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