Question

If the points $A\left(z\right),B(-z)$ and $C(z+1)$ are vertices of an equilateral triangle then

• A. area of triangle is $\left(\sqrt{3}/4\right)sq.units$
• B. $Re\left(z\right)=1/2$
• C. perimeter of the triangle is $3units$
• D. $Re\left(z\right)=-1/4$

Answer 1

Answer: (A), (C), (D)

The correct options are
A area of triangle is $\left(\sqrt{3}/4\right)sq.units$
C perimeter of the triangle is $3units$
D $Re\left(z\right)=-1/4$

As we know that if,

$A(z_1{)}^2+(z_2{)}^2+(z_3{)}^3=z_1{z}_2+z_2{z}_3+z_3{z}_1\Rightarrow z^2+(-z{)}^2+(z+1{)}^2=z(-z)+(-z\left)\right(z+1)+z(z+1)\Rightarrow 4z^2+2z+1=0$

$\therefore z=\frac{-1\pm \sqrt{3}i}{4}\Rightarrow Re\left(z\right)=\frac{-1}{4}$

$AB = z \, |z| = 2 \sqrt{\left(\dfrac{-1}{4}\right)^2 + \left( \pm \dfrac{\sqrt{3}}{4}\right)^2 } = 2 .\dfrac{1}{2} = 1$

$\therefore AB=BC=AC=1$ unit

$\therefore$ Area of equilateral $\Delta =\frac{\sqrt{3}}{4}(side{)}^2=\frac{\sqrt{3}}{4}uni{t}^2$

Perimeter $=3\times \left(side\right)=3\times 1unit=3units$

$\therefore$ Area = $\frac{\sqrt{3}}{4}$ SQ. units (A) option is correct.

$\therefore$ Perimeter = 3 units (C) option is correct.

$\therefore$ Re(z) = $\frac{-1}{4}$ unit (D) option is correct.

$\therefore$ A, C, D options are correct.