How do you solve #x^4+i=0#?

2 Answers
Apr 4, 2017

#= cis ((-pi)/8 ), cis ((3pi)/8 ), cis ((7pi)/8 ), cis ((11pi)/8 )#

Explanation:

#x^4 = - i#

#x = (- i)^(1/4)#

Using Euler's formula (or DeMoivre's theorem, if you like):

#-i = cos ((-pi)/2 + 2n pi) + i sin ((-pi)/2 + 2n pi) equiv cis ((-pi)/2 + 2n pi)#

#= e^(i ((-pi)/2 + 2 n pi))#

So:

#(-i)^(1/4) = e^(i ((-pi)/2 + 2 n pi)/4)#

#= e^(i ((-pi)/8 + (n pi)/2))#

#= cis ((-pi)/8 + (n pi)/2) #

Running through the #n#'s:

#= cis ((-pi)/8 ), cis ((3pi)/8 ), cis ((7pi)/8 ), cis ((11pi)/8 )#

And then they start repeating.

May 21, 2017

The four roots are:

#x=+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)#

#x=+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)#

Explanation:

Here's an alternative non-trigonometric method...

If #a, b# are real numbers with #b != 0#, then the square roots of #a+bi# are:

#+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)#

(See: https://socratic.org/s/aES256Hm)

Applying this to the square root of #-i = color(blue)(0)+(color(blue)(-1))i# we find that its square roots are:

#+-((sqrt((sqrt(color(blue)(0)^2+(color(blue)(-1))^2)+color(blue)(0))/2)) - (sqrt((sqrt(color(blue)(0)^2+(color(blue)(-1))^2)-color(blue)(0))/2))i)#

#=+-(sqrt(2)/2-sqrt(2)/2i)#

The square roots of #sqrt(2)/2-sqrt(2)/2i# are:

#+-((sqrt((sqrt((color(blue)(sqrt(2)/2))^2+(color(blue)(-sqrt(2)/2))^2)+color(blue)(sqrt(2)/2))/2)) - (sqrt((sqrt((color(blue)(sqrt(2)/2))^2+(color(blue)(-sqrt(2)/2))^2)-color(blue)(sqrt(2)/2))/2))i)#

#=+-((sqrt((sqrt(1/2+1/2)+sqrt(2)/2))/2)) - (sqrt((sqrt(1/2+1/2)-sqrt(2)/2)/2))i)#

#=+-((sqrt((1+sqrt(2)/2)/2)) - (sqrt((1-sqrt(2)/2)/2))i)#

#=+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)#

Rather than simplifying again, note that we can find the square roots of #-sqrt(2)/2+sqrt(2)/2i# by multiplying these by #i# to get:

#+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)#

So the fourth roots of #-i# (i.e. the roots of #x^4+i = 0#) are:

#+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)#

#+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)#