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Thread: 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 $\displaystyle (x + 2i)^{3} = 2$

    Take the cubed root of both sides

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

    Now subtract 2i from the left and the right

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

    Take the cubed root of both sides

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

    Now subtract 2i from the left and the right

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

    $\displaystyle (x+2i)^3 = 2$, then all the solutions are:
    $\displaystyle x+2i = \sqrt[3]{2},\sqrt[3]{2}\zeta,\sqrt[3]{2}\zeta^2$ where $\displaystyle \zeta$ is primitive root of unity.
    That means,
    $\displaystyle x = -2i + \sqrt[3]{2} \zeta^k$ for $\displaystyle k=0,1,2$.
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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 $\displaystyle x^{3} + 6x^{2}i - 12x - i - 2 = 0$ Could I use synthetic division somehow to simplify that knowing that $\displaystyle \sqrt[3]{2}-2i$ 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 $\displaystyle x^{3} + 6x^{2}i - 12x - i - 2 = 0$ Could I use synthetic division somehow to simplify that knowing that $\displaystyle \sqrt[3]{2}-2i$ is a factor? Or is there another way to factor it?
    No! Do not fact.

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

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