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

(i)
|z_1| = \sqrt{1+3} = 2

\mathrm{arg}z_1 = \mathrm{tan}^{-1}(\sqrt{3}) = \frac{\pi}{3}

\Rightarrow z_1 = 2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})

|z_2| = \sqrt{1+1} = \sqrt{2}

\mathrm{arg}z_2 = - \pi + \mathrm{tan}^{-1}(1) = - \frac{3\pi}{4}

\Rightarrow z_2 = \sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})

(ii)
\frac{z_1}{z_2} = \frac{2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})}{\sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})}

= \sqrt{2}[\mathrm{cos}(-\frac{11\pi}{12}) + i \mathrm{sin} (-\frac{11\pi}{12})]

(\frac{z_1}{z_2})^* = \sqrt{2}[\mathrm{cos}(\frac{11\pi}{12}) + i \mathrm{sin} (\frac{11\pi}{12})]

(iii)

(iv)
From (iii)
(x-1)^2 + \sqrt{3}^2 = 2^2

x = \pm 2

Since x > 0, x = 2

Thus, locus meets the positive real axis when x=2.

KS Comments:

The first two parts requires both students to be able to apply your knowledge of complex numbers formulas, expanding and simplifying them effectively. For (iii), the circle must cut through the origin, and student need to draw it clearly.

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