Results 1 to 8 of 8

Math Help - Complex numbers

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    14

    Complex numbers

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
     - i = \cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right).

    Now find the three cube roots. One of then is
    \cos \left( {\frac{{ - \pi }}{6}} \right) + i\sin \left( {\frac{{ - \pi }}{6}} \right).

    What are the other two?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    14
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    One of the cube roots is \cos{\left(-\frac{\pi}{6}\right)} + i\sin{\left(-\frac{\pi}{6}\right)}.

    All of the cube roots are evenly spaced around the unit circle.

    So they are all separated by an angle of \frac{2\pi}{3}.

    The radius is 1, so the other roots are at

    -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2}

    and

    \frac{\pi}{2} + \frac{2\pi}{3} = \frac{7\pi}{6} = -\frac{5\pi}{6}.


    Therefore the three roots are

    \textrm{cis}\,\left(-\frac{\pi}{6}\right), \textrm{cis}\,\left(\frac{\pi}{6}\right), \textrm{cis}\,\left(-\frac{5\pi}{6}\right).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    14
    Now I'm really confused!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by .::MadMaX::. View Post
    Now I'm really confused!
    Why are you cnfused?
    Those answers are identical as the ones you posted.
    The only difference is the last set is in a more mathematical form.
    That is some of us like answers to be -\pi<z\le \pi.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    14
    So I did good job?

    Thank you all,especially you Plato!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Quote Originally Posted by .::MadMaX::. View Post
    Now I'm really confused!
    I was having exactly the same confusion last night. Haha. See here.
    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: March 25th 2011, 11:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 25th 2009, 12:22 PM
  4. Replies: 2
    Last Post: February 7th 2009, 07:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum