Question

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

• A. 1
• B. $\omega$
• C. ${\omega }^2$
• D. $i\omega$, where $i=\sqrt{-1}$

Answer 1

The correct option is B $\omega$

To find : $\sqrt{\frac{1+{\omega }^2}{1+\omega }}$

$\omega$ is the cube root of unity

So, we have ${\omega }^3=1$

$\Rightarrow 1+\omega +{\omega }^2=0$

$\Rightarrow 1+\omega =-{\omega }^2$ and $1+{\omega }^2=-\omega$,

$\Rightarrow \sqrt{\frac{1+{\omega }^2}{1+\omega }}=\sqrt{\frac{w}{w^2}}=\sqrt{\frac{1}{\omega }}$

Now, ${\omega }^3=1$ $\Rightarrow \frac{1}{\omega }={\omega }^2$

$\therefore \sqrt{\frac{1}{\omega }}=\sqrt{{\omega }^2}=\omega$

Hence, B is correct.