Activity answers:
1. $z = -2 + 2i\sqrt{3}$. $|z| = \sqrt{4 + 12} = 4$. $z$ is in quadrant II; reference angle $= \arctan(2\sqrt{3}/2) = \pi/3$, so $\arg z = \pi - \pi/3 = 2\pi/3$. $z = 4(\cos(2\pi/3) + i\sin(2\pi/3))$. $z^4 = 4^4 (\cos(8\pi/3) + i\sin(8\pi/3)) = 256(\cos(2\pi/3) + i\sin(2\pi/3)) = 256(-1/2 + i\sqrt{3}/2) = -128 + 128i\sqrt{3}$.
2. Midpoint of $(1, 0)$ and $(0, -3)$: $(1/2, -3/2)$. Segment slope $= -3$; bisector slope $= 1/3$. Equation: $y + 3/2 = (1/3)(x - 1/2)$ $\Rightarrow$ $3y + 9/2 = x - 1/2$ $\Rightarrow$ $x - 3y = 5$.
3. $z_1/z_2 = (3/2)(\cos 30^{\circ} + i\sin 30^{\circ}) = (3/2)(\sqrt{3}/2 + i/2) = 3\sqrt{3}/4 + (3/4)i$.
4. $-16 = 16(\cos \pi + i\sin \pi)$. $z = 4(\cos(\pi/2 + k\pi) + i\sin(\pi/2 + k\pi))$, $k = 0, 1$. $z_0 = 4(\cos(\pi/2) + i\sin(\pi/2)) = 4i$; $z_1 = 4(\cos(3\pi/2) + i\sin(3\pi/2)) = -4i$. So $z = \pm 4i$.
5. The region is the quarter-disc obtained by intersecting (i) the closed disc of radius $1$ centred at $1 + i$ with (ii) the closed first-quadrant sector based at $1 + i$. Shape: a quarter disc, with its right angle at the vertex $1 + i$, bounded by the segment from $1 + i$ to $2 + i$ along the real-axis direction and from $1 + i$ to $1 + 2i$ along the imaginary-axis direction, with the curved arc joining $2 + i$ and $1 + 2i$.
Q1 (2 marks): $|z| = \sqrt{3 + 1} = 2$ [1]. $z$ lies in quadrant II ($\Re < 0$, $\Im > 0$); reference angle $\arctan(1/\sqrt{3}) = \pi/6$, so $\arg z = \pi - \pi/6 = 5\pi/6$. Therefore $z = 2(\cos(5\pi/6) + i\sin(5\pi/6))$ [1].
Q2 (3 marks): The condition $|z - 2| = |z - 2i|$ means $z$ is equidistant from $(2, 0)$ and $(0, 2)$; locus is the perpendicular bisector of the segment from $(2, 0)$ to $(0, 2)$ [1]. Midpoint $(1, 1)$; segment slope $-1$; bisector slope $1$; Cartesian equation $y = x$ [1]. Sketch: line $y = x$ on labelled axes, with the two fixed points marked [1].
Q3 (3 marks): $z_1 = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$, $z_2 = 2(\cos(-\pi/6) + i\sin(-\pi/6))$ [1]. $z_1^6 = (\sqrt{2})^6 (\cos(3\pi/2) + i\sin(3\pi/2)) = 8(-i) = -8i$. $z_2^4 = 16(\cos(-2\pi/3) + i\sin(-2\pi/3)) = 16(-1/2 - i\sqrt{3}/2) = -8 - 8i\sqrt{3}$ [1]. $\dfrac{z_1^6}{z_2^4} = \dfrac{-8i}{-8 - 8i\sqrt{3}} = \dfrac{-8i}{-8(1 + i\sqrt{3})} = \dfrac{i}{1 + i\sqrt{3}} = \dfrac{i(1 - i\sqrt{3})}{1 + 3} = \dfrac{i + \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4} + \dfrac{i}{4}$ [1].