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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