If theta in mathbb{R} and frac{1-i cos theta} | vedclass

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(A) Let $z = \frac{1-i \cos 	heta}{1+2 i \cos 	heta}$.Since $z$ is a real number,$z = \bar{z}$.$\frac{1-i \cos 	heta}{1+2 i \cos 	heta} = \frac{1+

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

Vedclass · Answer

(A) Let $z = \frac{1-i \cos 	heta}{1+2 i \cos 	heta}$.Since $z$ is a real number,$z = \bar{z}$.$\frac{1-i \cos 	heta}{1+2 i \cos 	heta} = \frac{1+i \cos 	heta}{1-2 i \cos 	heta}$.Cross-multiplying,we get:$(1-i \cos 	heta)(1-2 i \cos 	heta) = (1+i \cos 	heta)(1+2 i \cos 	heta)$.$1 - 2i \cos 	heta - i \cos 	heta + 2i^2 \cos^2 	heta = 1 + 2i \cos 	heta + i \cos 	heta + 2i^2 \cos^2 	heta$.Since $i^2 = -1$,this simplifies to:$1 - 3i \cos 	heta - 2 \cos^2 	heta = 1 + 3i \cos 	heta - 2 \cos^2 	heta$.Subtracting common terms from both sides:$-3i \cos 	heta = 3i \cos 	heta$.$6i \cos 	heta = 0$.Since $i 
eq 0$,we must have $\cos 	heta = 0$.Therefore,$	heta = (2n+1) \frac{\pi}{2}$ for $n \in I$.

If $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is a real number,then $\theta$ will be (where $I$ is the set of integers):

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