Let \alpha and \beta be two roots of the equation x2 + 2x + 2 = 0 , then \alpha ^{15} + \beta ^{15} is equal to :

Question

Accepted Answer

Answer: A. -256 Given equation, x2 + 2x + 2 = 0 	herefore x = {{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}} x = - 1 \pm i 	herefore \alpha = - 1 + i and \beta = - 1 - i Note : x + iy = r (cos	heta + isin	heta ) 	herefore (x + iy)n = rn (cosn	heta + isinn	heta ) 	herefore - 1 + i = \sqrt 2 [cos{{3\pi } \over 4} + isin{{3\pi } \over 4} ] \Rightarrow (- 1 + i)15 = {\left( {\sqrt 2 } ight)^{15}} [cos\left( {{{15.3\pi } \over 4}} ight) + i\sin \left( {{{15.3\pi } \over 4}} ight)] And -1 - i = \sqrt 2 \left[ {\cos \left( { - {{3\pi } \over 4}} ight) + i\sin \left( { - {{3\pi } \over 4}} ight)} ight] = \sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} ight] 	herefore (-1 - i)15 = {\left( {\sqrt 2 } ight)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} ight) - i\sin \left( {{{15.3\pi } \over 4}} ight)} ight] Now \alpha 15 + \beta 15 = (-1 + i)15 + (- 1 - i)15 = {\left( {\sqrt 2 } ight)^{15}} \left[ {2\cos \left( {{{15.3\pi } \over 4}} ight)} ight] = {\left( {\sqrt 2 } ight)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} ight)} ight] = {\left( {\sqrt 2 } ight)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} ight)} ight] = {\left( {\sqrt 2 } ight)^{15}} 	imes 2 	imes - {1 \over {\sqrt 2 }} = - {\left( {\sqrt 2 } ight)^{14}}.2 = - 27 	imes 2 = - 28 = - 256

Let $$\alpha $$ and $$\beta $$ be two roots of the equation x2 + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to :

Answer

Explanation

More JEE Math questions