Results 1 to 2 of 2

Thread: complex numbers

  1. #1
    Aug 2009

    complex numbers

    not sure if this is the right forum so sorry if i got it wrong. i'm having alot of trouble with this question. the numbers in brackets in front of the w's are meant to be subscripts

    if $\displaystyle w_n = \exp\left(2\pi i/n\right)$ show $\displaystyle w_n^n = 1$ and that

    $\displaystyle 1 + w_n +w_n^2 + w_n^3 + \dots + w_n^{n-1} = 0$

    and show

    $\displaystyle \left(x + y\cdot w_3 + z\cdot w_3^2\right)\left(x + y\cdot w_3^2 + z\cdot w_3\right) = x^2 + y^2 +z^2 - xy -yz -zx$

    sorry for the lack of tex in it. thanks so much for any help given!!


    Edit by Chris L T521: Reformatted question with $\displaystyle \text{\LaTeX}$. Please inform me if I have misinterpreted anything.
    Last edited by Chris L T521; Aug 13th 2009 at 04:18 PM. Reason: reformatted question with LaTeX.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member pankaj's Avatar
    Jul 2008
    New Delhi(India)
    $\displaystyle w^k_{n}=e^{\frac{2k\pi}{n}}=\left(e^{\frac{2\pi}{n }}\right)^k$

    If $\displaystyle k=n$,then,$\displaystyle w^n_{n}=\cos 2\pi+i\sin 2\pi=1+i.0=1$

    $\displaystyle 1+w_{n}+w^2_{n}+w^3_{n}+.....+w^{n-1}_{n}=\frac{1-w^n_{n}}{1-w_{n}}=\frac{1-1}{1-w_{n}}=0$

    For,$\displaystyle n=3$,anove results are,$\displaystyle w^3_{3}=1$ and $\displaystyle 1+w_{3}+w^2_{3}=0$

    $\displaystyle x^2+y^2w^3_{3}+z^2_{3}+xy(w_{3}+w^2_{3})+yz(w^2_{3 }+w^4_{3})+zx((w_{3}+w^2_{3})$

    $\displaystyle =x^2+y^2+z^2+xy(-1)+yz(w^2_{3}+w_{3})+zx(-1)$

    $\displaystyle =x^2+y^2+z^2-xy-yz-zx$
    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. Replies: 2
    Last Post: Feb 7th 2009, 06:12 PM
  4. Replies: 1
    Last Post: May 24th 2007, 03:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 24th 2007, 12:34 AM

Search Tags

/mathhelpforum @mathhelpforum