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

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Solution

$| a ω + b | = 1$

$\Rightarrow | a ω + b |^{2} = 1 \Rightarrow (a ω + b) (a ¯ ω + b) = 1$

$\Rightarrow a^{2} - a b + b^{2} = 1$

$\Rightarrow (a - b)^{2} + a b = 1$ ....(i)

when $(a - b)^{2} = 0$ and ab = 1, then (1, 1); (-1, -1)

when $(a - b)^{2} = 1$ and ab = 0, then (0, 1), (1, 0), (0, -1), (-1, 0)

Hence, 6 ordered pairs

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