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

(a)
w^2 = (a+bi)^2 = a^2 - b^2 + 2abi

\frac{w^2}{w^*} = \frac{a^2 - b^2 + 2abi}{a-bi} \times \frac{a+bi}{a+bi}

= \frac{1}{a^2 + b^2}(a^3 - 3ab^2 + 3a^2bi - b^3 i)

For \frac{w^2}{w^*} to be purely imaginary, Re{\frac{w^2}{w^*}} = 0

\Rightarrow a^3 - 3ab^2 = 0

a = 0 \mathrm{(NA)}~ \mathrm{or}~ a^2 - 3b^2 = 0

b= \pm \frac{a}{\sqrt{3}}

\therefore w = a \pm \frac{a}{\sqrt{3}}i

(bi)
z^5 = -32i = 2^5 e^{-i\frac{\pi}{2}}

z^5 = 2^5 e^{i(-\frac{\pi}{2}+2k\pi)} for k = 0, \pm 1, \pm 2

z = 2e^{i(\frac{-\pi + 4k \pi}{10})}

\Rightarrow |z| = 2

\mathrm{arg}(z) = - \frac{\pi}{10}, - \frac{\pi}{2}, - \frac{9 \pi}{10},  \frac{3 \pi}{10},  \frac{7 \pi}{10}

(bii)
z_1 = 2e^{i\frac{7\pi}{10}} and z_2 = 2e^{i\frac{-9\pi}{10}}

\angle OZ_2Z_2 = \frac{1}{2}(\pi - \frac{2\pi}{5}) = \frac{3\pi}{10}

\mathrm{arg}(Z_1 - Z_2) = \frac{3\pi}{10} + \frac{\pi}{10} = \frac{2\pi}{5}

By Cosine Rule,

|Z_1 - Z_2|^2 = 2^2 + 2^2 - 2(2)(2)\mathrm{cos} (\frac{2\pi}{5})

|Z_1 - Z_2|^2 = 8 - 8\mathrm{cos} (\frac{2\pi}{5})

|Z_1 - Z_2|^2 = 8[1 - \mathrm{cos} (\frac{2\pi}{5})]

|Z_1 - Z_2|^2 = 8[2\mathrm{sin^2} (\frac{\pi}{5})]

|Z_1 - Z_2|^2 = 16\mathrm{sin^2} (\frac{\pi}{5})

|Z_1 - Z_2| = 4\mathrm{sin} (\frac{\pi}{5})

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

The first part of the question is rather interesting, since we don’t use the traditional argument method to solve it here. But that approach works fine too.
The next part, students just need to be really carful with the root finding, its a standard tutorial question. What follows is a bit tedious, but with an aid of a simple diagram and understanding the properties of roots on an argand diagram, students should not struggle that much.

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