Let `alpha and beta` two roots of the equatins `x^(2)+2x+2=0.` then `alpha^(15)+beta^(15)` is equal to

Options

A. 256
B. 512
C. `-256`
D. `-512`

Answer

C

Explanation

We have, `x^(2)+2x+2=0`
`implies=x(-2+-sqrt(4-8))/(2)"["because"roots of"ax^(2)+bx+c=0"are given by"x=(-b+-sqrt(b^(2)-4ac))/(2a)"]"`
`impliesx=-1+-i`
`Let alpha=-1+i and beta=-1-i.`
Then, `alpha^(15)+beta^(15)=(-1+i)^(15)+(-1-i)^(15)`
`=-"["(1-i)^(15)+(1+i)^(15)"]"`
`=-[{sqrt2((1)/(sqrt2)-(i)/(sqrt2))}^(15)+{sqrt2((1)/(sqrt2)+(i)/(sqrt2))}^(15)]`
`{:[(,{sqrt2(cos""(pi)/(4)-isin""(pi)/(4))}^(15)),(,+{sqrt2(cos""(15pi)/(4)+i sin""(pi)/(4))}^(15)):}]`
`=-(sqrt2)^(15)[(cos""(15)/(n)-i sin""(15pi)/(4))+(cos""(15pi)/(4)+i sin""(15pi)/(4))]`
[using De' Moivre's theorem (`cos theta +- i sin theta)^(n)=cos n theta +-isin ntheta, n in Z`]
`=-(sqrt2)[2cos""(15pi)/(4)]=(sqrt2)^(15)[2xx(1)/(sqrt2)]`
`" "[becausecos""(15pi)/(4)=cos(4pi-(pi)/(4))=cos""(pi)/(4)=(1)/(sqrt2)]=(sqrt2)^(16)=-2^(8)=-256.`
Alternate Method
`alpha^(15)+beta^(16)=(-1+i)^(15)+(-1-i)^(15)`
`=-"["(1-i)^(15)+(1+i)^(15)"]"`
`=-[(1-i)^(16)/(1-i)+((1+i))/(1+i)]`
`=-[("["(1-i)^(2)"]"^(8))/(1-i)+("["(1+i)^(2)"]"^(8))/(1+i)]`
`=-[([1+i^(2)-2i]^(8))/(1-i)+([1+i^(2)+2i]^(8))/(1+i)]`
`=-2^(8)[(1)/(1-i)+(1)/(1+i)]" "[becausei^(4n)=1,nin Z]`
`=-256[(2)/(1-(i)^(2))]=-256[(2)/(2)]=-256`