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Math Help - about complex numbers

  1. #1
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    about complex numbers

    Hello everybody!

    I am trying to solve the following question: Find (-11-2i)^{1/3}

    I did the following: (-11-2i)^{1/3}=(a+bi) and I'll get a^{3}-3ab^{2}=-11 and 3a^{2}b-b^{3}=-2

    If I put a=1 I can solve this question. But I would like to know how to solve the system above.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Biscaim View Post
    I am trying to solve the following question:
    Find (-11-2i)^{1/3}
    There is no neat solution to that question.
    Let \theta=\arctan\left(\frac{2}{11}\right)-\pi.
    Then  \left(-11-2\imath\right)^{\frac{1}{3}}=\exp\left(\frac{\ln(5  )}{2}+\dfrac{\imath\theta}{3}\right)

    There is a good deal work to get to that answer.
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  3. #3
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    \displaystyle |-11 - 2i| = \sqrt{(-11)^2 + (-2)^2}

    \displaystyle = \sqrt{121 + 4}

    \displaystyle = \sqrt{125}.


    \displaystyle \arg{(-11 - 2i)} = \pi + \arg{(11 + 2i)}

    \displaystyle = \pi + \arctan{\left(\frac{2}{11}\right)}.


    So \displaystyle -11 - 2i = \sqrt{125}\,e^{i\left[\pi + \arctan{\left(\frac{2}{11}\right)}\right]}

    and therefore

    \displaystyle (-11 - 2i)^{\frac{1}{3}} = \left\{\sqrt{125}\,e^{i\left[\pi + \arctan{\left(\frac{2}{11}\right)}\right]}\right\}^{\frac{1}{3}}

    \displaystyle = \sqrt[6]{125}\,e^{i\left[\frac{\pi}{3} + \frac{1}{3}\arctan{\left(\frac{2}{11}\right)}\righ  t]}.
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