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

We calculate:

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

So, $w - z = -2i$.

Now, we compute:

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

To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:

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

Thus, $\frac{z}{w} = \frac{15}{17} + \frac{8}{17}i$.

We convert $1 + i$ into polar form:

First, calculate the modulus:

$r = \sqrt{1^2 + 1^2} = \sqrt{2},$

and the argument:

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

So, we write:

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

Using De Moivre's theorem:

$(1 + i)^{17} = \left(\sqrt{2}\right)^{17}\left(\cos\left(17 \cdot \frac{\pi}{4}\right) + i \sin\left(17 \cdot \frac{\pi}{4}\right)\right) = 2^{8.5}\left(\cos\frac{17\pi}{4} + i\sin\frac{17\pi}{4}\right).$

Now, $\frac{17\pi}{4}$ can be simplified by subtracting $4\pi$:

$\frac{17\pi}{4} - 4\pi = \frac{17\pi - 16\pi}{4} = \frac{\pi}{4}.$

Thus:

$(1 + i)^{17} = 2^{8.5}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = 2^{8.5}\left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right).$
Letting $2^{8.5} = 2^8 \cdot \sqrt{2} = 256\sqrt{2}$,

$= 256\sqrt{2}\frac{\sqrt{2}}{2} + 256\sqrt{2}\frac{\sqrt{2}}{2}i = 256 + 256i.$

Therefore, $(1 + i)^{17}= 256 + 256i$.

The equation given is:

$\frac{1}{z} + \frac{1}{\overline{z}} = 1.$

Multiplying through by $z \overline{z}$ gives:

$\overline{z} + z = z \overline{z}.$

Letting $z = x + iy$, we have:

$x - iy + x + iy = (x^2 + y^2)$ So:

$2x = x^2 + y^2.$

This can be rearranged to:

$y^2 = 2x - x^2.$

The resulting equation represents a sideways parabola.

Since $\omega = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$ represents a rotation by $\frac{\pi}{3}$ radians on the Argand plane, it follows that if $z_1$ is at some position, $z_2$ is simply $z_1$ rotated by that angle. Therefore, we conclude:

$z_2 = \omega z_1.$

From the relation established earlier:

$z_2 = \omega z_1.$

Thus:

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

Given that $|z_1| = 1$, it follows that:

$z_1^2 = |z_1|^2\omega = \omega^2.$

Let the roots be denoted by $z_1$ and $z_2$. The polynomial can be expressed as:

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

Using Vieta's formulas, we find that the sum is $a$ and the product is given as:

$z_1 z_2 = \omega^2 = b.$

Consequently, $z_1$ and $z_2$ satisfy the equation $z^2 - az - b = 0.$