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

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(D) Let $z = \frac{3+2 i \sin 	heta}{1-2 i \sin 	heta}$. To make $z$ real,the imaginary part of $z$ must be zero. Multiply the numerator and denomin

Vedclass · Accepted Answer

$	heta=n \pi$ for $n \in \mathbb{Z}$

Vedclass · Answer

(D) Let $z = \frac{3+2 i \sin 	heta}{1-2 i \sin 	heta}$. To make $z$ real,the imaginary part of $z$ must be zero. 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 $i^2 = -1$: $z = \frac{(3 - 4 \sin^2 	heta) + i(8 \sin 	heta)}{1 + 4 \sin^2 	heta}$ For $z$ to be real,the imaginary part must be zero: $\frac{8 \sin 	heta}{1 + 4 \sin^2 	heta} = 0$ $8 \sin 	heta = 0$ $\sin 	heta = 0$ Therefore,$	heta = n \pi$ for $n \in \mathbb{Z}$.

The values of $\theta$,for which $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is real are

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