what is its radius for the circle?

Question

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 ?

Answer 1

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,

$$r=\sqrt{\frac{2A}{|\Im(\omega)|}}.$$

Answer 2

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 $|z|$.

Answer 3

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 $r$.