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

Here we find $Zw$:

$Zw = (3 + i)(1 - i)$

Using the distributive property:

$= 3 - 3i + i - i^2 = 3 - 2i + 1 = 4 - 2i$

Thus, the answer is $4 - 2i$.

To compute $6/w$, we need to multiply by the conjugate of $w$:

$6/w = \frac{6}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{6(1 + i)}{1^2 + 1^2} = \frac{6(1 + i)}{2} = 3 + 3i$

Therefore, the answer is $3 + 3i$.

For $\beta = 1 - i\sqrt{3}$, we find the modulus:

$|\beta| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Now, to find the argument: $\arg(\beta) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$

Thus, in modulus-argument form, we have: $\beta = 2\text{cis}\left(-\frac{\pi}{3}\right)$

Using the properties of exponentiation in polar form:

From part (i), we have $\beta = 2\text{cis}\left(-\frac{\pi}{3}\right)$

Hence, $\beta^3 = (2)^3\text{cis}\left(3\left(-\frac{\pi}{3}\right)\right) = 8\text{cis}(-\pi) = -8$

So, $\beta^3$ in modulus-argument form is $8\text{cis}(-\pi)$.

$\beta^3 = -8$ can be expressed in the form $x + iy$ as:

$-8 + 0i$

Hence, the answer is $-8 + 0i$.

To sketch the region defined by the inequalities $|z - z| < 2$ and $|z - 1| \geq 1$:

1. The first inequality $|z - z| < 2$ essentially states that the points $z$ form a circle with radius 2 centered at $z$ itself, meaning there is no area to include since $z$ is a constant.

2. The second inequality $|z - 1| \geq 1$ defines the region outside or on the boundary of a circle centered at 1 with a radius of 1.

Sketching these depicts that there would be an empty region, since points cannot satisfy both conditions simultaneously.

Given the line $l$ makes an angle $\alpha$ with the positive real axis, the reflection of the point $P$ across line $l$ produces point $Q$. The argument of $z_1$, which is the angle $\arg(z_1)$, is less than $\alpha$. Because of the properties of reflection, the argument of $z_2$, or $\arg(z_2)$, is $\alpha + (\alpha - \arg(z_1))$, which simplifies to:

$\arg(z_2) = 2\alpha - \arg(z_1)$

Adding these: $\arg(z_1) + \arg(z_2) = \arg(z_1) + (2\alpha - \arg(z_1)) = 2\alpha.$

Thus, $\arg(z_1) + \arg(z_2) = 2\alpha$.