Let A = {theta in (0, 2pi) : frac{1+2i sin th | vedclass

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(C) Let $z = \frac{1+2i \sin 	heta}{1-i \sin 	heta}$.To make $z$ purely imaginary, its real part must be zero.Multiply the numerator and denominator

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$4\pi$

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(C) Let $z = \frac{1+2i \sin 	heta}{1-i \sin 	heta}$.To make $z$ purely imaginary, its real part must be zero.Multiply the numerator and denominator by the conjugate of the denominator $(1+i \sin 	heta)$:$z = \frac{(1+2i \sin 	heta)(1+i \sin 	heta)}{(1-i \sin 	heta)(1+i \sin 	heta)} = \frac{1 + i \sin 	heta + 2i \sin 	heta + 2i^2 \sin^2 	heta}{1 + \sin^2 	heta} = \frac{(1 - 2 \sin^2 	heta) + i(3 \sin 	heta)}{1 + \sin^2 	heta}$.For $z$ to be purely imaginary, $\operatorname{Re}(z) = 0$, so $\frac{1 - 2 \sin^2 	heta}{1 + \sin^2 	heta} = 0$.This implies $1 - 2 \sin^2 	heta = 0$, or $\sin^2 	heta = \frac{1}{2}$, which means $\sin 	heta = \pm \frac{1}{\sqrt{2}}$.In the interval $(0, 2\pi)$, the solutions for $	heta$ are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.The sum of these elements is $\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$.

Let $A = \{\theta \in (0, 2\pi) : \frac{1+2i \sin \theta}{1-i \sin \theta} \text{ is purely imaginary} \}$. Then the sum of the elements in $A$ is

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