How do you solve #z^4+z^2+1 = 0# ?

2 Answers
MattyMatty
Jan 24, 2018

#"See explanation"#

Explanation:

#"first put " y = z^2 "."#
#"so we have"#
#y^2 + y + 1 = 0#
#"and this is a quadratic equation"#
#"the discriminant is "1^2 - 4*1 = -3"#
#"so we have no real solutions, only complex ones"#
#y_{1,2} = (-1 pm sqrt(3) i)/2#
#=> z = pm sqrt (-1 pm sqrt(3) i)/ sqrt(2)#

George C.
Jan 27, 2018

#z = +-1/2+-sqrt(3)/2i#

Explanation:

Given:

#z^4+z^2+1 = 0#

We can complete the square to find:

#0 = z^4+z^2+1#

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

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

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

Then we can find the zeros of the quadratics by completing the square again:

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

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

#color(white)(0) = (2z-1)^2-(sqrt(3)i)^2#

#color(white)(0) = ((2z-1)-sqrt(3)i)((2z-1)+sqrt(3)i)#

#color(white)(0) = (2z-1-sqrt(3)i)(2z-1+sqrt(3)i)#

Hence:

#z = 1/2+-sqrt(3)/2i#

Similarly:

#0 = 4(z^2+z+1) = (2z+1-sqrt(3)i)(2z+1+sqrt(3)i)#

Hence:

#z = -1/2+-sqrt(3)/2i#

Alternatively, note that:

#(z^2-1)(z^4+z^2+1) = z^6-1#

Hence the zeros of #z^4+z^2+1# are the sixth roots of #1# apart from #+-1#.

Related questions
Impact of this question
18361 views around the world
You can reuse this answer
Creative Commons License