Question

If $\omega$ is imaginary cube root of unity, then arg(i $\omega$) + arg(i ${\omega }^2$) =

• A. 0
• B. π/2
• C. π
• D. -π

Answer 1

The correct option is D

-π

Basic concept

$\omega$ = $\frac{-1+\sqrt{3}i}{2}$ $\Rightarrow$ i$\omega$ = $\frac{-\sqrt{3}-i}{2}$ $\Rightarrow$ Arg(i $\omega$) = - $\frac{5\pi }{6}$

${\omega }^2$ = $\frac{-1-\sqrt{3}i}{2}$ $\Rightarrow$ i${\omega }^2$ = $\frac{\sqrt{3}-i}{2}$ $\Rightarrow$ $Arg\left(i{\omega }^2\right)$ = - $\frac{\pi }{6}$

∴ $Arg\left(i\omega \right)+Arg\left(i{\omega }^2\right)$ = -$\pi$