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Let $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $| z | = 2$ . If $z_{1}, z_{2}, z_{3}$ are in the clockwise sense and $z_{1} = 1 + i \sqrt{3},$ then:

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$z_{2} = 1 - i \sqrt{3}, z_{3} = - 2$

$z_{2} = 1 - i \sqrt{3}, z_{3} = 2$

$z_{2} = 1 + i \sqrt{3}, z_{3} = - 2$

$z_{2} = 1 + i \sqrt{3}, z_{3} = 2$

answer is A.

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Detailed Solution

$| z_{1} | = | z_{2} | = | z_{3} | = 2$

$\frac{z_{1}}{z_{2}} = |\frac{z_{1}}{z_{2}}| e^{2 π i / 3}$

$\Rightarrow$ $z_{2} = (1 + i \sqrt{3}) e^{- 2 π i / 3}$

$\Rightarrow$ $z_{2} = (1 + i \sqrt{3}) (- \frac{1}{2} - \frac{i \sqrt{3}}{2})$

$\Rightarrow$ $z_{2} = 1 - \sqrt{3} i$

Also $\frac{z_{3}}{z_{1}} = |\frac{z_{1}}{z_{2}}| e^{2 π i / 3}$

$\Rightarrow$ $z_{3} = (1 + i \sqrt{3}) (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) = - 2$

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