If frac{3-2 i sin theta}{1+2 i sin theta} is | vedclass

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(C) Let $z = \frac{3-2 i \sin 	heta}{1+2 i \sin 	heta}$.To rationalize,multiply the numerator and denominator by the conjugate of the denominator $(

Vedclass · Accepted Answer

$n \pi \pm \frac{\pi}{3}$

Vedclass · Answer

(C) Let $z = \frac{3-2 i \sin 	heta}{1+2 i \sin 	heta}$.To rationalize,multiply the numerator and denominator by the conjugate of the denominator $(1-2 i \sin 	heta)$:$z = \frac{(3-2 i \sin 	heta)(1-2 i \sin 	heta)}{(1+2 i \sin 	heta)(1-2 i \sin 	heta)}$$z = \frac{3 - 6 i \sin 	heta - 2 i \sin 	heta + 4 i^2 \sin^2 	heta}{1 + 4 \sin^2 	heta}$Since $i^2 = -1$,we have:$z = \frac{3 - 4 \sin^2 	heta - 8 i \sin 	heta}{1 + 4 \sin^2 	heta}$For $z$ to be a purely imaginary number,the real part must be zero:$\frac{3 - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta} = 0$$3 - 4 \sin^2 	heta = 0$$\sin^2 	heta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}ight)^2$$\sin^2 	heta = \sin^2 \left(\frac{\pi}{3}ight)$Therefore,the general solution is $	heta = n \pi \pm \frac{\pi}{3}$.

If $\frac{3-2 i \sin \theta}{1+2 i \sin \theta}$ is a purely imaginary number,then $\theta=$

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