How do you simplify #(-1)^(-1/8)#?
2 Answers
Deleted answer.
#(-1)^(-1/8) = sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2 i#
Explanation:
We can use the half angle formulas to find
#cos^2(theta/2) = 1/2 (1 + cos(theta))#
#sin^2(theta/2) = 1/2 (1 - cos(theta))#
Let
Then
#cos^2(pi/8) = 1/2 (1 + sqrt(2)/2) = (2 + sqrt(2)) / 4#
#sin^2(pi/8) = 1/2 (1 - sqrt(2)/2) = (2 - sqrt(2)) / 4#
So, since
#cos(pi/8) = sqrt((2+sqrt(2))/4) = sqrt(2+sqrt(2))/2#
#sin(pi/8) = sqrt((2-sqrt(2))/4) = sqrt(2-sqrt(2))/2#
Note that:
So
There are
So we find:
#(-1)^(-1/8)#
#= 1/(-1)^(1/8)#
#= 1/root(8)(-1)#
#= 1/(cos (pi/8) + i sin (pi/8))#
#=(cos(pi/8) - i sin(pi/8))/((cos(pi/8) - i sin(pi/8))(cos(pi/8) + i sin (pi/8))#
#=(cos(pi/8) - i sin(pi/8))/(cos^2(pi/8) + sin^2(pi/8))#
#=cos(pi/8) - i sin(pi/8)#
#= sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2 i#