The real value of theta for which the express | vedclass

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(B) Let $z = \frac{1 + i \cos 	heta}{1 - 2i \cos 	heta}$.To make $z$ a real number,we multiply the numerator and denominator by the conjugate of the

Vedclass · Accepted Answer

$(2n + 1)\frac{\pi}{2}$

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(B) Let $z = \frac{1 + i \cos 	heta}{1 - 2i \cos 	heta}$.To make $z$ a real number,we multiply the numerator and denominator by the conjugate of the denominator,which is $1 + 2i \cos 	heta$:$z = \frac{(1 + i \cos 	heta)(1 + 2i \cos 	heta)}{(1 - 2i \cos 	heta)(1 + 2i \cos 	heta)}$$z = \frac{1 + 2i \cos 	heta + i \cos 	heta + 2i^2 \cos^2 	heta}{1 + 4 \cos^2 	heta}$Since $i^2 = -1$,we have:$z = \frac{1 - 2 \cos^2 	heta + 3i \cos 	heta}{1 + 4 \cos^2 	heta}$For $z$ to be a real number,the imaginary part must be zero:$\frac{3 \cos 	heta}{1 + 4 \cos^2 	heta} = 0$This implies $\cos 	heta = 0$.The general solution for $\cos 	heta = 0$ is $	heta = (2n + 1)\frac{\pi}{2}$ for $n \in I$.Thus,the correct option is $(B)$.

The real value of $\theta$ for which the expression $\frac{1 + i \cos \theta}{1 - 2i \cos \theta}$ is a real number is $(n \in I)$:

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