Let $z = 5 - i.$ a) (i) Find $z^2$ in the form $x + iy$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2010 - Paper 1

First, we calculate:

$z + 2z = 5 - i + 2(5 - i) = 5 - i + 10 - 2i = 15 - 3i.$ Thus, in the form $x + iy$, we have:

$z + 2z = 15 - 3i,$ where $x = 15$ and $y = -3.$

To find $\frac{1}{z}$, we multiply the numerator and denominator by the conjugate:

$\frac{1}{z} = \frac{1}{5 - i} \cdot \frac{5 + i}{5 + i} = \frac{5 + i}{(5 - i)(5 + i)} = \frac{5 + i}{25 + 1} = \frac{5 + i}{26}.$ Thus, in the form $x + iy$, we have:

$\frac{1}{z} = \frac{5}{26} + \frac{1}{26} i,$ where $x = \frac{5}{26}$ and $y = \frac{1}{26}.$

To express $\sqrt{-3 - i}$ in modulus-argument form, we first calculate the modulus:

$r = |\sqrt{-3 - i}| = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}.$
Next, we find the argument:

$\theta = \tan^{-1}\left(\frac{-1}{-3}\right) = \tan^{-1}\left(\frac{1}{3}\right) + \pi = \theta + \pi.$ Therefore, in modulus-argument form:

$\sqrt{-3 - i} = \sqrt{10}\left(\cos(\theta + \pi) + i\sin(\theta + \pi)\right).$

Using De Moivre’s theorem, we know that:

$\left(\sqrt{-3 - i}\right)^{6} = r^{6}\left(\cos(6\theta) + i\sin(6\theta)\right).$ Since $r = \sqrt{10}$ which is positive, $r^{6} = 10^{3} = 1000.$ Moreover, to show it is a real number, we need:

$\sin(6\theta) = 0.$ Thus, $6\theta = n\pi$ for any integer n, leading to it being a real number. So, $\left(\sqrt{-3 - i}\right)^{6}$ is indeed a real number.

To sketch the regions:

1. For the inequality $1 \leq |z| \leq 2$, this represents the area between two circles centered at the origin with radii 1 and 2.
2. For the inequality $0 \leq z + z \leq 3$: Simplifying gives $0 \leq 2|z| \leq 3 \Rightarrow 0 \leq |z| \leq \frac{3}{2}.$

Plotting these constraints, the overlapping area is where both conditions hold true, which is the annular region between the circles of radius 1 and \frac{3}{2}.