Question

If z is any non -zero complex number, then area of the triangle formed by the complex numbers $z,\omega z,z+\omega z$ which represents the sides is

• A. $\frac{\sqrt{3}}{2}{z}^2$
• B. $\frac{3}{4}\left|z\right|$
• C. $\frac{\sqrt{3}}{4}\left|z^2\right|$
• D. None of these

Answer 1

Answer: (C)

$\frac{\sqrt{3}}{4}\left|z^2\right|$

Fact: $\omega$ $=\frac{-1+i\sqrt{3}}{2}=e^{n\pi \beta }$ and $\omega$ $z=|z|$ ...(i)which means \omega is a vector obtained by rotating vector z in anticlockwise direction through an angle ${120}^0$ Now $|z+\omega z|=\left|z(1+\omega )\right|=\left|z(-{\omega }^2)\right|=\left|z\right|...\left(ii\right)$ $\Rightarrow z,z+\omega z,\omega z$ are sides of equilateral $\Delta$ we know that area of equilateral triangle $=\frac{\sqrt{3}}{4}{\left(side\right)}^2$ $=\frac{\sqrt{3}}{4}{\left|z\right|}^2$