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Math Help - Complex no.

  1. #1
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    Complex no.

    If  \alpha = e^{i\frac{8\pi}{11}} . Then find  Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=

    Where Re = Real part of \alpha
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    Quote Originally Posted by jacks View Post
    If  \alpha = e^{i\frac{8\pi}{11}} . Then find  Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=

    Where Re = Real part of \alpha
    This seems pretty straight forward. What part are you having trouble with?
    Evaluating the sum?
    Finding the real part?
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  3. #3
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    As you need the real part of these then start with the fact that e^{i\theta} = \cos \theta + i\sin \theta
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    Actually you should end with that fact. Do the exponentiation first while it's in exponential form.
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    Here I Have Calculate Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5) and Getting

    \Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........  ............cos(\frac{40\pi}{11})

    Now How can I calculate that sum.....
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  6. #6
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    You can't get exact values, so you can leave it like that.
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    Quote Originally Posted by jacks View Post
    Here I Have Calculate Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5) and Getting

    \Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........  ............cos(\frac{40\pi}{11})

    Now How can I calculate that sum.....
    How did you get that?


    Edit: I now see!
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  8. #8
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    Quote Originally Posted by jacks View Post
    Here I Have Calculate Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5) and Getting

    \Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........  ............cos(\frac{40\pi}{11})

    Now How can I calculate that sum.....
    Let us denote \beta=8\pi/11 , we can use the following method:

    \alpha+\alpha^2+\ldots+\alpha^5=\dfrac{\alpha(\alp  ha^5-1)}{\alpha -1}=\dfrac{e^{i\beta}(e^{5\beta i}-1)}{e^{\beta i} -1}=

    e^{i\beta}\cdot \dfrac{e^{-i\beta/2}}{e^{i\beta/2}-e^{-i\beta/2}}\cdot \dfrac{1}{e^{-i5\beta/2}}\cdot (e^{i5\beta/2}-e^{-i5\beta/2})=

    e^{3i\beta}\cdot \dfrac{1}{2i\sin (\beta/2)}\cdot (2i\sin (5\beta/2))=(\cos 3\beta+i\sin 3\beta)\cdot \dfrac{\sin (5\beta/2)}{\sin (\beta/2)}=\ldots

    The rest is easy.

    Fernando Revilla
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  9. #9
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    Quote Originally Posted by jacks View Post
    Here I Have Calculate Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5) and Getting

    \Leftrightarrow \cos(\frac{8\pi}{11})+\cos(\frac{16\pi}{11})......  ..............\cos(\frac{40\pi}{11})

    Now How can I calculate that sum.....
    You can still do quite a bit of simplifying.

    Use some identities: \cos(-\theta)=\cos(\theta) ,\ \cos(\pi\pm\theta)=-\cos(\theta) ,\ \cos(2{\pi}n\pm\theta)=\cos(\theta)

    \cos({{8\pi}\over{11}})=\cos(\pi-{{3\pi}\over{11}})=-\cos({{3\pi}\over{11}})

    \cos({{16\pi}\over{11}})=\cos(\pi+{{5\pi}\over{11}  })=-\cos({{5\pi}\over{11}})

    \cos({{24\pi}\over{11}})=\cos(2\pi+{{2\pi}\over{11  }})=\cos({{2\pi}\over{11}})

    \cos({{32\pi}\over{11}})=\cos({{10\pi}\over{11}})=  \cos(\pi-{{\pi}\over{11}})=-\cos({{\pi}\over{11}})

    \cos({{40\pi}\over{11}})=\cos(4\pi-{{4\pi}\over{11}})=\cos({{4\pi}\over{11}})

    This result is quite interesting.

    Now look at the sum to product identity: \displaystyle \cos(\varphi)+\cos(\theta)=2\cos\!\left({{\varphi+  \theta}\over{2}}\right)\cos\!\left({{\varphi-\theta}\over{2}}\right)\,. This also can give an interesting result.
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