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.