$$\because$$ $${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }}$$ is purely imaginary

$$\therefore$$ $${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }} + {{1 + i\sin \alpha } \over {1 - 2i\sin \alpha }} = 0$$

$$ \Rightarrow 1 - 2{\sin ^2}\alpha = 0$$

$$\therefore$$ $$\alpha = {{5\pi } \over 4},\,{{7\pi } \over 4}$$

and $${{1 + i\cos \beta } \over {1 - 2i\cos \beta }}$$ is purely real

$${{1 + i\cos \beta } \over {1 - 2i\cos \beta }} - {{1 - i\cos \beta } \over {1 + 2i\cos \beta }} = 0$$

$$ \Rightarrow \cos \beta = 0$$

$$\therefore$$ $$\beta = {{3\pi } \over 2}$$

$$\therefore$$ $$S = \left\{ {\left( {{{5\pi } \over 2},{{3\pi } \over 2}} \right),\left( {{{7\pi } \over 4},{{3\pi } \over 2}} \right)} \right\}$$

$${Z_{\alpha \beta }} = 1 - i$$ and $${Z_{\alpha \beta }} = - 1 - i$$

$$\therefore$$ $$\sum\limits_{(\alpha ,\beta ) \in S} {\left( {i{Z_{\alpha \beta }} + {1 \over {i{{\overline Z }_{\alpha \beta }}}}} \right) = i( - 2i) + {1 \over i}\left[ {{1 \over {1 + i}} + {1 \over { - 1 + i}}} \right]} $$

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