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(C) Given $\pi < \alpha, \beta < 2\pi$.For $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ to be purely imaginary,its real part must be zero.$\frac{(1-i \s

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(C) Given $\pi For $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ to be purely imaginary,its real part must be zero.$\frac{(1-i \sin \alpha)(1-2i \sin \alpha)}{1+4 \sin^2 \alpha} = \frac{1 - 2\sin^2 \alpha - 3i \sin \alpha}{1+4 \sin^2 \alpha}$.Setting the real part to zero: $1 - 2\sin^2 \alpha = 0 \Rightarrow \sin^2 \alpha = \frac{1}{2}$.Since $\pi For $\frac{1+i \cos \beta}{1-2i \cos \beta}$ to be purely real,its imaginary part must be zero.$\frac{(1+i \cos \beta)(1+2i \cos \beta)}{1+4 \cos^2 \beta} = \frac{1 - 2\cos^2 \beta + 3i \cos \beta}{1+4 \cos^2 \beta}$.Setting the imaginary part to zero: $3 \cos \beta = 0 \Rightarrow \cos \beta = 0$.Since $\pi For $\alpha = \frac{5\pi}{4}$,$\sin 2\alpha = \sin \frac{5\pi}{2} = 1$. For $\alpha = \frac{7\pi}{4}$,$\sin 2\alpha = \sin \frac{7\pi}{2} = -1$.For $\beta = \frac{3\pi}{2}$,$\cos 2\beta = \cos 3\pi = -1$.Thus,$Z_1 = 1-i$ and $Z_2 = -1-i$.For $Z = 1-i$,$iZ + \frac{1}{i\bar{Z}} = i(1-i) + \frac{1}{i(1+i)} = (1+i) + \frac{1}{i-1} = 1+i + \frac{-1-i}{2} = \frac{1+i}{2}$.For $Z = -1-i$,$iZ + \frac{1}{i\bar{Z}} = i(-1-i) + \frac{1}{i(-1+i)} = (1-i) + \frac{1}{-i-1} = 1-i + \frac{i-1}{2} = \frac{1-i}{2}$.Sum $= \frac{1+i}{2} + \frac{1-i}{2} = 1$.

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