Question

What is $\sqrt{\frac{1+{\omega }^2}{1+\omega }}$ equal to, where $\omega$ is the cube root of unity?

Answer 1

Since $\omega$ being the cube root of unity we have ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$.

So, $1+{\omega }^2=-\omega$.....(1) and $1+\omega =-{\omega }^2$.......(2).

Now

$\sqrt{\frac{1+{\omega }^2}{1+\omega }}$

$=\sqrt{\frac{-\omega }{-{\omega }^2}}$ {Using (1) and (2)]

$=\sqrt{\frac{1}{\omega }}$.