what is its radius for the circle? If z lies on the circle centred at origin, the area of the triangle is $4 \sqrt 3$ sq. unit. And $z, \omega z,z+\omega z$ are the vertices of triangle. What is the radius ?
 A: The area of the triangle is the half of the cross product of two sides, obtained by
$$2A=|\Im((z+\omega z-z)(z+\omega z-\omega z)^*)|=|\Im(\omega zz^*)|=|z^2||\Im(\omega)|.$$
Hence,
$$r=\sqrt{\frac{2A}{|\Im(\omega)|}}.$$
A: The sides of the triangle have lengths
$$|z-\omega z|=|z(1-\omega)|=|1-\omega||z|$$
$$|z-(\omega z+z)|=|\omega z|=|\omega||z|$$
$$|\omega z+z-\omega z|=|z|$$
Use Heron's formula
$$A^2=\frac{|1-\omega|+|\omega|+1}{2} |z|^4 \left(\frac{|1-\omega|+|\omega|+1}{2}-|1-\omega|\right)\\ \cdot\left(\frac{|1-\omega|+|\omega|+1}{2}-|\omega|\right)\left(\frac{|1-\omega|+|\omega|+1}{2}-1\right)$$
You know $A$. Solve for $|z|$.
A: You don't need all that.  A little geometric reasoning works.
Render $\omega$ as either complex conjugate cube root of unity.  Then $1+\omega=-\overline\omega$ where $\overline\omega$ is the second complex conjugate cube root of unity.  Plot this out on the complex plane, for an arbitrary value of $z$, and you see that the origin and the triangle vertices are collectively the vertices of a rhombus whose sides are $r=|z|$.  The angles of this rhombus measure $60°$.  Its area, which is twice that of the original triangle because the long side of the triangle is a diagonal of the rhombus, is:
$A=2(4\sqrt{3})=r^2\sin(60°)$
Of course the trigonometric function value is just $(\sqrt{3})/2$, so you easily solve for $r$.
