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

To find $w - z$, we compute:

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

So, the answer is:

$w - z = -2i.$

To compute $\frac{z}{w}$, we have:

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

Multiplying numerator and denominator by the conjugate $(4 + i)$ gives:

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

Therefore,

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

To express $1 + i$ in polar form, we find its magnitude and angle:

• The magnitude $r$ is given by:

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

• The angle $\theta$ can be calculated as:

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

Thus, we have:

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

Using the De Moivre's Theorem, we can express

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

Calculating:

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

Next, we find the angles:

$17 \times \frac{\pi}{4} = \frac{17\pi}{4} = 4\pi + \frac{\pi}{4} = \frac{\pi}{4}$

Thus,

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

Finally, we have:

$(1 + i)^{17} = 256\sqrt{2}\left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = 256(1 + i).$

The expression

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

can be manipulated to describe a circle in the complex plane. In polar coordinates, this represents a circle centered at rac{1}{2} with a radius of $\frac{1}{2}$. The locus of $P$, as $z$ varies while satisfying the given equation, will thus trace out a circle.

Given $\omega = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$, this represents a transformation in the Argand diagram that scales and rotates the vector $a$. Therefore, the relationship is established:

$z_2 = \omega a$

where $z_2$ denotes the complex number obtained by rotating and scaling $a$.

From the problem's context, we know both $z_1$ and $z_2$ are derived directly as results of the equilateral triangle properties. Given $z_1$ can be expressed as an application of $z_2$, we will have:

$z_1 z_2 = z_1(\omega a) = a\cdot k = a^2.$

This derives from the relationships established in the equilateral triangles.

The equation $z^2 - az + a^2 = 0$ can be solved using the quadratic formula. The discriminant is given by:

$D = b^2 - 4ac = (-a)^2 - 4(1)(a^2) = a^2 - 4a^2 = -3a^2.$

Since the discriminant is negative, the solutions must take the form of complex roots, which aligns with the previously established roots $z_1$ and $z_2$.