Let w = frac{sqrt{3} + i}{2} and P = {w^n : n | vedclass

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(C) Given $w = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Thus, $P = \{e^{in\pi/6} : n =

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$(C) \frac{2\pi}{3}$

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(C) Given $w = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Thus, $P = \{e^{in\pi/6} : n = 1, 2, 3, \ldots\}$.$H_1 = \{z : \operatorname{Re}(z) > 1/2\}$. For $z = e^{in\pi/6} = \cos(n\pi/6) + i \sin(n\pi/6)$, $\operatorname{Re}(z) = \cos(n\pi/6) > 1/2$ implies $n\pi/6 \in (0, \pi/3) \cup (5\pi/3, 2\pi)$. For $n \in Z^+$, this gives $n = 1$ $(z_1 = e^{i\pi/6} = \frac{\sqrt{3}+i}{2})$ and $n = 11$ $(z_1 = e^{i11\pi/6} = \frac{\sqrt{3}-i}{2})$.$H_2 = \{z : \operatorname{Re}(z) The possible angles $\angle z_1 O z_2$ are the differences in arguments: $|\arg(z_1) - \arg(z_2)|$.Possible arguments for $z_1$ are $\pm \pi/6$. Possible arguments for $z_2$ are $\pm 5\pi/6$.The differences are $|5\pi/6 - \pi/6| = 4\pi/6 = 2\pi/3$, $|-5\pi/6 - \pi/6| = |-\pi| = \pi$, $|5\pi/6 - (-\pi/6)| = \pi$, and $|-5\pi/6 - (-\pi/6)| = |-4\pi/6| = 2\pi/3$.Thus, the possible values are $2\pi/3$ and $\pi$. Comparing with the options, $2\pi/3$ is present.

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