Complex roots of polynomial

[funnybabe]

Hello,

I'm not sure if I'm putting this in the right thread but I am a little stuck on this problem. the question is find the complex roots of z^6-2z^3-3.

I subbed in y=z^3 and used the quadratic formula and got 3 and -1 as roots. now this is where I am stuck. since there was no i part, I am not sure how to write it up.

it is similar to a homework question where we used e^i(some number)

anyways, if anyone could show me how to get this answer, that would be great and how to write it up properly

 

[Quaid]

find the complex roots of z^6 - 2z^3 - 3

I subbed in y=z^3 and used the quadratic formula and got 3 and -1 as roots

this is where I am stuck

You found that y = -1 or y = 3

Now reverse the substitution z^3 = y, using those two y values to obtain six z values.

Looks like two of the six roots are Real; the other four have imaginary parts.