JEE Maths›Complex Numbers & Quadratic Equations›Visual Solution
Visual SolutionPYQ 2024Standard
Question
If ω\omega is a cube root of unity, then 1+ω+ω21 + \omega + \omega^2 equals:
(A)11
(B)−1-1
(C)00
(D)ω\omega
Solution Path
z2+z+1=0z^2 + z + 1 = 0 has roots ω,ω2\omega, \omega^2. Sum of roots =−1= -1, so 1+ω+ω2=01 + \omega + \omega^2 = 0.
01Question Setup
1/4
Find the value of 1+ω+ω21 + \omega + \omega^2 where ω\omega is a cube root of unity.
1+ω+ω2=  ?1 + \omega + \omega^2 = \;?
02Cube Roots of Unity
2/4
z3=1z^3 = 1 factors as (z−1)(z2+z+1)=0(z-1)(z^2+z+1) = 0. Three roots: 1,ω,ω21, \omega, \omega^2 on the unit circle.
z3−1=(z−1)(z2+z+1)z^3 - 1 = (z-1)(z^2+z+1)
03Sum IdentityKEY INSIGHT
3/4
From z2+z+1=0z^2+z+1=0: ω+ω2=−1\omega + \omega^2 = -1. Therefore 1+ω+ω2=01 + \omega + \omega^2 = 0.
1+ω+ω2=01 + \omega + \omega^2 = 0
04Final Answer
4/4
The sum of all cube roots of unity is always zero.
0\boxed{0}
Concepts from this question2 concepts unlocked
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De Moivre's Theorem

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(cos \theta + i sin \theta)^n = cos n\theta + i sin n\theta

(eiθ)n=einθ(e^{i\theta})^n = e^{in\theta}

Powers of complex numbers become trivial. Core tool for roots of unity problems.

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Euler Form Multiplication

EASY

Multiplying in Euler form = adding exponents

eiα⋅eiβ=ei(α+β)e^{i\alpha} \cdot e^{i\beta} = e^{i(\alpha + \beta)}

Turns complex multiplication into simple addition - saves 2-3 minutes per question

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