Question

If $\omega $ is the non-real cube root of unity, then the number of ordered pairs of integers $\left(a , b\right)$ , such that $\left|a \omega + b\right|=1$ , is equal to

Answer 1

We have, $\left|a \omega + b\right|^{2}=1$
$\Rightarrow \left(a \omega + b\right)\left(a \overset{-}{\omega } + b\right)=1$
$\Rightarrow a^{2}+ab\left(\omega + \overset{-}{\omega }\right)+b^{2}=1$
$\Rightarrow a^{2}-ab+b^{2}=1$
$\Rightarrow \left(a - b\right)^{2}+ab=1\ldots ..\left(i\right)$
$\left(A s, \, 1 + \omega + \left(\omega \right)^{2} = 0\right)$
When $\left(a - b\right)^{2}=0$ and $ab=1$ then $\left(1 , 1\right);\left(- 1 , - 1\right)$
When $\left(a - b\right)^{2}=1$ and $ab=0$ then $\left(0 , 1\right);\left(1 , 0\right);\left(0 , - 1\right);\left(- 1 , 0\right)$
Hence, $\left(0 , 1\right);\left(1 , 0\right);\left(0 , - 1\right);\left(- 1 , 0\right);\left(1 , 1\right);\left(- 1 , - 1\right)$ i.e., $6$ ordered pairs.