QJEEM22Ma28bc: JEE Main 2022 Mathematics - Class 11 JEEM22Ma28bc Solution: Let S be the set of all , for which the complex number is…

$\because$ ${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }}$ 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 ${Z_{\alpha \beta }} = - 1 - i$

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

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