QJEEM19Mad126: JEE Main 2019 Mathematics - Class 11 JEEM19Mad126 Solution: Let and be two roots of the equation x 2 + 2x + 2 = 0 , then +…

Given equation,

x² + 2x + 2 = 0

$\therefore$   x = ${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$

x = $-$ 1 $\pm$ i

$\therefore$    $\alpha$ = $-$ 1 + i

and $\beta$ = $-$ 1 $-$ i

Note :

x + iy = r (cos $\theta$ + isin $\theta$ )

$\therefore$   (x + iy)ⁿ = rⁿ (cosn $\theta$ + isinn $\theta$ )

$\therefore$    $-$ 1 + i = $\sqrt 2$ [cos ${{3\pi } \over 4}$ + isin ${{3\pi } \over 4}$ ]

$\Rightarrow$   ( $-$ 1 + i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}$ [cos $\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$ ]

And $-$ 1 $-$ i = $\sqrt 2$ $\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$

= $\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$

$\therefore$   ( $-$ 1 $-$ i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$

Now

$\alpha$ ¹⁵ + $\beta$ ¹⁵

= ( $-$ 1 + i)¹⁵ + ( $-$ 1 $-$ i)¹⁵

=   ${\left( {\sqrt 2 } \right)^{15}}$ $\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$

= $- {\left( {\sqrt 2 } \right)^{14}}.2$

= $-$ 2⁷ $\times$ 2

= $-$ 2⁸

= $-$ 256