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15 Questions around this concept.
Let ω be a complex number such that
where if
then k is equal to :
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
If the second root is represented by ⍵, then the third root will be represented by ⍵2 (we can check that by squaring the second root, we get the third root)
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 = ⍵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
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.
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