The area of the triangle whose vertices are $i, \omega$ and $\omega^2$ is (Where $\omega$ is a complex cube root of unity other than $1$,$i$ is an imaginary number) . . . . . . sq.units

  • A
    $\frac{3 \sqrt{3}}{4}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\frac{3 \sqrt{3}}{2}$
  • D
    $\frac{\sqrt{3}}{4}$

Explore More

Similar Questions

If $z$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,then $\bar{z} \omega =$

Let $S=S_1 \cap S_2 \cap S_3$,where $S_1=\{z \in \mathbb{C}:|z|<4\}$,$S_2=\{z \in \mathbb{C}: \operatorname{Im}[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}]>0\}$,and $S_3=\{z \in \mathbb{C}: \operatorname{Re} z>0\}$.
$1.$ Area of $S=$
$(A) \frac{10 \pi}{3} \quad (B) \frac{20 \pi}{3} \quad (C) \frac{16 \pi}{3} \quad (D) \frac{32 \pi}{3}$
$2.$ $\min _{z \in S}|1-3 i-z|=$
$(A) \frac{2-\sqrt{3}}{2} \quad (B) \frac{2+\sqrt{3}}{2} \quad (C) \frac{3-\sqrt{3}}{2} \quad (D) \frac{3+\sqrt{3}}{2}$

For $a, b, c, d \in R$, if $z_1 = a + ib$ and $z_2 = c + id$ are such that $|z_1| = |z_2| = 1$ and $\operatorname{Re}(z_1 \bar{z}_2) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfy

Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = \{ z \in \mathbb{C} : \operatorname{Re}(a z^2 + bz) = a \text{ and } \operatorname{Re}(b z^2 + az) = b \}$ is equal to

If $P$ is a complex number whose modulus is $1$,then the equation $\left(\frac{1+iz}{1-iz}\right)^4=P$ has

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo