Question

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

• A. $\frac{3\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{4}$
• C. 1
• D. 2

Answer 1

The correct option is A

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

$z_1+z_2+z_3=0\Rightarrow \overline{z_1}+\overline{z_2}+\overline{z_3}=0...\left(1\right)$
$\Rightarrow \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0(\because |z_1|=|z_2|=|z_3|=1)\Rightarrow z_2{z}_3+z_3{z}_1+z_1{z}_2=0...\left(2\right)$
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 $z^3-d=0\left(3\right)$
[using (1)and (2)]
If $\alpha$ is a root of (3), then other two roots of (3) are $\alpha w$ and $\alpha w^2$
( ${\alpha }^3-d=0\Rightarrow |\alpha {|}^3=|d|=1\Rightarrow |\alpha |=1$ )
Let $z_1=\alpha ,z_2=\alpha wand{z}_3=\alpha w^2$
Area of $\Delta$ whose vertices are $z_1,z_2$ and $z_3$ is

$\Delta =-\frac{1}{4i}\left|\begin{array}{ccc}\alpha & \overline{\alpha }& 1\\ \alpha w& \overline{\alpha }{w}^2& 1\\ \alpha w^2& \overline{\alpha }w& 1\end{array}\right|=\frac{i}{4}\left(\alpha \overline{\alpha }\right)\left|\begin{array}{ccc}1& 1& 1\\ w& w^2& 1\\ w^2& w& 1\end{array}\right|=\frac{3}{4}|w-w^2|=\frac{3}{4}|\frac{-1}{2}+\frac{\sqrt{3}i}{2}-(\frac{-1}{2}+\frac{-\sqrt{3}i}{2}\left)\right|=\frac{3\sqrt{3}}{4}$