The set of all values of theta such that frac | vedclass

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(A) Let $z = \frac{1-i \cos 	heta}{1+2 i \sin 	heta}$.To make $z$ purely imaginary,the real part of $z$ must be $0$.Multiply the numerator and denom

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$\left\{n \pi+(-1)^n \frac{\pi}{4}, n \in \mathbb{Z}\right\}$

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(A) Let $z = \frac{1-i \cos 	heta}{1+2 i \sin 	heta}$.To make $z$ purely imaginary,the real part of $z$ must be $0$.Multiply the numerator and denominator by the conjugate of the denominator: $1-2i \sin 	heta$.$z = \frac{(1-i \cos 	heta)(1-2i \sin 	heta)}{(1+2i \sin 	heta)(1-2i \sin 	heta)} = \frac{1 - 2i \sin 	heta - i \cos 	heta + 2i^2 \sin 	heta \cos 	heta}{1 + 4 \sin^2 	heta}$.Since $i^2 = -1$,$z = \frac{(1 - 2 \sin 	heta \cos 	heta) - i(2 \sin 	heta + \cos 	heta)}{1 + 4 \sin^2 	heta}$.The real part is $\frac{1 - 2 \sin 	heta \cos 	heta}{1 + 4 \sin^2 	heta} = \frac{1 - \sin(2 	heta)}{1 + 4 \sin^2 	heta}$.Setting the real part to $0$,we get $1 - \sin(2 	heta) = 0$,which implies $\sin(2 	heta) = 1$.Thus,$2 	heta = 2n \pi + \frac{\pi}{2}$,which simplifies to $	heta = n \pi + \frac{\pi}{4}$ for $n \in \mathbb{Z}$.

The set of all values of $\theta$ such that $\frac{1-i \cos \theta}{1+2 i \sin \theta}$ is purely imaginary is

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