Question

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

• A. sum of possible $z$ is $\frac{1}{2}$
• B. Sum of possible $z$ is $1$
• C. Product of possible $z$ is $\frac{1}{4}$
• D. Product of possible $z$ is $\frac{1}{2}$

Answer 1

Answer: (A), (C)

Product of possible $z$ is $\frac{1}{4}$

Triangle $ABC$ is equilateral. Hence,
$z^2+(-z{)}^2+(1-z{)}^2=z(-z)+z(1-z)+(-z)(1-z)$
$\Rightarrow 3z^2-2z+1=-z^2$
$\Rightarrow 4z^2-2z+1=0$
sum of roots $=\frac{2}{4}=\frac{1}{2}$
and product of roots is $=\frac{1}{4}$

Alternate Solution:
Using rotation of $A$ about $B$ with an angle of $\pi /3$, we get $C$
$\therefore \frac{1-z-(-z)}{z-(-z)}=e^{\pm i\pi /3}\left[\text{equilateral triangle}\right]\Rightarrow \frac{1}{2z}=-{\omega }^2,-\omega \Rightarrow z=-\frac{\omega }{2},-\frac{{\omega }^2}{2}$
$\therefore$ sum of the values of $z$
$=-\frac{1}{2}(\omega +{\omega }^2)=\frac{1}{2}$
and product of the values of $z=\frac{1}{4}$