(C) Given,$\alpha^{2}-\alpha+1=0$. Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) Given,$\alpha^{2}-\alpha+1=0$. Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3}=-1$. Therefore,$\alpha^{6}=1$. We need to find $\alpha^{2011}$. Since $2011 = 6 \times 335 + 1$,we have $\alpha^{2011} = (\alpha^{6})^{335} \times \alpha^{1} = (1)^{335} \times \alpha = \alpha$. Thus,$\alpha^{2011} = \alpha$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

Difficult

If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$,then $\alpha^{2011}$ is equal to

KCET 2012 All KCET

A
$-\alpha$
B
$\alpha^{2}$
C
$\alpha$
D
$1$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

4-1.Complex numbers

Subtopic

De Moivre's theorem and Roots of unity

Previous Year

KCET 2012 Paper All KCET years →

The product of the four values of the complex number $(1+i)^{3/4}$ is

DifficultAP EAMCET 2025

View Solution

If $\omega$ is a complex cube root of unity and $(1+\omega)^7=A+B \omega$,then the values of $A$ and $B$ are,respectively.

DifficultMHT CET 2021

View Solution

If $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$,then the value of $x^3 + y^3 + z^3$ is equal to

MediumIIT 1970

View Solution

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is

MediumTS EAMCET 2019

View Solution

If $Z \neq 0$ is a complex number such that $Z^2 + Z|Z| + |Z|^2 = 0$,then $Z$ is in the set (Here $\omega$ is a complex cube root of unity).

MediumAP EAMCET 2017

View Solution

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

If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$,then $\alpha^{2011}$ is equal to

Similar Questions

The product of the four values of the complex number $(1+i)^{3/4}$ is

If $\omega$ is a complex cube root of unity and $(1+\omega)^7=A+B \omega$,then the values of $A$ and $B$ are,respectively.

If $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$,then the value of $x^3 + y^3 + z^3$ is equal to

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is

If $Z \neq 0$ is a complex number such that $Z^2 + Z|Z| + |Z|^2 = 0$,then $Z$ is in the set (Here $\omega$ is a complex cube root of unity).

Vedclass Test Series

Exam Paper Generator

Online Exam Module