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Math Help - Solving a polynomial with complex numbers

  1. #1
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    Solving a polynomial with complex numbers

    Can I get help with this question?

    z^6 + (i+1)z^3 + i =0
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  2. #2
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    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).
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    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?
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  4. #4
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    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?
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  5. #5
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    Re: Solving a polynomial with complex numbers

    yes thank you
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  6. #6
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    Re: Solving a polynomial with complex numbers

    Quote Originally Posted by frigid View Post
    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.
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