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Math Help - Complex number equation

  1. #1
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    Question Complex number equation

    Hello can you please help me to solve this equation.

    z^3+8=0

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  2. #2
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    Re: Complex number equation

    Use the sum of two cubes: a^3+b^3 = (a+b)(a^2-ab+b^2)
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  3. #3
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    Re: Complex number equation

    Quote Originally Posted by depecheSoul View Post
    Hello can you please help me to solve this equation.

    z^3+8=0

    An alternative...

    \displaystyle \begin{align*} z^3 + 8 &= 0 \\ z^3 &= -8 \\ z^3 &= 8e^{i\pi} \\ z &= \left(8e^{i\pi}\right)^{\frac{1}{3}} \\ z &= 2e^{\frac{i\pi}{3}} \end{align*}

    Since this is a cubic, there are three solutions, all evenly spaced around a circle. So their angles are separated by \displaystyle \frac{2\pi}{3}, which means the three solutions are \displaystyle 2e^{-\frac{i\pi}{3}}, 2e^{\frac{i\pi}{3}} and \displaystyle 2e^{i\pi}.

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