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Use Residue Theorems or Laurent Series to evaluate integral

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral using any method:
    C (z10) / (z - (1/2))(z10 + 2), where C : |z| = 1​

    2. Relevant equations
    C f(z) dz = 2πi*(Σki=1 Resp_i f(z)

    3. The attempt at a solution
    Rewrote the function as (1/(z-(1/2)))*(1/(1+(2/z^10))). Not sure if Laurent series expansion is the best choice for this problem but I ended up getting: (Σn=0 (1/2)n / zn+1)*(Σn=0 (-1)n 2n/(z10)n)

    I get stuck at this point but i tried working out the series and get: 1 - 2/z12 + 1/z23+ ...

    So would the residue just be 1 and the ∫C f(z) dz = 2πi?

    **sorry in advance for my formatting**
     
  2. jcsd
  3. Jun 4, 2015 #2

    vela

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    What are the poles and where are they located?
     
  4. Jun 4, 2015 #3
    Not sure if this is correct but would one pole be +1/2 (Used this function (1/(z-(1/2))) for my reasoning)?
     
  5. Jun 4, 2015 #4

    vela

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    You just need to find where the denominator vanishes, so ##z=1/2## is definitely one pole. Where are the rest?
     
  6. Jun 4, 2015 #5
    That's where I'm kind of stuck. For the other function in the denominator, (1/(1+2/z^10)), it doesn't go to 0 even if z = 0?
     
  7. Jun 4, 2015 #6

    vela

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    Oh, look at the integrand before you messed with it. In other words, when does the denominator of ##\frac{z^{10}}{(z-1/2)(z^{10}+2)}## vanish?
     
  8. Jun 4, 2015 #7
    Would the poles be z = 1/2, (-2)^(1/10) ?
     
  9. Jun 4, 2015 #8

    vela

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    Expand on what you mean by (-2)^(1/10). Remember, you're working with complex numbers, so taking fractional powers is a little more involved. Consider solving the equation ##z^{10} = -2e^{2\pi n i}## where ##n \in \mathbb{Z}##.
     
  10. Jun 4, 2015 #9
    Went to office hours today, it turns out for this question, it wasn't limited to just the residue theorems and Laurent series expansion methods. I tried using Cauchy integral formula, where z0 = 1/2 and f(z) = (z^10 / (z^10 + 2)) and got 2πi / 2049 as the final answer.
     
  11. Jun 4, 2015 #10

    vela

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    Regardless of which method you end up using, you still need to determine where the poles are. Just make sure you know how to do that.
     
  12. Jun 4, 2015 #11

    Ray Vickson

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    The number ##-2## has ten 10th roots!
     
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