Question

lf $z$ lies on the circle centered at origin. lf area of the triangle whose vertices are $z$, $\omega z$ and $z+\omega z$, where $\omega$ is the cube root of unity, is $4\sqrt{3}$ sq. units.

The radius of the circle is?

• A. $1$ units
• B. $2$ units
• C. $3$ units
• D. $4$ units

Answer 1

Answer: (D)

$4$ units

We know that,

Angle between $1,w,$ and $w^2$ is ${120}^0$

Therefore,

Angle between $z,zw$ is also ${120}^0$

Hence, it forms a isosceles triangle with side length $|z|,|zw|$ and $|z+wz|$ and equal angles will be ${30}^0$.

Therefore, the area of isosceles triangle will be

$\frac{1}{2}\times 2\times |z|\cos {30}^0\times |z|\sin {30}^0=4\sqrt{3}$

$\Rightarrow \frac{|}{z}{|}^{2}4=4$

$\Rightarrow |z|=r=4$