z = frac{3 + 2i sin theta}{1 - 2i sin theta}, | vedclass

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Question

(B) For a complex number $z$ to be purely imaginary,its real part must be zero,i.e.,$	ext{Re}(z) = 0$.Given $z = \frac{3 + 2i \sin 	heta}{1 - 2i \si

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$n \pi \pm \frac{\pi}{3}$,where $n \in Z$

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(B) For a complex number $z$ to be purely imaginary,its real part must be zero,i.e.,$	ext{Re}(z) = 0$.Given $z = \frac{3 + 2i \sin 	heta}{1 - 2i \sin 	heta}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin 	heta)$:$z = \frac{(3 + 2i \sin 	heta)(1 + 2i \sin 	heta)}{(1 - 2i \sin 	heta)(1 + 2i \sin 	heta)}$$z = \frac{3 + 6i \sin 	heta + 2i \sin 	heta + 4i^2 \sin^2 	heta}{1^2 + (2 \sin 	heta)^2}$Since $i^2 = -1$,we have:$z = \frac{3 + 8i \sin 	heta - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta}$$z = \frac{3 - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta} + i \frac{8 \sin 	heta}{1 + 4 \sin^2 	heta}$For $z$ to be purely imaginary,$	ext{Re}(z) = 0$:$\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)$Therefore,$	heta = n \pi \pm \frac{\pi}{3}$,where $n \in Z$.

$z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}, \quad (i = \sqrt{-1})$ will be purely imaginary if $\theta =$

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