How do you simplify #(-1)^(-1/8)#?

2 Answers
Guilherme N.
Dec 28, 2015

Deleted answer.

George C.
Dec 28, 2015

#(-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(pi/8)# and #sin(pi/8)#, which we use later...

#cos^2(theta/2) = 1/2 (1 + cos(theta))#

#sin^2(theta/2) = 1/2 (1 - cos(theta))#

Let #theta = pi/4#.

Then #cos(theta) = sqrt(2)/2# and we find:

#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 #pi/8# is in Q1 we want the positive square roots:

#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: #e^(i pi) = -1#

So #(-1)^(1/8) = (e^(i pi))^(1/8) = e^(i pi/8) = cos(pi/8) + i sin(pi/8)#

There are #7# other #8#th roots of #-1#, but this is the principal one.

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#

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