Let $z = 3 + i$ and $w = 1 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2005 - Paper 1

To find $2w$, where $w = 1 - i$:

$2w = 2(1 - i) = 2 - 2i.$ Thus, $2w = 2 - 2i$.

To calculate $6/w$:

6/w & = 6/(1 - i) \\ & = 6(1 + i)/(1 - i)(1 + i) \\ & = (6 + 6i)/(1 + 1) \\ & = (6 + 6i)/2 \\ & = 3 + 3i. \end{align*}$$ So, $6/w = 3 + 3i$.

Given eta = 1 - i oot{3}:

• The modulus is:

oot{3})^2) = ext{sqr}(1 + 3) = ext{sqr}(4) = 2.$$ - The argument is: $$ ext{arg}(eta) = an^{-1} igg( rac{- oot{3}}{1} igg) = -rac{ ext{pi}}{3}.$$ So, in modulus-argument form, $eta = 2 ext{cis}(-rac{ ext{pi}}{3})$.

To express eta^3, we calculate:

• The modulus will be:

|eta^3| = |eta|^3 = 2^3 = 8.

• The argument is:

ext{arg}(eta^3) = 3 imes ext{arg}(eta) = 3 imes igg(-rac{ ext{pi}}{3}igg) = - ext{pi}.

Thus, eta^3 = 8 ext{cis}(- ext{pi}).

From the previous step:

eta^3 = 8 ext{cis}(- ext{pi}) = 8 ( ext{cos}(- ext{pi}) + i ext{sin}(- ext{pi})) = 8(-1 + 0i) = -8.

Hence, eta^3 = -8 + 0i.

To sketch the region defined by the inequalities $|z - z| < 2$ and $|z - 1| ext{ at least 1}$, consider:

1. The region defined by $|z - z| < 2$ is a circle around the point $z$ with radius 2.
2. The region $|z - 1| > 1$ represents an annulus centered at 1 with an outer radius of 1 and inner radius also of 1.

The overlap of these regions can be represented graphically.

In this geometric reflection, since $P$ is reflected across the line $l$, the angles $ext{arg}(z_1)$ and $ext{arg}(z_2)$ are equal to the angle $heta$. Thus, when combined, it forms:

$| ext{arg}(z_1)| + | ext{arg}(z_2)| = heta + heta = 2 heta.$