If $Z_{1}, Z_{2}, Z_{3}$ are complex numbers such that $| Z_{1} | = | Z_{2} | = | Z_{3} | = 1 a n d Z_{1} + Z_{2} + Z_{3} = 0$ , then area of triangle whose vertices $Z_{1}, Z_{2}$ and $Z_{3}$ is

2

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1

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\frac{\sqrt{3}}{4}

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\frac{3 \sqrt{3}}{4}

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Solution

The correct option is D $\frac{3 \sqrt{3}}{4}$
$| Z_{1} + Z_{2} |^{2} + | Z_{1} - Z_{2} |^{2} = 2 (| Z_{1} |^{2} + | Z_{2} |^{2}) \Rightarrow | - Z_{3} |^{2} + | Z_{1} - Z_{2} |^{2} = 2 (1 + 1) \Rightarrow | Z_{1} - Z_{2} |^{2} = 3 \Rightarrow | Z_{1} - Z_{2} | = \sqrt{3}$
Similarly, $| Z_{2} - Z_{3} |^{2} = 3 \Rightarrow | Z_{1} - Z_{2} | = \sqrt{3}$
$∴ Z_{1}, Z_{2}, Z_{3}$ forms an equilateral triangle with area $= \frac{\sqrt{3}}{4} a^{2} = \frac{3 \sqrt{3}}{4} s q . u n i t s$

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