(A) The cube roots of unity are the solutions to the equation $z^3 = 1$.These roots are given by $1$,$\omega = \frac{-1 + i\sqrt{3}}{2}$,and $\omega^2

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(A) The cube roots of unity are the solutions to the equation $z^3 = 1$.These roots are given by $1$,$\omega = \frac{-1 + i\sqrt{3}}{2}$,and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Comparing these with the given options,the correct value is $\frac{-1 + i\sqrt{3}}{2}$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers De Moivre's theorem and Roots of unity

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One of the cube roots of unity is

A
$\frac{-1 + i\sqrt{3}}{2}$
B
$\frac{1 + i\sqrt{3}}{2}$
C
$\frac{1 - i\sqrt{3}}{2}$
D
$\frac{\sqrt{3} - i}{2}$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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De Moivre's theorem and Roots of unity

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

Easy

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If the complex cube roots of $(-i)$ are $\alpha, \beta, \gamma$,then $\alpha^2+\beta^2+\gamma^2=$

DifficultTS EAMCET 2019

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For $n \in Z^{+}$,$(1+\sin \theta+i \cos \theta)^n+(1+\sin \theta-i \cos \theta)^n=$

DifficultAP EAMCET 2018

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If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^n+\beta^n = \ldots \cos \left(\frac{n\pi}{3}\right)$ for any $n \in N$.

DifficultAP EAMCET 2020

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If $1, \omega, \omega^{2}$ are three cube roots of unity,then $(1-\omega+\omega^{2})(1+\omega-\omega^{2})$ is

MediumKCET 2015

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One of the cube roots of unity is

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If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

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