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

(i) Locus of z is a circle with centre (-5, 1) and radius 4.

Graph for 4(ai)
Graph for 4(ai)

By Pythagoras’ Theorem, x^{2} + x^{2} = 16 = 2 \sqrt{2}
z = -5 + 2 \sqrt{2} + i(1 - 2 \sqrt{2}), z = -5 - 2 \sqrt{2} + i(1 + 2 \sqrt{2})

(i) |w| = 2, arg(w) = -\frac{\pi}{6} \Rightarrow w = 2 e^{-i \frac{\pi}{6}}
w^{6} = (2 e^{-i \frac{\pi}{6}})^{6} = 64 e^{-i\pi} =64 e^{i\pi}

(ii) \frac{w^{n}}{w^{\ast}} = \frac{(2 e^{-i \frac{\pi}{6} } )^{n}}{2 e^{i \frac{\pi}{6}}} = 2^{n-1} e^{- \frac{i(n+1) \pi}{6}}

arg(\frac{w^{n}}{w^{\ast}}) = - \frac{i(n+1) \pi}{6}

For \frac{w^{n}}{w^{\ast}} to be real, - \frac{i(n+1) \pi}{6} = -\pi, -2\pi, -3\pi, ...

\therefore, \mathrm{smallest} \thinspace n = 5, 11, 17

Personal Comments:
Students should use a compass and projector to draw the locus in (ai). The locus in (aii) is not required to be drawn, should students want to draw, I suggest they draw a new figure for it and not construct on that of (ai). Students that struggle to be neat, can always use the graph paper provided, but please do label the question properly. For (bi), some students forgot that 0 \le \theta \textless 2 \pi was stated by question and did not change it.

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