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Question

If ω is the imaginary cube root of unity, then find the member of ordered pairs of integers (a, b) such that  |aω+b|=1.___


Solution

|aω+b|=1

|aω+b|2=1(aω+b)(a¯ω+b)=1

a2ab+b2=1

(ab)2+ab=1     ....(i)

when (ab)2=0 and ab = 1, then (1, 1); (-1, -1) 

when (ab)2=1 and ab = 0, then (0, 1), (1, 0), (0, -1), (-1, 0) 

Hence, 6 ordered pairs

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