Find roots of a complex no? #X^5=1#

1 Answer
Mar 7, 2018

#X=1#

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

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

Explanation:

Given:

#X^5=1#

We find:

#0 = X^5-1 = (X-1)(X^4+X^3+X^2+X+1)#

So one root is #X=1#

For the other roots, solve:

#0 = X^4+X^3+X^2+X+1#

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

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

#color(white)(0)=X^2(((X+1/X)+1/2)^2-5/4)#

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

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

If #X^2+1/2(1-sqrt(5))X+1 = 0# then:

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

#color(white)(X) = 1/4(sqrt(5)-1)+-1/4sqrt((6-2sqrt(5))-16))#

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

If #X^2+1/2(1+sqrt(5))X+1 = 0# then:

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

#color(white)(X) = 1/4(sqrt(5)+1)+-1/4sqrt((6+2sqrt(5))-16))#

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