Angle Subtended at Origin by Elements in Complex Subsets

To find the possible values of the angle $\angle z_1 O z_2$, we begin by expressing the given complex number $w$ in polar (Euler's) form: $w = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$

The set $P$ is given by: $P = \{w^n : n = 1, 2, 3, \ldots\}$ Since $w^{12} = e^{i 2\pi} = 1$, the sequence $w^n$ is periodic with period $12$. Thus, $P$ has $12$ distinct elements: $w^n = e^{i n\pi/6} = \cos\left(\frac{n\pi}{6}\right) + i \sin\left(\frac{n\pi}{6}\right), \quad \text{for } n \in \{1, 2, 3, \ldots, 12\}$

Next, we identify the elements belonging to the subsets $H_1$ and $H_2$.

1. Determining the set $P \cap H_1$:

The set $H_1$ consists of complex numbers $z$ such that $\text{Re}(z) > \frac{1}{2}$. For $z = w^n$: $\text{Re}(w^n) = \cos\left(\frac{n\pi}{6}\right) > \frac{1}{2}$ This inequality holds when: $-\frac{\pi}{3} < \frac{n\pi}{6} < \frac{\pi}{3} \pmod{2\pi}$ Checking for integer values of $n \in \{1, 2, \ldots, 12\}$:

• For $n = 1$: $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} > \frac{1}{2}$
• For $n = 2$: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \ngtr \frac{1}{2}$
• For $n = 11$: $\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} > \frac{1}{2}$
• For $n = 12$: $\cos(2\pi) = 1 > \frac{1}{2}$

Thus, the possible arguments $\theta_1 = \arg(z_1)$ for $z_1 \in P \cap H_1$ are: $\theta_1 \in \left\{ -\frac{\pi}{6}, 0, \frac{\pi}{6} \right\}$

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2. Determining the set $P \cap H_2$:

The set $H_2$ consists of complex numbers $z$ such that $\text{Re}(z) < -\frac{1}{2}$. For $z = w^n$: $\text{Re}(w^n) = \cos\left(\frac{n\pi}{6}\right) < -\frac{1}{2}$ This inequality holds when: $\frac{2\pi}{3} < \frac{n\pi}{6} < \frac{4\pi}{3} \pmod{2\pi}$ Checking for integer values of $n \in \{1, 2, \ldots, 12\}$:

• For $n = 4$: $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \nless -\frac{1}{2}$
• For $n = 5$: $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} < -\frac{1}{2}$
• For $n = 6$: $\cos(\pi) = -1 < -\frac{1}{2}$
• For $n = 7$: $\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} < -\frac{1}{2}$
• For $n = 8$: $\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \nless -\frac{1}{2}$

Thus, the possible arguments $\theta_2 = \arg(z_2)$ for $z_2 \in P \cap H_2$ are: $\theta_2 \in \left\{ \frac{5\pi}{6}, \pi, \frac{7\pi}{6} \right\}$

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3. Finding the angle $\angle z_1 O z_2$:

The angle between the lines connecting the origin $O$ to $z_1$ and $z_2$ is the acute or obtuse geometric angle $\angle z_1 O z_2 \in [0, \pi]$, given by: $\angle z_1 O z_2 = \min\left(|\theta_2 - \theta_1|, 2\pi - |\theta_2 - \theta_1|\right)$

Evaluating all possible pairings between $\theta_1 \in \left\{ -\frac{\pi}{6}, 0, \frac{\pi}{6} \right\}$ and $\theta_2 \in \left\{ \frac{5\pi}{6}, \pi, \frac{7\pi}{6} \right\}$:

1. If $\theta_1 = \frac{\pi}{6}$:

   • For $\theta_2 = \frac{5\pi}{6}$: $\angle z_1 O z_2 = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$
   • For $\theta_2 = \pi$: $\angle z_1 O z_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
   • For $\theta_2 = \frac{7\pi}{6}$: $\angle z_1 O z_2 = \frac{7\pi}{6} - \frac{\pi}{6} = \pi$

2. If $\theta_1 = 0$:

   • For $\theta_2 = \frac{5\pi}{6}$: $\angle z_1 O z_2 = \frac{5\pi}{6} - 0 = \frac{5\pi}{6}$
   • For $\theta_2 = \pi$: $\angle z_1 O z_2 = \pi - 0 = \pi$
   • For $\theta_2 = \frac{7\pi}{6}$: $\angle z_1 O z_2 = 2\pi - \frac{7\pi}{6} = \frac{5\pi}{6}$

3. If $\theta_1 = -\frac{\pi}{6}$:

   • For $\theta_2 = \frac{5\pi}{6}$: $\angle z_1 O z_2 = \frac{5\pi}{6} - \left(-\frac{\pi}{6}\right) = \pi$
   • For $\theta_2 = \pi$: $\angle z_1 O z_2 = 2\pi - \left(\pi - \left(-\frac{\pi}{6}\right)\right) = 2\pi - \frac{7\pi}{6} = \frac{5\pi}{6}$
   • For $\theta_2 = \frac{7\pi}{6} \equiv -\frac{5\pi}{6}$: $\angle z_1 O z_2 = \left| -\frac{5\pi}{6} - \left(-\frac{\pi}{6}\right) \right| = \frac{4\pi}{6} = \frac{2\pi}{3}$

Hence, the possible values for $\angle z_1 O z_2$ are: $\left\{ \frac{2\pi}{3}, \frac{5\pi}{6}, \pi \right\}$

Matching with the given options, the possible values are $\frac{2\pi}{3}$ and $\frac{5\pi}{6}$.

Therefore, the correct options are C and D.