QJEEM23Ma8be3: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8be3 Solution: Let and then and are roots of the equation.

${\left( {1 - \sqrt 3 i} \right)^{200}}$

$= {\left[ {2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)} \right]^{200}}$

$= {2^{200}}{\left( {\cos {\pi \over 3} - i\sin {\pi \over 3}} \right)^{200}}$

$= {2^{200}}\left( {\cos {{200\pi } \over 3} - i\sin {{200\pi } \over 3}} \right)$

$= {2^{200}}\left( {\cos \left( {66\pi + {{2\pi } \over 3}} \right) - i\sin \left( {66\theta + {{2\pi } \over 3}} \right)} \right)$

$= {2^{200}}\left( {\cos {{2\pi } \over 3} - i\sin {{2\pi } \over 3}} \right)$

$= {2^{200}}\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right)$

$= {2^{199}}\left( { - 1 - \sqrt 3 i} \right)$

$= {2^{199}}\left( {p + iq} \right)$

$\therefore$ p = $-$ 1 and q = $- \sqrt3$

Now, $p - q + {q^2} = - 1 + \sqrt 3 + 3 = 2 + \sqrt 3 = \alpha$

and $p + q + {q^2} = - 1 - \sqrt 3 + 3 = 2 - \sqrt 3 = \beta$

$\therefore$ $\alpha + \beta = 4$

$\alpha \beta = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 4 - 3 = 1$

$\therefore$ Quadratic equation is

${x^2} - (\alpha + \beta )x + \alpha \beta = 0$

$\Rightarrow {x^2} - 4x + 1 = 0$