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# Cube roots of unity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• 15 Questions around this concept.

## Solve by difficulty

Let ω be a complex number such that

where   if

then k is equal to :

## Concepts Covered - 1

Cube roots of unity

Let z be the cube root of unity (1)

So, z3 = 1

⇒ z3 - 1 = 0

⇒ (z - 1)(z2 + z + 1) = 0

⇒ z - 1 = 0  or z2 + z + 1 = 0

$\\\mathrm{\therefore \;z=1\;\;or\;\;z=\frac{-1\pm\sqrt{(1-4)}}{2}=\frac{-1\pm i\sqrt{3}}{2}} \\\mathrm{Therefore, \;\mathbf{z=1},\;\mathbf{z=\frac{-1+ i\sqrt{3}}{2}}\;\;and\;\;\mathbf{z=\frac{-1- i\sqrt{3}}{2}}}$

If the second root is represented by , then the third root will be represented by (we can check that by squaring the second root, we get the third root)

$\\\mathrm{\mathbf{\omega=\frac{-1+ i\sqrt{3}}{2}},\;\;\omega^2=\mathbf{\frac{-1- i\sqrt{3}}{2}}}$

So, 1, ⍵, ⍵2 are cube roots of unity and ⍵, ⍵2 are the non-real complex root of unity.

Properties of Cube roots of unity

i) 1 + ⍵ + ⍵2 = 0 and ⍵3 = 1 (Using sum and produc of roots relations for the equation z3 - 1 = 0)

ii) To find ⍵n , first we write ⍵ in multiple of 3 with remainder being 0 or 1 or 2.

Now ⍵n = ⍵3q + r = (⍵3)q·⍵r  = ⍵(Where r is from 0,1,2)

Eg, ⍵121 = ⍵3.40 + 1 = (⍵3)40·⍵1  = ⍵

iii) |⍵| = |⍵2 | = 1, arg(⍵) = 2π/3, arg(⍵2) = 4π/3 or -2π/3

iv) We can see that ⍵ and ⍵2  differ by the minus sign of imaginary part hence $\overline{\omega}=\omega^2$

v) Cube roots of -1 are -1, -⍵, -⍵2

vi) The cube roots of unity when represented on the complex plane has its point on vertices of triangle circumscribed by a unit circle whose one vertices lies on the     +ve X-axis.

## Study it with Videos

Cube roots of unity

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