Question

The area of the triangle whose vertices are $i,\alpha ,\beta$ where $i=\sqrt{-1}$ and $\alpha ,\beta$ are the nonreal cuberoots of unity, is

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{3\sqrt{3}}{4}$
• C. 0
• D. $\frac{\sqrt{3}}{4}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{4}$

Hence the vertices are
$1,w,w^2$
Now these form an equilateral triangle with sides length of
$|w|=|w^2|=1$ unit.
Thus, the required area is
$\frac{\sqrt{3}{a}^2}{4}$
Now $a=1,$
Therefore, the area is $\frac{\sqrt{3}}{4}$
Hence, option 'D' is correct.