Question

The cube roots of unity when represented on the Argand diagram from the vertices of an equilateral triangle.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A True

Let $w$ be the cube root of unity. $\therefore w^3=1\&1+w+w^2=0$where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
The vertices of triangle are $1,w\&w^2$
$a=|1-w|=|1+\frac{-1+i\sqrt{3}}{2}|=\left|\frac{1+i\sqrt{3}}{2}\right|=1b=|w-w^2|=|\frac{-1+i\sqrt{3}}{2}+\frac{-1-i\sqrt{3}}{2}|=1c=|1-w^2|=|1+\frac{-1-i\sqrt{3}}{2}|=\left|\frac{1-i\sqrt{3}}{2}\right|=1$
Since $a=b=c$, therefore triangle is an equilateral triangle.
Ans: A