# Let $A = \left\{ \theta \in \left (-\frac{\pi}{2}, \pi \right ): \frac{3+2 i\sin\theta}{1-2i\sin\theta} \textup{\;is\;purely\;imaginary}\right \}$ . Then the sum of the elements in A is:Option 1)$\frac{5\pi}{6}$Option 2)$\pi$Option 3)$\frac{3\pi}{4}$Option 4)$\frac{2\pi}{3}$

Purely Imaginary Complex Number -

$z=x+iy, \boldsymbol{x=0}, y\epsilon R$

& i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

from the concept if z is purely img, then real part of z is = 0

$Z=\frac{3+2i\sin \theta }{1-2i\sin \theta }$

factorize Z

$Z=\frac{3+2i\sin \theta }{1-2i\sin \theta }\times \frac{1+2i\sin \theta }{1+2i\sin \left ( \theta \right )}$

$Z=\left ( \frac{3-4\sin ^{2}\theta }{1+4\sin ^{2}\theta } \right )+i\left ( \frac{8\sin \theta }{1+4\sin ^{2}\theta } \right )$

$Re\left ( z \right )=0$

So,

$\frac{3-4\sin ^{2}\left ( \theta \right )}{1+4\sin ^{2}\left ( \theta \right )}=0\Rightarrow \sin ^{2}\theta =\frac{3}{4}$

$\Rightarrow \sin \left ( \theta \right )=\pm \frac{\sqrt{3}}{2}\Rightarrow \theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}$

$\because \theta \equiv \left ( -\frac{\pi }{2} ,\pi \right )$

The sum of the element in A is

$-\frac{\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$

Option 1)

$\frac{5\pi}{6}$

Option 2)

$\pi$

Option 3)

$\frac{3\pi}{4}$

Option 4)

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

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