Question

Let $\omega \left(\omega \neq 1\right)$ is a cube root of unity, such that $\left(1+\omega^{2}\right)^{8}$ , where $a,b\in R,$ then $\left|a + b\right|$ is equal to

• A. $1$
• B. $3$
• C. $0$
• D. $2$

Answer 1

Answer: (D)

We have, $\left(- \omega \right)^{8}=a+b\omega $
$\Rightarrow \omega^{2}=a+b \omega .$
$\Rightarrow -\frac{1}{2}-i\frac{\sqrt{3}}{2}=a+b\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)$
Comparing real and imaginary parts
$a=-1,b=-1$
$\Rightarrow \left|a + b\right|=2$