Let $w = \frac{\sqrt{3}+i}{2}$ and $P = \{w^{n} : n = 1, 2, 3, \ldots\}$. Further $H_1 = \{z \in \mathbb{C} : \operatorname{Re} z > \frac{1}{2}\}$ and $H_2 = \{z \in \mathbb{C} : \operatorname{Re} z < -\frac{1}{2}\}$, where $\mathbb{C}$ is the set of all complex numbers. If $z_1 \in P \cap H_1$, $z_2 \in P \cap H_2$ and $O$ represents the origin, then $\angle z_1 O z_2 =$

Let w = (√3+i)/2 and P = {wⁿ : n = 1, 2, 3, …}. Further H₁ = {z ∈ C : Re z > 1/2} and H₂ = {z ∈ C : Re z < -1/2}, where C is the set of all complex numbers. If z₁ ∈ P ∩ H₁, z₂ ∈ P ∩ H₂ and O represents the origin, then ∠ z₁ O z₂ = $w = \dfrac{\sqrt{3}+i}{2},\qquad P = \{w^n : n = 1, 2, 3, \ldots\}$ $H_1 : \operatorname{Re} z > \tfrac{1}{2},\qquad H_2 : \operatorname{Re} z < -\tfrac{1}{2}$ $\text{To find: } \angle z_1 O z_2,\quad z_1 \in P \cap H_1,\ \ z_2 \in P \cap H_2$ $|w| = \sqrt{\tfrac{3}{4} + \tfrac{1}{4}} = 1$ $\arg w = \tan^{-1}\!\dfrac{1/2}{\sqrt{3}/2} = \dfrac{\pi}{6}$ $w = e^{i\pi/6} \Rightarrow P = \{\,e^{i n\pi/6} : n = 1, 2, 3, \ldots\,\}$ $\operatorname{Re} z = \cos\theta > \tfrac{1}{2} \Leftrightarrow \theta \in \left(-\tfrac{\pi}{3}, \tfrac{\pi}{3}\right)$ $P \cap H_1 = \{w,\ w^{11},\ w^{12}\} = \{e^{i\pi/6},\ e^{i11\pi/6},\ 1\}$ $\operatorname{Re} z = \cos\theta < -\tfrac{1}{2} \Leftrightarrow \theta \in \left(\tfrac{2\pi}{3}, \tfrac{4\pi}{3}\right)$ $P \cap H_2 = \{w^5,\ w^6,\ w^7\} = \{e^{i5\pi/6},\ e^{i\pi},\ e^{i7\pi/6}\}$ $z_1 = w = e^{i\pi/6},\qquad z_2 = w^5 = e^{i5\pi/6}$ $\angle z_1 O z_2 = \dfrac{5\pi}{6} - \dfrac{\pi}{6} = \dfrac{4\pi}{6} = \dfrac{2\pi}{3}$ $\angle z_1 O z_2 = \dfrac{2\pi}{3} \Rightarrow \text{option (3)}$