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Math Help - complex numbers, help

  1. #1
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    complex numbers, help

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

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

    I have tried to square the first equation and get,


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

    and i can't seem to go any further from here, any help appreciated.
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  2. #2
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    Re: complex numbers, help

    It's true for all z.
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  3. #3
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    Re: complex numbers, help

    Okay, thank you,

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

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

    but i am not sure how to put this together to get  \alpha^{2} + \alpha -1 = 0
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  4. #4
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    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.
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  5. #5
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    Re: complex numbers, help

    Okay,

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

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

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

     z^{-r} = 1 * z^{-r}  = z^{5} z^{-r} = z^{5-r}

    b)

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

     \frac{ 1(z^{5}-1}{z-1}  = 0
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  6. #6
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    Re: complex numbers, help

    Quote Originally Posted by Tweety View Post
    Okay,
    1) Z is the complex number  e^{\frac{2\pi i }{5}
    c) Let  \alpha = z+z^{-1}. show that  \alpha^{2} + \alpha -1 = 0 . Also show that  \alpha = 2cos\frac{2\pi}{5} and hence find an expression for  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.
    z=\exp \left( {\frac{{2\pi i}}{5}} \right).

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

    Now try the question.
    Last edited by Plato; November 14th 2012 at 09:49 AM.
    Thanks from Tweety
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  7. #7
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    Re: complex numbers, help

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

    Sorry.
    Thanks from Tweety
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  8. #8
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    Re: complex numbers, help

    i am actually stil not able to compute part 'c'
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  9. #9
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    Re: complex numbers, help

    Quote Originally Posted by Tweety View Post
    i am actually stil not able to compute part 'c'
    If \alpha  = 2\cos \left( {\frac{{2\pi }}{5}} \right) can't you show that \alpha^2+\alpha-1=0~?
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  10. #10
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    Re: complex numbers, help

    Quote Originally Posted by Plato View Post
    If \alpha  = 2\cos \left( {\frac{{2\pi }}{5}} \right) can't you show that \alpha^2+\alpha-1=0~?
    BUt I dont understand how alpha equals that ?
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  11. #11
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    Re: complex numbers, help

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

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

    Recall that \cos is an even function and \sin is an odd function.
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  12. #12
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    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
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