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Complex number

  1. Mar 14, 2009 #1
    I'm looking at a problem involving complex numbers and a proof. It shows the solution too, but I don't get how they did a certain step.
    At one step they end up with this: (NOTE: the sigma should have 'n=0' on bottom and infinity on top, but I don't know how to do that in latex. If someone knows, please share):
    [tex]Im(e^{i\theta}\sum(re^{2i\theta})^{n})[/tex]
    Then in the next step they get from the above to the below:
    [tex]Im(e^{i\theta}\frac{1}{1-re^{2i\theta}})[/tex]
    If someone could please explain how they are equal I would greatly appreciate it.

    I didn't include the default template for threads on this section of the forum because I don't think it really applies to this question, if I'm wrong then I appologize. I didn't really attempt much of a solution, other than just writing out the sums to see if I could do anything with that... no luck.
     
  2. jcsd
  3. Mar 14, 2009 #2
    The way to do limits on the sum is \sum^{\infty}_{0}. You can use ^{} or _{} to put things at top or bottom of anything you want, generally.

    The reason that they can get rid of the sum is that it is a geometric series, which has the explicit formula
    [itex]\sum^{\infty}_{k=0}ar^{k} = \frac{a}{1-r}[/itex]
    When [itex]r<1[/itex] (do you see why r must be less than 1?).

    Cf. http://en.wikipedia.org/wiki/Geometric_series#Formula
     
  4. Mar 14, 2009 #3
    Of course! It even says 0 < r < 1 in the problem, lol, can't believe I overlooked that.

    Thank you
     
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