How do you find all the real and complex roots of #z^3 +1 = 0#?

1 Answer
Jan 15, 2016

Answer:

Factor then solve the remaining quadratic using the quadratic formula to find:

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

Explanation:

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

Solve #z^2-z+1 = 0# using the quadratic formula with #a=1#, #b=-1# and #c=1#:

#z = (-b+-sqrt(b^2-4ac))/(2a) = (1+-sqrt(-3))/2 = 1/2+-sqrt(3)/2i#

Hence #z^3+1 = 0# has roots #z=-1#, #z=1/2+-sqrt(3)/2i#