A value of theta for which frac{2 + 3i sin th | vedclass

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Question

(B) For a complex number to be purely imaginary,its real part must be equal to $0$.Given expression: $Z = \frac{2 + 3i \sin 	heta}{1 - 2i \sin 	heta

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a complex number to be purely imaginary,its real part must be equal to $0$.Given expression: $Z = \frac{2 + 3i \sin 	heta}{1 - 2i \sin 	heta}$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)}$$Z = \frac{2 + 4i \sin 	heta + 3i \sin 	heta + 6i^2 \sin^2 	heta}{1 - (2i \sin 	heta)^2}$Since $i^2 = -1$:$Z = \frac{2 + 7i \sin 	heta - 6 \sin^2 	heta}{1 + 4 \sin^2 	heta}$$Z = \frac{2 - 6 \sin^2 	heta}{1 + 4 \sin^2 	heta} + i \frac{7 \sin 	heta}{1 + 4 \sin^2 	heta}$Setting the real part to $0$:$\frac{2 - 6 \sin^2 	heta}{1 + 4 \sin^2 	heta} = 0$$2 - 6 \sin^2 	heta = 0$$6 \sin^2 	heta = 2$$\sin^2 	heta = \frac{1}{3}$$\sin 	heta = \pm \frac{1}{\sqrt{3}}$Therefore,$	heta = \sin^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

$A$ value of $\theta$ for which $\frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$ is purely imaginary,is:

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