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Math Help - complex numbers

  1. #1
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    complex numbers

    how do i solve (x+2i)^3=2
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  2. #2
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    Ok so we have (x + 2i)^{3} = 2

    Take the cubed root of both sides

    x + 2i = \sqrt[3]{2}

    Now subtract 2i from the left and the right

    x = \sqrt[3]{2} - 2i
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  3. #3
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    Quote Originally Posted by SnipedYou View Post
    Ok so we have (x + 2i)^{3} = 2

    Take the cubed root of both sides

    x + 2i = \sqrt[3]{2}

    Now subtract 2i from the left and the right

    x = \sqrt[3]{2} - 2i
    What about the other solutions?

    (x+2i)^3 = 2, then all the solutions are:
    x+2i = \sqrt[3]{2},\sqrt[3]{2}\zeta,\sqrt[3]{2}\zeta^2 where \zeta is primitive root of unity.
    That means,
    x = -2i + \sqrt[3]{2} \zeta^k for k=0,1,2.
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  4. #4
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    I'm searching for them, but I saw that one right when I started trying to solve it.

    I'm stuck here x^{3} + 6x^{2}i - 12x - i - 2 = 0 Could I use synthetic division somehow to simplify that knowing that \sqrt[3]{2}-2i is a factor? Or is there another way to factor it?
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  5. #5
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    Quote Originally Posted by SnipedYou View Post
    I'm searching for them, but I saw that one right when I started trying to solve it.

    I'm stuck here x^{3} + 6x^{2}i - 12x - i - 2 = 0 Could I use synthetic division somehow to simplify that knowing that \sqrt[3]{2}-2i is a factor? Or is there another way to factor it?
    No! Do not fact.

    The equation,
    x^3 = a^3
    So,
    x^3 - a^3 = 0
    (x-a)(x^2+xa+a^2)=0
    So, x-a=0 \mbox{ and }x^2+xa+a^2 = 0
    So, x = a, x = \frac{-a\pm \sqrt{a^2 - 4a^2}}{2}

    So in the equation,
    (x+2i)^3 = a^3 where a=\sqrt[3]{2}.
    Now use the result above.
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