Let A = left{ theta in left( -frac{pi}{2}, pi | vedclass

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(D) Let $z = \frac{3 + 2i \sin 	heta}{1 - 2i \sin 	heta}$.Multiplying the numerator and denominator by the conjugate of the denominator: $z = \frac{

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$\frac{2\pi}{3}$

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(D) Let $z = \frac{3 + 2i \sin 	heta}{1 - 2i \sin 	heta}$.Multiplying the numerator and denominator by the conjugate of the denominator: $z = \frac{(3 + 2i \sin 	heta)(1 + 2i \sin 	heta)}{(1 - 2i \sin 	heta)(1 + 2i \sin 	heta)}$$z = \frac{3 + 6i \sin 	heta + 2i \sin 	heta - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta} = \frac{(3 - 4 \sin^2 	heta) + 8i \sin 	heta}{1 + 4 \sin^2 	heta}$.For $z$ to be purely imaginary,the real part must be zero:$3 - 4 \sin^2 	heta = 0 \implies \sin^2 	heta = \frac{3}{4} \implies \sin 	heta = \pm \frac{\sqrt{3}}{2}$.Given $	heta \in \left( -\frac{\pi}{2}, \pi ight)$:If $\sin 	heta = \frac{\sqrt{3}}{2}$,then $	heta = \frac{\pi}{3}$ or $	heta = \frac{2\pi}{3}$.If $\sin 	heta = -\frac{\sqrt{3}}{2}$,then $	heta = -\frac{\pi}{3}$.The set $A = \left\{ -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} ight\}$.The sum of the elements is $-\frac{\pi}{3} + \frac{\pi}{3} + \frac{2\pi}{3} = \frac{2\pi}{3}$.

Let $A = \left\{ \theta \in \left( -\frac{\pi}{2}, \pi \right) : \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \text{ is purely imaginary} \right\}$. Then the sum of the elements in $A$ is

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