Results 1 to 2 of 2

Math Help - complex numbers factorisation

  1. #1
    Super Member
    Joined
    Aug 2009
    Posts
    639

    complex numbers factorisation

    how do you find k for
    l 1- exp (ix) l = k l sin ( x/2) l for all real x


    i tried to square both sides of the equation and i got:

    1- ( cos (^2) x + sin(^2) x ) -2i sin x( 1- cos x) -2 cos x = k^2 sin(^ 2) ( x/2)

    factorising the sin (^2) ( x/2) terms to one side,
    sin (^2) (x/2) ( 4 - 4 i sin x) -2 = k(^2) sin(^2) ( x/2)

    and then i got stuck..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
     |1 - e^{ix} |= |e^{\frac{ix}{2}} ( e^{-\frac{ix}{2}} - e^{\frac{ix}{2}} ) |

     = |e^{\frac{ix}{2}} |  |e^{-\frac{ix}{2}} - e^{\frac{ix}{2}} |

     = (1) |-2i \sin(\frac{x}{2})| = |-2i| |\sin(\frac{x}{2})|

     = 2 |\sin(\frac{x}{2})|


    By comparing the multiple ,  k = 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integrate without complex factorisation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 6th 2011, 07:28 AM
  2. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  3. More Complex Factorisation
    Posted in the Algebra Forum
    Replies: 8
    Last Post: December 4th 2009, 04:36 PM
  4. Complex Factorisation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2008, 08:08 AM
  5. Replies: 1
    Last Post: May 24th 2007, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum