The expression frac{3 + 2isin theta}{1 - 2isi | vedclass

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

Vedclass · Accepted Answer

$n\pi$

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(C) Let $z = \frac{3 + 2i\sin 	heta}{1 - 2i\sin 	heta}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i\sin 	heta)$:$z = \frac{(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 + 4\sin^2 	heta}$Since $i^2 = -1$,we have:$z = \frac{3 - 4\sin^2 	heta + 8i\sin 	heta}{1 + 4\sin^2 	heta} = \left( \frac{3 - 4\sin^2 	heta}{1 + 4\sin^2 	heta} ight) + i\left( \frac{8\sin 	heta}{1 + 4\sin^2 	heta} ight)$For $z$ to be real,the imaginary part must be zero:$	ext{Im}(z) = \frac{8\sin 	heta}{1 + 4\sin^2 	heta} = 0$This implies $\sin 	heta = 0$.The general solution for $\sin 	heta = 0$ is $	heta = n\pi$,where $n$ is an integer.

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

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