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

Calculating $w - z$, we substitute the values of $w$ and $z$:

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

To find rac{z}{w}, we use the values of $z$ and $w$:

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

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

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

Thus, we conclude:

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

To express $1 + i$ in polar form, we first calculate the magnitude $r$ and argument $heta$:

The magnitude is given as:

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

Next, we find the argument:

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

Thus, we can write:

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

Using De Moivre's Theorem, we obtain:

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

Calculating the magnitude, we have:

$\left(\sqrt{2}\right)^{17} = 2^{8.5} = 256\sqrt{2}.$

Now determining the angle:

$17 \cdot \frac{\pi}{4} = \frac{17\pi}{4} = 4\pi + \frac{\pi}{4} = 2\pi + \frac{\pi}{4}.$

This results in:

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

Thus, we compute:

$256\sqrt{2} \left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = 256(1 + i) = 256 + 256i.$

Given the equation:

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

multiplying through by $z\bar{z}$, we get:

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

This indicates that $z$ and its conjugate sum up to their product. Thus, in the Argand diagram, $P$ represents all points where the sum of the points in the circles intersects at radius 1.

As $z$ varies, the equation describes a circle on the Argand plane centered at rac{1}{2}, with radius $\frac{1}{2}$. Hence, the geometric nature of the locus of point $P$ is a circle with center $(\frac{1}{2}, 0)$ and a radius of $\frac{1}{2}$.

Given that $\omega = \text{cos }\left(\frac{\pi}{3}\right) + i\text{sin }\left(\frac{\pi}{3}\right)$ represents a rotation of rac{\pi}{3} radians in the complex plane. Thus, if $z_1$ is represented as a point in the plane, multiplying it by $\omega$ rotates it to produce $z_2$. Hence,

$z_2 = \omega z_1.$

Using the relationship from part (i), we note:

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

Since $|\omega| = 1$, we have:

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

Thus:

$z_1 z_2 = \bar{z}^2$ is shown by substituting the known forms of $z_1$ and $z_2.$

Since we know $z_1$ and $z_2$ are related through the quadratic relationship where the product of the roots $z_1 z_2 = c^2$ and their sum is:

$z_1 + z_2 = a.$

By applying the quadratic formula, if the coefficients match this relation, then $z_1$ and $z_2$ are indeed the roots of:

$z^2 - az - c^2 = 0.$