$ are roots of the quadratic, we can use Vieta's formulas:

• The sum of the roots $z_1 + z_2 = a$

• The product of the roots $z_1 z_2 = b^2$.

Both fit the form since the roots satisfy:

$z^2 - (z_1 + z_2)z + z_1 z_2 = 0.$