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

To find $w - z$, we compute:

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

We find rac{z}{w} by substituting the values of $z$ and $w$:

rac{z}{w} = rac{4 + i}{4 - i}.

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

= rac{(4 + i)(4 + i)}{(4 - i)(4 + i)} = rac{16 + 8i - 1}{16 + 1} = rac{15 + 8i}{17} = rac{15}{17} + rac{8}{17}i.

To express $1 + i$ in the desired form, we compute its magnitude:

r = rac{ ext{sqrt}(1^2 + 1^2)} = ext{sqrt}(2),

and the angle $heta$ is given by:

heta = an^{-1}rac{1}{1} = rac{ ext{pi}}{4}.

Thus,

1 + i = ext{sqrt}(2)ig( ext{cos} rac{ ext{pi}}{4} + i ext{sin} rac{ ext{pi}}{4}ig).

Using De Moivre's theorem:

(1 + i)^{17} = ( ext{sqrt}(2))^{17}ig( ext{cos}(rac{17 ext{pi}}{4}) + i ext{sin}(rac{17 ext{pi}}{4})ig).

Evaluating $ext{sqrt}(2)^{17} = 2^{8.5} = 256 ext{sqrt}(2)$, the angle simplifies to rac{17 ext{pi}}{4} - 4 ext{pi} = rac{5 ext{pi}}{4}.

Thus,

= 256 ext{sqrt}(2)ig(-rac{1}{ ext{sqrt}(2)} - irac{1}{ ext{sqrt}(2)}ig) = -128 - 128i.

For the given locus condition:

rac{1}{z} + rac{1}{ar{z}} = 1,

this can be rewritten in terms of $x$ and $y$ (where $z = x + iy$) leading to a specific geometrical representation in the Argand diagram. This means all points $P$ will trace a circle or a line depending on how $z$ satisfies the equation. This represents the relationship between the complex number and its conjugate.

Given that $O$, $P$, and $R$ form a regular equilateral triangle, the rotation indicated by ar{ heta} about $z_1$ effectively gives the position of $z_2$. Therefore,

z_2 = ar{ heta} z_1, confirming that $z_2$ is derived from rotating $z_1$ by angle rac{ heta}{3}.

Substituting the earlier results, we observe:

z_1 z_2 = z_1 (ar{ heta} z_1) = ar{ heta} z_1^2,

Now since ar{z} = e^{-z_1} would relate, establishing that

z_1 z_2 = ar{z}^2..

Utilizing Vieta's formulas here, we assert roots of a quadratic equation generated hence yields that:

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

Here, $z_1 + z_2 = a$ and identifying with the conjugate shifts provide the correct format as required.