# Solving a polynomial with complex numbers

• Nov 24th 2012, 10:05 PM
frigid
Solving a polynomial with complex numbers
Can I get help with this question?

z^6 + (i+1)z^3 + i =0
• Nov 24th 2012, 10:15 PM
Deveno
Re: Solving a polynomial with complex numbers
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).
• Nov 24th 2012, 10:32 PM
frigid
Re: Solving a polynomial with complex numbers
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?
• Nov 24th 2012, 10:33 PM
MarkFL
Re: Solving a polynomial with complex numbers
Another approach:

I would factor as follows:

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

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

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

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

Can you proceed from here?
• Nov 24th 2012, 10:45 PM
frigid
Re: Solving a polynomial with complex numbers
yes thank you
• Nov 25th 2012, 02:01 AM
Deveno
Re: Solving a polynomial with complex numbers
Quote:

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:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(i+1) \pm \sqrt{(i+1)^2 - 4i}}{2}$

$= \frac{-(i+1) \pm \sqrt{2i - 4i}}{2} = \frac{-(i+1) \pm \sqrt{-2i}}{2}$

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

this gives:

$w = \frac{-(i+1) \pm (-1+i)}{2} \text{, so } w = -1 \text{ or } -i$.

since w = z3, we are solving $z^3 = -1 \iff z^3 + 1 = 0 \text{ or }z^3 = -i \iff z^3 + i = 0$

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