Results 1 to 4 of 4

Thread: Find the absolute value

  1. #1
    Senior Member
    Joined
    Jan 2010
    Posts
    273

    Find the absolute value

    Find the absolute value and the argument $\displaystyle \theta$ in the range $\displaystyle 0 <= \theta < 2\pi$ for $\displaystyle (1 - i\sqrt{3})^9 $


    I dun remember how to do this type of question, I wonder if someone can show me the process
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    Quote Originally Posted by wopashui View Post
    Find the absolute value and the argument $\displaystyle \theta$ in the range $\displaystyle 0 <= \theta < 2\pi$ for $\displaystyle (1 - i\sqrt{3})^9 $


    I dun remember how to do this type of question, I wonder if someone can show me the process
    Use polar coordinates.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    78
    $\displaystyle 1-i\sqrt{3}=2e^{-\frac{\pi i}{3}}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,793
    Thanks
    3035
    And then $\displaystyle (1- i\sqrt{3})^9= 2^9 e^{9\left(\frac{\pi i}{3}\right)}= 2^9 e^{3\pi i}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: May 23rd 2010, 10:59 PM
  2. Find the absolute max and absolute min
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Nov 22nd 2009, 03:38 AM
  3. Replies: 2
    Last Post: Nov 8th 2009, 01:52 PM
  4. Find the absolute minimum
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Mar 23rd 2009, 06:41 PM
  5. Find the absolute maximum and absolute minimum
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Dec 12th 2008, 09:46 AM

Search Tags


/mathhelpforum @mathhelpforum