How do you find all the real and complex roots of #z^7+z^4+4z^3+4=0#?

1 Answer
mason m
Jan 19, 2016

#z=-1,-1+-i,1+-i,1/2+-sqrt3/2i#

Explanation:

Factor by grouping:

#z^4(z^3+1)+4(z^3+1)=0#

#(z^4+4)(z^3+1)=0#

First, #(z^3+1)# can be easily split apart using the sum of cubes identity.

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

The other part, #(z^4+4)#, is more difficult to factor. We can try to turn it into a difference of squares.

#z^4+4=(z^4+4z^2+4)-4z^2#
#color(white)(xxxx)=(z^2+2)^2-(2z)^2#
#color(white)(xxxx)=(z^2+2z+2)(z^2-2z+2)#

Now, we have #3# quadratic factors and one linear factor:

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

Solving each of these via the quadratic formula or completing the square gives the seven distinct values of #z#:

#z=-1,-1+-i,1+-i,1/2+-sqrt3/2i#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1925 views around the world
You can reuse this answer
Creative Commons License