Suppose $\omega$ is a cube root of unity with $\omega \neq 1$. Suppose $P$ and Q are the points on the complex plane defined by $\omega$ and $\omega^2$. If $O$ is the origin, then what is the angle between OP and OQ ?

• A

  $60^{\circ}$
• B

  $90^{\circ}$
• Correct answer
  C

  $120^{\circ}$
• D

  $150^{\circ}$

Solution

$\begin{aligned} & \because \omega=\frac{-1}{2}+i \frac{\sqrt{3}}{2} \\ & \& \omega^2=\frac{-1}{2}-i \frac{\sqrt{3}}{2} \\ & -\frac{1}{2}+\frac{\sqrt{3}}{2} i \\ & (\because \omega \neq 1)\end{aligned}$
$P$ and $Q$ are points on complex plane. Angle between $O P$ and $O Q$ is $m_1$ for line $O P$, $\begin{aligned} & m_1=\frac{\frac{\sqrt{3}}{2}-0}{\frac{-1}{2}-0} \\ & \Rightarrow m_1=-\sqrt{3} \end{aligned}$ $m_2$ for line $O Q$ $\begin{aligned} & m_2=\frac{\frac{-\sqrt{3}}{2}-0}{\frac{-1}{2}-0} \\ & \Rightarrow m_2=\sqrt{3} \end{aligned}$ $\begin{aligned} & \theta=\tan ^{-1}\left[\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}\right] \\ & =\tan ^{-1}\left[\frac{-2 \sqrt{3}}{-2}\right]=\pi-\tan ^{-1} \sqrt{3}=\pi-\tan ^{-1} \tan \frac{\pi}{3} \\ & \theta=\pi-\frac{\pi}{3} \\ & \theta=120^{\circ}\end{aligned}$

Hence, the correct answer is option: 'C'.