Question

Let $z_1,z_2$ and $z_1+z_2$ represents three points $A,B$ and $C$ in Argand plane. If $|z_1|=|z_2|=|z_1+z_2|=0$, then prove that the area of triangle $ABC$ is $\frac{\sqrt{3}}{4}|z_1{|}^2$.

Answer 1

R.E.F image

Let $\left|z_1\right|=\left|z_2\right|=|z_1+z_2|=\left|z\right|$

$\Rightarrow$ A,B,C lie on circle of radius $\left|z\right|$

and center of origin

$|z_1+z_2|=\sqrt{{\left|z\right|}^2+{\left|z\right|}^2+2\times {\left|z\right|}^{2}cos\Theta }$

$\left|z\right|=\sqrt{2}\left|z\right|\sqrt{1+cos\Theta }$

$\Rightarrow 1=\sqrt{2}\sqrt{2}cos\Theta /2$

$\Rightarrow \Theta ={120}^{\circ }$

Clearly quadrilateral OACB is a rhombus

$\Rightarrow ar\left(△ABC\right)=ar\left(△OAB\right)$

$=ar\frac{1}{2}{\left|z\right|}^{2}sin\Theta$

$=\frac{1}{2}{\left|z_2\right|}^2\frac{\sqrt{3}}{2}[\because \left|z\right|=\left|z_2\right|]$

$=\frac{\sqrt{3}}{4}{\left|z_2\right|}^2$