All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$z^2 = -8i$

$z^2 = 8e^{i(-\frac{\pi}{2}+2k\pi}$

$z = 2\sqrt{2}e^{-i\frac{\pi+4k\pi}{4}}$

$z = 2\sqrt{2}e^{-i\frac{\pi}{4}}, \mathrm{~or~} 2\sqrt{2}e^{i\frac{3\pi}{4}}$

$z = 2 - 2i, ~ -2+2i$

(ii)
Observe we let $z = 2w+4$, then

$(2w+4)^2 = -8i$

$4w^2 + 16w + 16 + 8i = 0$

$w^2 + 4w + 4 + 2i = 0$

$\Rightarrow w = \frac{-4+2-2i}{2}, \mathrm{~or~} \frac{-4-2+2i}{2}$

$\therefore, w = -1-i, \mathrm{~or~} -3+i$

(iii)

(iv)
Both perpendicular bisector are parallel to each other, so there will be no point of intersections.

Students should be cautious to leave their answers in cartesian form in (i). For (ii), students can solve by otherwise methods, that is, to use the quadratic formula $\frac{-b\pm \sqrt{}{b^2-4ac}}{2a}$