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Math Help - more complex variable help

  1. #1
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    more complex variable help

    Show that for any z \in C an Z not equal to 1 one has

    1 + z +...+z^n = (1-z^(n+1))/(1-z)

    and use it to show that

    1 + \sum from k=1 to n of cos(k \theta) = 1/2 + (sin((2n+1) \theta/2))/(2sin( \theta/2))
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  2. #2
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    Quote Originally Posted by LCopper2010 View Post
    Show that for any z \in C an Z not equal to 1 one has

    1 + z +...+z^n = (1-z^(n+1))/(1-z)

    and use it to show that

    1 + \sum from k=1 to n of cos(k \theta) = 1/2 + (sin((2n+1) \theta/2))/(2sin( \theta/2))
    \frac{1}{2}+\sum_{k=1}^{n}e^{iku}=\frac{e^{i(n+1)u  }-1}{e^{iu}-1}-\frac{1}{2}=\frac{2(e^{i(n+1)u}-1)-(e^{iu}-1)}{2(e^{iu}-1)}=\frac{1}{2} \left( \frac{2e^{i(n+1)u}-e^{iu}-1}{e^{iu}-1}\right)

    Now multiply the numerator and denominaotor by e^{-\frac{u}{2}} to get

    \frac{1}{2} \left( \frac{2e^{i(n+\frac{1}{2})u}-e^{i\frac{u}{2}}-e^{-i\frac{u}{2}}}{e^{i\frac{u}{2}}-e^{i\frac{u}{2}}}\right)=\frac{1}{2}\left( \frac{2(cos[(n+\frac{1}{2})u]+i\sin[(n+\frac{2}{2})u])-2\cos(\frac{u}{2})}{2i\sin(\frac{u}{2})}\right)

    Now just equate the real and immaginary parts to get your result
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