Question 408 of 455MediumMCQ
Let \[\alpha \] and \[\beta \] be two roots of the equation x2 + 2x + 2 = 0 , then \[\alpha ^{15}\] + \[\beta ^{15}\] is equal to :
Solutions · 1
Official solution
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
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