Results 1 to 3 of 3

Thread: about complex numbers

  1. #1
    Newbie
    Joined
    Mar 2009
    From
    São Paulo- Brazil
    Posts
    22

    about complex numbers

    Hello everybody!

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

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

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

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1
    Quote Originally Posted by Biscaim View Post
    I am trying to solve the following question:
    Find $\displaystyle (-11-2i)^{1/3}$
    There is no neat solution to that question.
    Let $\displaystyle \theta=\arctan\left(\frac{2}{11}\right)-\pi$.
    Then $\displaystyle \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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    $\displaystyle \displaystyle |-11 - 2i| = \sqrt{(-11)^2 + (-2)^2}$

    $\displaystyle \displaystyle = \sqrt{121 + 4}$

    $\displaystyle \displaystyle = \sqrt{125}$.


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

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


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

    and therefore

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

    $\displaystyle \displaystyle = \sqrt[6]{125}\,e^{i\left[\frac{\pi}{3} + \frac{1}{3}\arctan{\left(\frac{2}{11}\right)}\righ t]}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: Mar 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Aug 25th 2009, 11:22 AM
  4. Replies: 2
    Last Post: Feb 7th 2009, 06:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum