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Thread: Summation of complex numbers

  1. #1
    Senior Member I-Think's Avatar
    Apr 2009

    Summation of complex numbers

    Given that$\displaystyle w_n=3^{-n}cos2n\theta$ for $\displaystyle n=1,2,3$... use De Moivre's Theorem to show that

    $\displaystyle 1+w_1+w_2+w_3+...+w_{N-1}=\frac{9-3cos2\theta+3^{-N+1}cos2(N-1)\theta-3^{N+2}cos2N\theta}{10-6cos2\theta}$

    I am really hopelessly lost on this question.
    Help is greatly appreciated.
    Thanks in advance.
    Last edited by I-Think; Nov 2nd 2009 at 02:03 PM.
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    Let $\displaystyle S_1=1+\frac{1}{3}\cos 2\theta+\frac{1}{3^2}\cos 4\theta+\ldots+\frac{1}{3^n}\cos 2n\theta$

    and $\displaystyle S_2=\frac{1}{3}\sin 2\theta+\frac{1}{3^2}\sin 4\theta+\ldots+\frac{1}{3^n}\sin 2n\theta$

    Then $\displaystyle S_1+iS_2=1+\frac{1}{3}(\cos 2\theta+i\sin 2\theta)+\frac{1}{3^2}(\cos 4\theta+i\sin 4\theta)+\ldots+\frac{1}{3^n}(\cos 2n\theta+i\sin 2n\theta)$

    Let $\displaystyle z=\frac{1}{3}(\cos 2\theta+i\sin 2\theta)$

    Then $\displaystyle S_1+iS_2=1+z+z^2+\ldots+z^n$

    Now use the sum of a geometric progression and put the right hand member in the form $\displaystyle A+Bi$. Then $\displaystyle S_1=A, \ S_2=B$
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