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Homework Help: Complex variables, De moivres formula

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    z is a complex number different from 1 and n >= 1 is an integer

    [tex] 1 + z + z^2+ ..... + z^n = \frac{z^{n+1} - 1}{z-1} [/tex]

    show that:
    [tex] \sin(\theta) + \sin(2 \theta)+ .... \sin(n \theta) = \frac{ \sin(n \theta/2) \sin((n+1) \theta / 2)}{\sin(\theta / 2)}[/tex]



    3. The attempt at a solution

    First I Im both sides and De Moivres formula

    [tex] Im( 1 + z + z^2+ ..... + z^n) = r \sin( \theta) + r^2 \sin(2 \theta) + ... r^n \sin(n \theta) [/tex]

    [tex] Im( \frac{z^{n+1} - 1}{z-1} ) = \frac{r^n \sin((n+1) \theta)}{\sin(\theta)} [/tex]
    1. The problem statement, all variables and given/known data

    Can anyone give me a hint from here or tell me if I am on the wrong track
     
  2. jcsd
  3. Sep 22, 2010 #2

    hunt_mat

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    Almost, use [tex]z=re^{i\theta}[/tex] in your complex geometric progression and use De Morevre's theorem and take the imaginary part to obtain the solution.
     
  4. Sep 23, 2010 #3
    Thanks, I get the left side, but how do I simplefy the right side?
    Left side

    [tex] Im( 1 + r e^{ \theta} + ... + r^n e^{ \theta}) [/tex]
    [tex] = sin( \theta) + ..... sin( n \theta) [/tex]

    right side

    [tex] Im( \frac{ r^{n+1} e^{(n+1) \theta} -1}{r e^{ \theta} -1}) [/tex]
    How do I take the imaginary part from this?
     
  5. Sep 23, 2010 #4

    hunt_mat

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    The RHS side should be (from your LaTeX)
    [tex]
    \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}
    [/tex]
    Multiply by the complex conjugate of the denominator and take the imaginary part.
     
  6. Sep 23, 2010 #5

    ehild

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    Choose r=1.

    ehild
     
  7. Sep 23, 2010 #6
    Thanks, but not sure if I understand
    [tex] Im( \frac{ r^{n+1} e^{ i (n+1) \theta} -1}{r e^{i \theta} -1}) [/tex]

    The complex conjugate [tex] r e ^{i (- \theta)} [/tex]

    [tex] Im( \frac{ r^{n+2} e^{ i n \theta} - r e^{i (- \theta)}{r^2 -r e^{i (- \theta)}}) [/tex]

    Something wrong with my Latex, but what I have done i multiplied the denominator and the numerator with [tex] r e ^{i (-\theta} [/tex] and I still have a complet denominator.
     
    Last edited: Sep 23, 2010
  8. Sep 23, 2010 #7

    hunt_mat

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    What we're saying is that:
    [tex]
    \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\cdot (e^{-i\theta}-1)/(e^{-i\theta}-1})
    [/tex]
    The denominator should now be a real number and you are free to choose the imaginary part at your leisure.
     
    Last edited: Sep 23, 2010
  9. Sep 23, 2010 #8

    hunt_mat

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    I can't get the Latex to work but the fraction should read as:
    [tex]
    \frac{e^{-i\theta}-1}}{e^{-i\theta}-1}}
    [/tex]
     
  10. Sep 24, 2010 #9
    Thanks^for all the help, I get it now.
    ----------------------------------------------
    [tex] (e^{i \theta} -1)(e^{- i \theta} -1) [/tex]
    [tex]= (1 - e^{i \theta} -e^{- i \theta} +1) [/tex]
    [tex]= 1 - cos( \theta) - i sin( \theta) - cos( - \theta) - i sin( - \theta) +1 [/tex]
    [tex]= 2 - 2cos(\theta) [/tex]

    ---------------------------------------
    [tex] cos( \theta) = cos( - \theta) [/tex]
    [tex] sin(- \theta) = - sin( \theta) [/tex]
     
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