Class 11Mathematics1 Complex Numbers and Quadratic Equations: Complex NumbersDe Moivre's theorem and Roots of unity
DifficultIf $1, \omega, \omega^2$ are the three cube roots of unity,then $(3 + \omega^2 + \omega^4)^6 = $
- A$64$
- B$729$
- C$2$
- D$0$
Explore More
Similar Questions
The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2018}+z^{2017}+1=0$ satisfy the equation
The value of $(1 - \omega + \omega^2)(1 - \omega^2 + \omega)^6$,where $\omega, \omega^2$ are the complex cube roots of unity.
Medium
View SolutionIf $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1$ where $m, n \in N$,then the greatest common divisor of the least values of $m$ and $n$ is:
If $\alpha, \beta \in \mathbb{C}$ are distinct roots of the equation $x^2 - x + 1 = 0$,then $\alpha^{101} + \beta^{107}$ is equal to:
If $Z = r(\cos \theta + i \sin \theta), (\theta \neq -\pi / 2)$ is a solution of $x^3 = i$,then $r^9(\cos \theta + i \sin \theta)^9 = $
Vedclass 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