Results 1 to 12 of 12
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By a tutor

Thread: complex numbers, help

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    631

    complex numbers, help

    Let $\displaystyle \alpha = z+z^{-1}. $

    show that $\displaystyle \alpha^{2} + \alpha -1 = 0 $

    I have tried to square the first equation and get,


    $\displaystyle \alpha^{2} = z^{2} + z^{-2} +2 $

    and i can't seem to go any further from here, any help appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2008
    From
    UK
    Posts
    484
    Thanks
    66

    Re: complex numbers, help

    It's true for all z.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Re: complex numbers, help

    Okay, thank you,

    I know $\displaystyle \alpha = z+ z^{-1} = z + z^{10} $

    $\displaystyle \alpha^{2} = z^{2} + z^{-2} +2 = z^{2} + z^{9} +2 $

    but i am not sure how to put this together to get $\displaystyle \alpha^{2} + \alpha -1 = 0 $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2008
    From
    UK
    Posts
    484
    Thanks
    66

    Re: complex numbers, help

    I hope your teacher set your task in writing. If so please post it exactly as it was given to you.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Re: complex numbers, help

    Okay,

    1) Z is the complex number $\displaystyle e^{\frac{2\pi i }{5} $

    a) State the value of $\displaystyle z^{5} $ , hence show that $\displaystyle z^{-r} = z^{5-r} $ for all $\displaystyle r \in Z $
    b) use the formula for the sum of a geometric series to show that $\displaystyle \sum_{r=0}^{4} z^{r} = 0 $
    c) Let $\displaystyle \alpha = z+z^{-1}. $ show that $\displaystyle \alpha^{2} + \alpha -1 = 0 $. Also show that $\displaystyle \alpha = 2cos\frac{2\pi}{5} $ and hence find an expression for $\displaystyle cos\frac{2\pi}{5} $ in surd form.

    i have done all parts except part c, am stuck on that bit

    my solutons for a and b

    $\displaystyle z^{5} = 1 $ recognising z as an 5th root of unity?

    $\displaystyle z^{-r} = 1 * z^{-r} = z^{5} z^{-r} = z^{5-r} $

    b)

    $\displaystyle S_{n} = \frac{a(r^{n}-1)}{r-1} $

    $\displaystyle \frac{ 1(z^{5}-1}{z-1} = 0 $
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2827
    Awards
    1

    Re: complex numbers, help

    Quote Originally Posted by Tweety View Post
    Okay,
    1) Z is the complex number $\displaystyle e^{\frac{2\pi i }{5} $
    c) Let $\displaystyle \alpha = z+z^{-1}. $ show that $\displaystyle \alpha^{2} + \alpha -1 = 0 $. Also show that $\displaystyle \alpha = 2cos\frac{2\pi}{5} $ and hence find an expression for $\displaystyle cos\frac{2\pi}{5} $ in surd form.
    I think that I have written this before. But again, posting the complete problem to begin with saves a lot of time.

    Let us use a more convenient notation.
    $\displaystyle z=\exp \left( {\frac{{2\pi i}}{5}} \right)$.

    Here are useful facts for any nonzero complex number $\displaystyle w$.
    If $\displaystyle w=\exp(i\theta)$ then $\displaystyle w^{-1}=\exp(-i\theta)$ and$\displaystyle w+w^{-1}=2\cos(\theta)$

    Now try the question.
    Last edited by Plato; Nov 14th 2012 at 09:49 AM.
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jan 2008
    From
    UK
    Posts
    484
    Thanks
    66

    Re: complex numbers, help

    In post 2 I meant it's NOT true for all z.

    Sorry.
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Re: complex numbers, help

    i am actually stil not able to compute part 'c'
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2827
    Awards
    1

    Re: complex numbers, help

    Quote Originally Posted by Tweety View Post
    i am actually stil not able to compute part 'c'
    If $\displaystyle \alpha = 2\cos \left( {\frac{{2\pi }}{5}} \right)$ can't you show that $\displaystyle \alpha^2+\alpha-1=0~?$
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Re: complex numbers, help

    Quote Originally Posted by Plato View Post
    If $\displaystyle \alpha = 2\cos \left( {\frac{{2\pi }}{5}} \right)$ can't you show that $\displaystyle \alpha^2+\alpha-1=0~?$
    BUt I dont understand how alpha equals that ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2827
    Awards
    1

    Re: complex numbers, help

    Quote Originally Posted by Tweety View Post
    BUt I dont understand how alpha equals that ?
    I posted it above.
    If $\displaystyle w=\exp(i\theta)$ then $\displaystyle w^{-1}=\exp(-i\theta)$ and$\displaystyle w+w^{-1}=2\cos(\theta)$

    $\displaystyle \cos(\theta)+i\sin(\theta)+\cos(-\theta)+i\sin(-\theta)=$
    $\displaystyle [\cos(\theta)+\cos(-\theta)]+i[\sin(\theta)+\sin(-\theta)]=$
    $\displaystyle [\cos(\theta)+\cos(\theta)]+i[\sin(\theta)-\sin(\theta)]=$
    $\displaystyle 2\cos(\theta)$

    Recall that $\displaystyle \cos$ is an even function and $\displaystyle \sin$ is an odd function.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Nov 2012
    From
    Moldova
    Posts
    1

    Re: complex numbers, help

    Hello!
    Would try to help you!
    I'm not sure, but what i got is
    Rectangualr form a^2*z^2+1/z^2+2
    Polar form a^2*z^2+1/z^2+2
    check it here Online Complex Numbers Calculator
    I used Complex Numbers Calculator
    Good luck
    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