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

Answer: 6

We have, ${\left|a\omega +b\right|}^2=1$
$\Rightarrow \left(a\omega +b\right)\left(a\stackrel{-}{\omega }+b\right)=1$
$\Rightarrow a^2+ab\left(\omega +\stackrel{-}{\omega }\right)+b^2=1$
$\Rightarrow a^2-ab+b^2=1$
$\Rightarrow {\left(a-b\right)}^2+ab=1\dots ..\left(i\right)$
$\left(As,1+\omega +{\left(\omega \right)}^2=0\right)$
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.