Question

The area of triangle formed by complex numbers $z$ ,$\omega zandz+\omega z$ as its sides is $\sqrt{k}$ sq. units, given that $|z{|}^2=4$.find k

Answer 1

Given complex numbers $z,wz$ and $z+wz$.

since,

$|z|=|wz|$................[$|w|=1$]

Also $z(1+w)=z(-w^2)$

Hence, $|z|=|z+wz|$..........[$|-w^2|=1$]

So the given triangle is an equilateral triangle with side $|z|$.

Also $|z{|}^2=4⟹|z|=2$

Hence area of equilateral triangle$=\frac{\sqrt{3}}{4}{2}^2=\sqrt{3}$

Hence, $k=3$.