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

Now, we calculate:

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

Simplifying this, we get:

$w - z = 4 - i - 4 - i = -2i.$

Therefore, our final answer is:

$w - z = 0 - 2i.$

To find the quotient, we divide $z$ by $w$:

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

To simplify this expression, we multiply the numerator and the 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, in the form $x + iy$, we have:

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

To express $1 + i$ in the required polar form, we first calculate the modulus $r$:

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

Now, we find the argument $heta$:

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

Thus, we can write:

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

Using De Moivre's theorem:

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

Calculating the modulus:

$\left(\sqrt{2}\right)^{17} = 2^{\frac{17}{2}} = 128 \sqrt{2}.$

Finding the angles:

$17 \cdot \frac{\pi}{4} = \frac{17\pi}{4} = 4\pi + \frac{\pi}{4} = \frac{\pi}{4} \quad (\text{since we subtract } 4\pi)$

Thus, we write:

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

Putting it all together:

$(1 + i)^{17} = 128 \sqrt{2}\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = 128(1 + i) = 128 + 128i.$

The condition $\frac{1}{z} + \frac{1}{\bar{z}} = 1$ can be rewritten as: $\frac{z + \bar{z}}{z \bar{z}} = 1.$

This implies that:

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

This represents a relationship that geometrically describes the set of points $P$ on the Argand plane which satisfy the condition of being an intersecting line or circle.
As $z$ varies, $P$ describes a curve that can be interpreted based on its distance from the origin, revealing a specific geometrical locus.

Since $riangle OPR$ and $riangle OQR$ are equilateral triangles with unit sides, the point $R$ can be derived from point $P$ by rotating it by rac{\pi}{3} radians, corresponding to the multiplication by a complex exponential:

$z_2 = \omega z_1,$ where $heta = \frac{\pi}{3}$.

To show that $z_1 z_2 = \bar{z}^2$, we substitute:

$z_2 = \omega z_1 = \left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)z_1.$

Thus, $z_1 z_2 = z_1\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)z_1= z_1^2 \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right).$

Given that $|z_1| = 1$, we know that ar{z} = \frac{1}{z}, hence:

$z_1 z_2 = \bar{z}^2 = |z|^2 = 1.$

To show that $z_1$ and $z_2$ are the roots of the polynomial:

$z^2 - az - a^2 = 0,$ we apply Viète's formulas.
Since $z_1$ and $z_2$ are the roots, we have:

$z_1 + z_2 = a ext{ and } z_1 z_2 = -a^2.$

Substituting these results demonstrates that indeed, $z_1$ and $z_2$ are legitimate roots of the given equation.