If $1, \omega, \omega^2$ are the cube roots of unity,then their product is
- A$0$
- B$\omega$
- C$-1$
- D$1$
Explore More
Similar Questions
If $\omega$ is a complex cube root of unity,then $(1 + \omega - \omega^2)(1 - \omega + \omega^2) = $
Easy
View SolutionLet $z_k = \cos \left(\frac{2k\pi}{10}\right) + i \sin \left(\frac{2k\pi}{10}\right); k = 1, 2, \ldots, 9$.
List-$I$ List-$II$ $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$ $1.$ True $Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers. $2.$ False $R.$ $\frac{|1-z_1||1-z_2| \ldots |1-z_9|}{10}$ equals $3.$ $1$ $S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals $4.$ $2$
Codes: $P \quad Q \quad R \quad S$
| List-$I$ | List-$II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$ | $1.$ True |
| $Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{|1-z_1||1-z_2| \ldots |1-z_9|}{10}$ equals | $3.$ $1$ |
| $S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $P \quad Q \quad R \quad S$
If $1, \omega, \omega^2$ are the cube roots of unity,then $(1 - 2\omega + \omega^2)^6$ is equal to
Medium
View Solution${\left( \frac{-1 + i\sqrt{3}}{2} \right)^{20}} + {\left( \frac{-1 - i\sqrt{3}}{2} \right)^{20}} = $
Medium
View SolutionIf $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$,then $(\bar{z})^{100}$ lies in
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo