If $| z_{1} | = | z_{2} | = | z_{3} | = 1$ and $z_{1} + z_{2} + z_{3} = 0$
then area of the triangle whose vertices are $z_{1}, z_{2}, z_{3}$ is

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

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

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Solution

The correct option is A
$\frac{3 \sqrt{3}}{4}$

$z_{1} + z_{2} + z_{3} = 0 \Rightarrow_{1} +_{2} +_{3} = 0 . . . (1)$
$\Rightarrow \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} = 0 (∵ | z_{1} | = | z_{2} | = | z_{3} | = 1) \Rightarrow z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2} = 0 . . . (2)$
Let $z_{1} z_{2} z_{3} = d | z_{1} | | z_{2} | | z_{3} | = | d | = 1$
Note that $z_{1}, z_{2}, z_{3}$ are roots of the equation
$(z - z_{1}) (z - z_{2}) (z - z_{3}) = 0$
or $z^{3} - d = 0 (3)$
[using (1)and (2)]
If $α$ is a root of (3), then other two roots of (3) are $α w$ and $α w^{2}$
Also, $α^{3} - d = 0 \Rightarrow | α |^{3} = | d | = 1 \Rightarrow | α | = 1$
Let $z_{1} = α, z_{2} = α w a n d z_{3} = α w^{2}$
Area of $Δ$ whose vertices are $z_{1}, z_{2}$ and $z_{3}$ is

$Δ = - \frac{1}{4 i} ∣ ∣ ∣ ∣ ∣ \begin{matrix} α & ¯ α & 1 α w & ¯ α w^{2} & 1 α w^{2} & ¯ α w & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{i}{4} (α ¯ α) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 w & w^{2} & 1 w^{2} & w & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{3}{4} | w - w^{2} | = \frac{3}{4} ∣ ∣ ∣ \frac{- 1}{2} + \frac{\sqrt{3} i}{2} - (\frac{- 1}{2} + \frac{- \sqrt{3} i}{2}) ∣ ∣ ∣ = \frac{3 \sqrt{3}}{4}$

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If |z1|=|z2|=|z3|=1 and z1+z2+z3=0 then area of the triangle whose vertices are z1,z2,z3 is

If $| z_{1} | = | z_{2} | = | z_{3} | = 1$ and $z_{1} + z_{2} + z_{3} = 0$
then area of the triangle whose vertices are $z_{1}, z_{2}, z_{3}$ is