frac{3 + 2isin theta}{1 - 2isin theta} will b | vedclass

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Question

(C) complex number is purely imaginary if its real part is $0$. Multiply the numerator and denominator by the conjugate of the denominator: $\frac{3 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) complex number is purely imaginary if its real part is $0$. Multiply the numerator and denominator by the conjugate of the denominator: $\frac{3 + 2i\sin 	heta}{1 - 2i\sin 	heta} 	imes \frac{1 + 2i\sin 	heta}{1 + 2i\sin 	heta} = \frac{3 + 6i\sin 	heta + 2i\sin 	heta + 4i^2\sin^2 	heta}{1^2 + (2\sin 	heta)^2} = \frac{(3 - 4\sin^2 	heta) + i(8\sin 	heta)}{1 + 4\sin^2 	heta}$. For the expression to be purely imaginary,the real part must be $0$: $\frac{3 - 4\sin^2 	heta}{1 + 4\sin^2 	heta} = 0$ $3 - 4\sin^2 	heta = 0$ $\sin^2 	heta = \frac{3}{4}$ $\sin 	heta = \pm \frac{\sqrt{3}}{2} = \sin \left( \pm \frac{\pi}{3} ight)$ Therefore,$	heta = n\pi \pm \frac{\pi}{3}$.

$\frac{3 + 2i\sin \theta}{1 - 2i\sin \theta}$ will be purely imaginary,if $\theta = $ [Where $n$ is an integer]

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