QJEEM22Mada2a: JEE Main 2022 Mathematics - Class 11 JEEM22Mada2a Solution: The real part of the complex number is equal to :

Given,

${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$

$= {{{{(1 + 2i)}^2}{{(1 - 2i)}^2}{{(1 + 2i)}^6}} \over {(3 + 2i)(4 + 6i)}}$

$= {{{{(1 - 4{i^2})}^2}{{(1 + 2i)}^6}} \over {12 + 18i + 8i + 12{i^2}}}$

$= {{{{(1 + 5)}^2}{{\left[ {{{(1 + 2i)}^2}} \right]}^3}} \over {12 + 26i - 12}}$

$= {{25{{(1 + 4{i^2} + 4i)}^3}} \over {26i}}$

$= {{25{{(1 - 4 + 4i)}^3}} \over {26i}}$

$= {{25{{( - 3 + 4i)}^3}} \over {26i}}$

$= {{25} \over {26i}}\left[ {{{( - 3)}^3} + {{(4i)}^3} + 3\,.\,{{( - 3)}^2}\,.\,4i + 3( - 3)\,.\,{{(4i)}^2}} \right]$

$= {{25} \over {26i}}( - 27 - 64i + 108i + 144)$

$= {{25} \over {26i}}(117 + 44i)$

$= {{25i} \over {26{i^2}}}(117 + 44i)$

$= {{25i} \over { - 26}}(117 + 44i)$

$= {{25 \times 117i} \over { - 26}} - {{25 \times 44{i^2}} \over {26}}$

$= {{25 \times 117i} \over { - 26}} + {{22 \times 25} \over {13}}$

$= {{25 \times 117i} \over { - 26}} + {{550} \over {13}}$

$\therefore$ Real part $= {{550} \over {13}}$