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Thread: Fun with complex numbers

  1. #1
    Senior Member I-Think's Avatar
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    Fun with complex numbers

    By considering $\displaystyle {\Sigma_{n=1}^{N}}$$\displaystyle z^{2n-1}$ where $\displaystyle z=e^i\theta$, show that
    $\displaystyle \Sigma_{n=1}^{N} cos(2n-1)\theta=\frac{2N\theta}{2sin\theta}$

    My attempt
    Let X be equal to the sum
    $\displaystyle X=z+z^3+...+z^{2N-1}$
    $\displaystyle \frac{X}{z^2}=z^{-1}+x+...+z^{2N-3}$

    $\displaystyle X-\frac{X}{z^2}=z^{2N-1}-z^-1$
    $\displaystyle X=\frac{z^{2N+1}-z}{z^2-1}$

    Now
    $\displaystyle \Sigma_{n=1}^{N}cos(2n-1)\theta + \Sigma_{n=1}^{N}isin(2n-1)\theta=X$

    $\displaystyle X=\frac{cos(2N+1)\theta+isin(2N+1)\theta-cos\theta-isin\theta}{cos2\theta+isin2\theta-1}$

    Rationalizing with $\displaystyle (cos2\theta-1)-isin2\theta)$

    numerator of$\displaystyle X=cos(2N+1)\theta{cos2\theta}$$\displaystyle -cos(2N-1)\theta+isin(2N+1)\theta{cos2\theta}$$\displaystyle -isin(2N+1)\theta-cos\theta{cos2\theta}$$\displaystyle +cos\theta-icos2\theta{sin2\theta}+isin\theta-icos(2N+1)\theta{sin2\theta}$$\displaystyle +sin{2N+1)\theta{sin2\theta}+icos\theta{sin2\theta }-sin\theta{sin2\theta}}$

    Collecting the real terms
    $\displaystyle cos(2N+1)\theta{cos2\theta}+sin(2N+1)\theta{sin2\t heta}-(cos\theta{cos2\theta}+sin\theta{sin2\theta}$$\displaystyle -cos(2N+1)\theta-cos(2N-1)\theta+cos\theta$
    which becomes
    $\displaystyle cos(2N-1)\theta-cos(2N-1)\theta-cos\theta+cos\theta=0$
    Needless to say, I've hit a little snag, so help is politely requested.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by I-Think View Post
    By considering $\displaystyle {\Sigma_{n=1}^{N}}$$\displaystyle z^{2n-1}$ where $\displaystyle z=e^i\theta$, show that
    $\displaystyle \Sigma_{n=1}^{N} cos(2n-1)\theta=\frac{2N\theta}{2sin\theta}$

    My attempt
    Let X be equal to the sum
    $\displaystyle X=z+z^3+...+z^{2N-1}$
    $\displaystyle \frac{X}{z^2}=z^{-1}+x+...+z^{2N-3}$

    $\displaystyle X-\frac{X}{z^2}=z^{2N-1}-z^-1$
    $\displaystyle X=\frac{z^{2N+1}-z}{z^2-1}$

    Now
    $\displaystyle \Sigma_{n=1}^{N}cos(2n-1)\theta + \Sigma_{n=1}^{N}isin(2n-1)\theta=X$

    $\displaystyle X=\frac{cos(2N+1)\theta+isin(2N+1)\theta-cos\theta-isin\theta}{cos2\theta+isin2\theta-1}$

    Rationalizing with $\displaystyle (cos2\theta-1)-isin2\theta)$

    numerator of$\displaystyle X=cos(2N+1)\theta{cos2\theta}$$\displaystyle -cos(2N-1)\theta+isin(2N+1)\theta{cos2\theta}$$\displaystyle -isin(2N+1)\theta-cos\theta{cos2\theta}$$\displaystyle +cos\theta-icos2\theta{sin2\theta}+isin\theta-icos(2N+1)\theta{sin2\theta}$$\displaystyle +sin{2N+1)\theta{sin2\theta}+icos\theta{sin2\theta }-sin\theta{sin2\theta}}$

    Collecting the real terms
    $\displaystyle cos(2N+1)\theta{cos2\theta}+sin(2N+1)\theta{sin2\t heta}-(cos\theta{cos2\theta}+sin\theta{sin2\theta}$$\displaystyle -cos(2N+1)\theta-cos(2N-1)\theta+cos\theta$
    which becomes
    $\displaystyle cos(2N-1)\theta-cos(2N-1)\theta-cos\theta+cos\theta=0$
    Needless to say, I've hit a little snag, so help is politely requested.
    Note that

    $\displaystyle \sum_{n = 1}^N{z^{2n - 1}} = \sum_{n = 1}^N{e^{i(2n - 1)\theta}}$

    which is a geometric series, with $\displaystyle N$ terms, $\displaystyle a = e^{i\theta}$ and $\displaystyle r = e^{2i\theta}$.


    So $\displaystyle \sum_{n = 1}^N{e^{i(2n - 1)}} = \frac{e^{i\theta}(1 - e^{2Ni\theta})}{1 - e^{2i\theta}}$.


    Now convert this to real and imaginary parts and everything should fall into place.
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