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Find the real part

  1. Oct 7, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If [itex]\large α = e^{i\frac{8∏}{11}}[/itex], then find [itex]Re(α+α^{2}+α^{3}+α^{4}+α^{5})[/itex]


    2. Relevant equations


    3. The attempt at a solution
    [itex]\large e^{i\frac{8∏}{11}}+e^{i\frac{16∏}{11}}............+e^{i\frac{40∏}{11}}[/itex]

    [itex]cos \frac{8∏}{11}+isin \frac{8∏}{11}..........[/itex]

    Since I am interested only in real part so now I have to find the value of

    [itex]cosθ+cos2θ........cos5θ[/itex]

    where [itex]θ= \frac{8∏}{11}[/itex]

    I think some trigonometry must be applied since it seems to me sum of a trigonometrical series.
     
  2. jcsd
  3. Oct 7, 2012 #2

    ehild

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    Is not

    [itex]α+α^{2}+α^{3}+α^{4}+α^{5}[/itex]

    a geometric series?

    hild
     
  4. Oct 7, 2012 #3

    Ray Vickson

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    You have found the real part; it is a sum of 5 terms. What is wrong with that answer?

    RGV
     
  5. Oct 8, 2012 #4

    utkarshakash

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    Hey I have found the answer but not completely. I have to find the value of cosθ+cos2θ.....
    which I dont know how to solve
     
  6. Oct 8, 2012 #5

    Mentallic

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    Well as people have already mentioned, the answer IS
    [itex]\cos(8\pi/11)+\cos(16\pi/11)+...+\cos(40\pi/11)[/itex]
    but that's messy, and this question has been cleverly constructed so that there is a nice answer.

    [tex]\alpha+\alpha^2+...+\alpha^5[/tex]
    [tex]=1+\alpha+\alpha^2+...+\alpha^5-1[/tex]

    [tex]=\frac{1-\alpha^6}{1-\alpha}-1[/tex]

    Now, notice that

    [tex]\alpha^6=e^{48\pi i/11}=e^{4\pi i/11}=\left(e^{8\pi i/11}\right)^{1/2}=\alpha^{1/2}[/tex]

    So we can now turn the expression into

    [tex]=\frac{1-\alpha^{1/2}}{1-\alpha}-1[/tex]

    [tex]=\frac{1-\alpha^{1/2}}{(1-\alpha^{1/2})(1+\alpha^{1/2})}-1[/tex]

    I'm sure you can finish it off from here :smile:
     
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