JEE maths#19

Algebra Level 4

If ω \omega is a primitive cube root of unity, then find the number of ordered pairs of integers ( a , b ) (a,b) such that ∣ a ω + b ∣ = 1 |a \omega + b| = 1 .

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The answer is 6.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Rahil Sehgal
Apr 3, 2017

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Aman Bhandare
Apr 4, 2017

Using 1+ω+ω^2=0 and the fact that modulus of any root of unity equals 1, one can figure out all possible ordered pairs which are (1,0), (-1,0) , (1,1) , (-1,-1) , (0,1) , (0,-1)

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