The value of theta, for which frac{2+3i sin t | vedclass

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

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$\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(C) Let $z = \frac{2+3i \sin 	heta}{1-2i \sin 	heta}$.To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(1+2i \sin 	heta)$:$z = \frac{(2+3i \sin 	heta)(1+2i \sin 	heta)}{(1-2i \sin 	heta)(1+2i \sin 	heta)}$$z = \frac{2 + 4i \sin 	heta + 3i \sin 	heta + 6i^2 \sin^2 	heta}{1 + 4 \sin^2 	heta}$Since $i^2 = -1$, we have:$z = \frac{(2 - 6 \sin^2 	heta) + i(7 \sin 	heta)}{1 + 4 \sin^2 	heta}$For $z$ to be purely imaginary, the real part must be zero:$\operatorname{Re}(z) = \frac{2 - 6 \sin^2 	heta}{1 + 4 \sin^2 	heta} = 0$$2 - 6 \sin^2 	heta = 0$$\sin^2 	heta = \frac{2}{6} = \frac{1}{3}$$\sin 	heta = \frac{1}{\sqrt{3}}$$	heta = \sin^{-1}\left(\frac{1}{\sqrt{3}}ight)$

The value of $\theta$, for which $\frac{2+3i \sin \theta}{1-2i \sin \theta}$ is purely imaginary, where $i=\sqrt{-1}$, is

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