Let S be the set of all $$(\alpha, \beta)$$, $$\pi < \alpha, \beta < 2\pi$$, for which the complex number $$\frac{1-i\sin\alpha}{1+2i\sin\alpha}$$ is purely imaginary and $$\frac{1+i\cos\beta}{1-2i\cos\beta}$$ is purely real. Let $$Z_{\alpha\beta} = \sin 2\alpha + i\cos 2\beta, (\alpha,\beta) \in S$$. Then $$\sum_{(\alpha,\beta)\in S}(iZ_{\alpha\beta} + \frac{1}{i\bar{Z}_{\alpha\beta}})$$ is equal to

$$\frac{1-i\sin\alpha}{1+2i\sin\alpha}$$

is purely imaginary.

Rationalizing,
$$\frac{(1-i\sin\alpha)(1-2i\sin\alpha)}{1+4\sin^2\alpha}$$

$$\frac{1-2\sin^2\alpha-3i\sin\alpha}{1+4\sin^2\alpha}$$

Purely imagina$$ry(\Rightarrow)$$
$$1-2\sin^2\alpha=0$$
$$\sin^2\alpha=\frac{1}{2}$$

Since$$(\pi<\alpha<2\pi),$$
$$\alpha=\frac{5\pi}{4},\frac{7\pi}{4}$$

Similarly, $$\frac{1+i\cos\beta}{1-2i\cos\beta}$$

purely real gives
$$\cos\beta=0$$

and in$$((\pi,2\pi)),$$
$$\beta=\frac{3\pi}{2}$$
$$Z_{\alpha\beta}=\sin2\alpha+i\cos2\beta$$

Since
$$\cos2\beta=\cos3\pi=-1$$

we get
$$Z=\sin2\alpha-i$$

Thus:

• for $$(\alpha=\frac{5\pi}{4}),(Z=1-i)$$
• for $$(\alpha=\frac{7\pi}{4}),(Z=-1-i)$$

Compute
$$iZ+\frac{1}{i\overline{Z}}$$

For (Z=1-i):
$$1+i+\frac{1}{-1+i}=\frac{1+i}{2}$$

For (Z=-1-i):
$$1-i+\frac{1}{-1-i}=\frac{1-i}{2}$$

Hence

$$\sum_{(\alpha,\beta)\in S}^{ }\left(iZ_{\alpha\beta}+\frac{1}{i\overline{Z}_{\alpha\beta}}\right)$$

$$\frac{1+i}{2}+\frac{1-i}{2}=1$$