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Math Help - nth roots of a complex number

  1. #1
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    nth roots of a complex number

    Find all solutions of the equation

    x^3+125=0
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  2. #2
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    x^3+125=0

    Wouldn't that be -5 ?

    If you must have it in complex number form, a + bi, that could be -5 + 0i.

    A trivial answer to a trivial question, I hope.

    But if I am wrong, perhaps more mysteries will be revealed by the next respondent.
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  3. #3
    Jen
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    Quote Originally Posted by The Math Man View Post
    Find all solutions of the equation

    x^3+125=0
    a^3+b^3=(a+b)(a^2-ab+b^2)

    a=x, b=5 After that you can use the quadratic formula to get the two complex roots.

    Hope that helps.
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  4. #4
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    more roots

    Thanks, Jen.

    Would it be too much for you to show the steps to the roots?

    I don't recall ever being asked to go beyond factoring an expressions of this sort and it would be enlightening to see an example.
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  5. #5
    Jen
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    Quote Originally Posted by Bradley View Post
    Thanks, Jen.

    Would it be too much for you to show the steps to the roots?

    I don't recall ever being asked to go beyond factoring an expressions of this sort and it would be enlightening to see an example.
    x^3+125=x^3+5^3=(x+5)(x^2-5x+25)

    So one root is x=-5.

    Now using the quadratic formula we get

    x=\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(25)}}{2(1)}=\frac{5 \pm \sqrt{-75}}{2}=\frac{5 \pm 5i\sqrt{3}}{2}

    Since these are zero's x-\frac{5 \pm 5i\sqrt{3}}{2} are factors.

    \left( x+5 \right) \left(x- \frac{5 + 5i\sqrt{3}}{2}\right) \left( x- \frac{5 - 5i\sqrt{3}}{2}\right)
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