How do you find all the real and complex roots of #z^5 + 1 = 0#?

1 Answer
George C.
Mar 16, 2018

#z = -1#

#z = 1/4(sqrt(5)-1)+-1/4sqrt(10+2sqrt(5))i#

#z = 1/4(-sqrt(5)-1)+-1/4sqrt(10-2sqrt(5))i#

Explanation:

#0 = z^5+1#

#color(white)(0) = (z+1)(z^4-z^3+z^2-z+1)#

#color(white)(0) = (z+1)z^2(z^2-z+1-1/z+1/z^2)#

#color(white)(0) = (z+1)z^2((z+1/z)^2+(z+1/z)-1)#

#color(white)(0) = (z+1)z^2(((z+1/z)+1/2)^2-(sqrt(5)/2)^2)#

#color(white)(0) = (z+1)z^2(z+1/z+1/2-sqrt(5)/2)(z+1/z+1/2+sqrt(5)/2)#

#color(white)(0) = (z+1)(z^2+(1/2-sqrt(5)/2)z+1)(z^2+(1/2+sqrt(5)/2)z+1)#

So:

#0 = z+1 rarr z = -1#

or:

#0 = z^2+(1/2-sqrt(5)/2)z+1#

and hence:

#z = 1/2(-1/2+sqrt(5)/2+-sqrt((1/2-sqrt(5)/2)^2-4))#

#color(white)(z) = 1/4(sqrt(5)-1)+-1/4sqrt(10+2sqrt(5))i#

or:

#0 = z^2+(1/2+sqrt(5)/2)z+1#

and hence:

#z = 1/2(-1/2-sqrt(5)/2+-sqrt((1/2+sqrt(5)/2)^2-4))#

#color(white)(z) = 1/4(-sqrt(5)-1)+-1/4sqrt(10-2sqrt(5))i#

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