Thread: Solving a polynomial with complex numbers

Can I get help with this question?

z^6 + (i+1)z^3 + i =0

let w = z³, and use the quadratic formula to solve for w. then find the cube roots of w (there will be 3 for each solution you find for w, use demoivre).

How do I use the quadratic formula on w^2 + (i +1)w+ i?

Would it be a=1,b=(i +1) and c=i?

Another approach:

I would factor as follows:









Can you proceed from here?

yes thank you

Originally Posted by frigid

How do I use the quadratic formula on w^2 + (i +1)w+ i?

Would it be a=1,b=(i +1) and c=i?

yes, the quadratic formula works in ANY field, as long as 1+1 ≠ 0.

so we have:





the two square roots of -2i are: -1+i and 1-i (check this by squaring them).

this gives:

.

since w = z³, we are solving

which is exactly where MarkFL2's solution leads you to.