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Complex Analysis Series Question

  1. Jul 13, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    Let 0 < r < 1. Show that

    from n=1 to n=∞ of Σ(r^ncos(n*theta)) = (rcos(theta)-r^2)/(1-2rcos(theta)+r^2)

    Hint. This is an example of the statement that sometimes the fastest path to a “real” fact is via complex numbers. Let z = reiθ. Then, since r = |z|, and 0 < r < 1, the series n=1 to n=∞ Σz^n converges to 1/(1−z).


    2. Relevant equations
    z = x+iy
    Not really sure what else yet...

    3. The attempt at a solution

    So here, I understand the first step is to look at the sum from n=0 to n=∞ of z^n and that we know converges to 1/(1-z) when |z|<1.
    My question is how do I split this up into real and imaginary forms to solve this problem.

    Do I split it up in the summation such that from n=0 to n=∞ Σ (x+iy)^n = the sum from n=0 to n=∞ Σx^n + the sum from n=1 to n=∞ Σ(iy)^n and then find the solution that way?
     
  2. jcsd
  3. Jul 13, 2015 #2

    fzero

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    Because of the exponential factor, ##z^n = (r e^{i\theta})^n## simplifies in such a way that it is very easy to compute the real and imaginary parts.
     
  4. Jul 13, 2015 #3

    RJLiberator

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    r^n * e^(i*θ*n)

    r^n is just r^n
    e^(i*θ*n) simplifies to cos(θn)+isin(θn)

    And then I'd take the summations of each of these?
     
  5. Jul 13, 2015 #4

    fzero

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    Yes, you can see that ##r^n\cos n\theta## is what appeared in the original sum.
     
  6. Jul 14, 2015 #5

    RUber

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    This exercise wants you to sum the complex form, then take the real part of the result.
    ## \cos(\theta n ) = \text{Re}\{e^{i\theta n}\}##
    ## \sum \text{Re}\{ z \} = \text{Re} \left\{ \sum z \right\}##
     
  7. Jul 14, 2015 #6

    RJLiberator

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    Hm.

    I'm sort of understanding, yes.

    I see now that the Sum of (r^n*cos(n*theta)) is what we originally wanted indeed.
    But how do I make the jump now from the sum of (r^ncos(n*theta)) to equalling (rcos(theta)-r^2)/(1-2rcos(theta)+r^2) ?

    Is this just how it is defined, am I missing some understanding of series here?
     
  8. Jul 14, 2015 #7

    RUber

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    Look back at post #2.
    First you need to replace ##r\cos(n\theta)## with its complex form. Then take the sum.

    After that, you will have something that looks like ##\sum z^n## which you already know how to simplify.

    The last challenge is extracting the real part of your result. Multiply by the complex conjugate of the denominator and manipulate it into the form you want.
     
  9. Jul 14, 2015 #8

    RUber

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    Oh, and as I have been working through this, I noticed that you have the wrong sum formula.
    ##\sum_{n=0}^\infty z^n = \frac{1}{1-z}##
    So starting at n=1, you have to subtract z^0 away.
    Giving
    ##\sum_{n=1}^\infty z^n = \frac{1}{1-z}-\frac{1-z}{1-z} = \frac{z}{1-z}##.
     
  10. Jul 14, 2015 #9

    RJLiberator

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    rcos(n*theta) = re^(i*n) as complex form.

    So when we replace it with this, we take the sum from n=0 to n=infinity of (re^(i*n)) and see the answer: (-r+r*e^(i*theta)*e^i)/(-1+e^i)

    I'm not really sure where the complex conjugate comes into play here.
    I understand the strategy you are advising, but it seems that I am missing a basic point in understanding actual series and how to compute them.

    Also, based on your last post, is this a typo then in my "hint" of the problem? The series converges to z/(1-z)?
     

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  11. Jul 14, 2015 #10

    RUber

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    Yes the hint is wrong.

    I am not sure what you did when you said:
    Let ##z = r e ^{i\theta}## then put it into the sum formula.
    The complex conjugate of ##1-r e^{i\theta}## is ##1-r e^{-i\theta}##.
    Once your denominator is real, taking the real part of the fraction is easy.
     
  12. Jul 14, 2015 #11

    RJLiberator

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    Eureka!
    A big help was fixing the hint's error to z/(1-z).
    Also, the complex conjugate worked beautifully.

    Without your help, I would have been lost here.

    I now understand this problem pretty well.

    Thank you.
     
  13. Jul 14, 2015 #12

    RUber

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    Glad to help.
     
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