Question

If $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$ 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 $\frac{3\sqrt{3}}{4}$

Given:-

$\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$and

$z_1+z_2+z_3=0$

To find the area of triangle whose vertices are$z_1,z_2,z_3$

${|z_1+z_2|}^2+{|z_1-z_2|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|+{\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|$

${\left|{-z}_3\right|}^2+{|z_1-z_2|}^2=2({\left|z_1\right|}^2+{\left|z_2\right|}^2)$

$1+{|z_1-z_2|}^2=2(1+1)$

$1+{|z_1-z_2|}^2=4$

${|z_1-z_2|}^2=3$

$|z_1-z_2|=\sqrt{3}$

Similarly,$|z_3-z_2|=|z_3-z_1|=\sqrt{3}$

Thus, the$△$ is an equilateral triangle whose wide length is$\sqrt{3}$

Hence, the area of equilateral triangle is$\frac{\sqrt{3}}{4}{a}^2$

Where a is the length of the side of $△$

area of triangle$=\frac{\sqrt{3}}{4}{a}^2$

$=\frac{\sqrt{3}}{4}{\left(\sqrt{3}\right)}^2$

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

Hence, the area of$△$ is$\frac{3\sqrt{3}}{4}$