The real value of alpha for which frac{1-i si | vedclass

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(C) Let $z = \frac{1-i \sin \alpha}{1+2 i \sin \alpha}$.To make $z$ purely real,its imaginary part must be zero.Multiply the numerator and denominator

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$n \pi, n \in N$

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(C) Let $z = \frac{1-i \sin \alpha}{1+2 i \sin \alpha}$.To make $z$ purely real,its imaginary part must be zero.Multiply the numerator and denominator by the conjugate of the denominator $(1-2i \sin \alpha)$:$z = \frac{(1-i \sin \alpha)(1-2i \sin \alpha)}{(1+2i \sin \alpha)(1-2i \sin \alpha)}$$z = \frac{1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha}$Since $i^2 = -1$,we have:$z = \frac{1 - 3i \sin \alpha - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} = \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - i \frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$For $z$ to be purely real,the imaginary part must be $0$:$-\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0$$\Rightarrow 3 \sin \alpha = 0$$\Rightarrow \sin \alpha = 0$$\Rightarrow \alpha = n \pi, n \in N$.

The real value of $\alpha$ for which $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is

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