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Is this line integral correct?

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    C1(0) dz / (z * sin2(z))

    2. Relevant equations

    Residue Theorem material

    3. The attempt at a solution

    z * sin2(z)
    = z * (1/2 - cos(2z)/2)
    = z * [1/2 - (1/2)∑(-1)n(2z)n/(2n)! ]
    = z3 + ...

    ---> z * sin2(z) has a zero of order 3 at z = 0
    ---> 1/(z * sin2(z)) has a pole of order 3 at z = 0

    The circle C1(0) encloses z = 0 (its only singularity), so we have

    2πi * Res[f, 0] = 2πi * 1/(3-1)! * limz-->0 d2/dz2 (z-0)3 1/(z * sin2(z)),

    which I don't want to calculate if there's an easier way to do the problem. Is there an easier way? I'm too cool for calculating double derivatives.
     
  2. jcsd
  3. Jan 15, 2012 #2

    Dick

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    You could expand it in a Laurent series around z=0 and look for the coefficient of the 1/z term if you are so dead set against finding derivatives.
     
  4. Jan 15, 2012 #3
    Ah, I see! So perform long division using the series from the trig identity I invoked?
     
  5. Jan 15, 2012 #4

    Dick

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    Factor out all of the z's like you did so you should be left with something like z^3*(1-g(z)). Then use the expansion 1/(1-g(z))=1+g(z)+g(z)^2+... Remember?
     
    Last edited: Jan 15, 2012
  6. Jan 16, 2012 #5
    I got 1/3. That seem legit?
     
  7. Jan 16, 2012 #6

    Dick

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    That's what I got. Not sure it makes it legit.
     
  8. Jan 16, 2012 #7
    Awesome, brah. Now let's look at ∫C f(z), where f(z) = dz/(z-1)2(z2+4) and C = C4(0). The integrand clearly has 3 singularities, all of which lie inside C. They are z = 1, which has order 1; z = 2i, which has order 2; and z = -2i, which has order 2. The residue is therefore

    2πi * (Res[f, 1] + Res[f, 2i], Res[f, -2i]).​

    But hang on a sec, brah. For Res[f, 1] I got limz-->1 (z-1) * 1/(z-1)2(z2+4) = ∞. Wat do?
     
  9. Jan 16, 2012 #8

    Dick

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    The pole at z=1 has order 2. The other two have order 1. Why do you think otherwise?
     
  10. Jan 16, 2012 #9
    Let me redeem myself. To find the order of the pole z=1, I need to write (z-1)2(z2+4) in the form ∑an(z-1)n. Is this supposed to be incredibly obvious? I can't figure it out.
     
  11. Jan 16, 2012 #10

    Dick

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    No. You are supposed to write f(z)=g(z)/(z-1)^n where is g(z) is analytic in a neighborhood of z=1 and g(1) is not zero. What are n and g(z)? Yes, it's supposed to be pretty obvious.
     
  12. Jan 16, 2012 #11
    A punctured neighborhood of z=1?
     
  13. Jan 16, 2012 #12

    Dick

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    Ok. Punctured. It might have a removable singularity at z=1.
     
  14. Jan 16, 2012 #13
    You must be referring to g(z) = 1/(z2+4), n = 2.
     
  15. Jan 16, 2012 #14

    Dick

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    Sure. Pole of order 2.
     
  16. Jan 16, 2012 #15
    Ah, I understand the reasoning now.

    Since g(z) = 1/(z2+4) is analytic in a neighborhood about z=1, it will equal some power series centered at z=1:

    g(z) = ∑an(z-1)n. ​

    Thus
    g(z)/(z-1)2
    = (z-1)-2 (a0 + a1(z-1) + a2(z-2)2 + ...)
    = a0/(z-1)2 + ....

    Thus f(z) = g(z)/(z-1)2 has a finite number of terms of the form a-k/(z-1)k (with k>0), the first of them being being k = 2. Is that basically right?
     
  17. Jan 16, 2012 #16

    Dick

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    Sure. The point is near z=1 f(z) behaves like a0/(z-1)^2 plus small corrections.
     
  18. Jan 16, 2012 #17
    Sweet deal. (Maybe I should actually buy the textbook and stop sleeping in class.)

    I have another question, though. (See this as an opportunity to become an even higher contributor to the homework help sub-forum.)

    Am I correct that z-2csc(z) has a pole of order 3 at z=0?

    For z2sin(z)
    = z2(z - z3/3! + z5/5! - z7/7! + z9/9! + ...)
    = z3 - z5/3! + ...
    undeniably has a zero of order 3 at z=0.


    And in that case,
    Res[z-2csc(z)] = 1/(2-1)! limz-->0 ((z3 * z-2csc(z))'' = limz-->0 (z*sin2(z) - 2cos(z)sin(z) + 2z*cos2(z))/sin3(z) ????????????
     
  19. Jan 16, 2012 #18

    Dick

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    Sure again. Now you can either use l'Hopital on the trig stuff to find the limit, or you can use the Laurent series trick like in the first problem.
     
    Last edited: Jan 16, 2012
  20. Jan 16, 2012 #19
    Sweet deal. Again, how do I solve z6+1=0?

    I tried this:

    z6+1=0 <---> (z3+i)(z3-i)=0


    z3+i
    = (x+iy)3+i = 0
    <---> (x2+2xyi - y2)(x+iy) + i = (x3 - 3xy2) + i * (3x2y - y3 + 1) = 0
    <---> x2 = 3y2; 3x2y - y3 = -1
    <---> y = (1/8)-1/3 = -1/2

    So two of our zeroes are √3/2 -1/2 * i and -√3/2 - 1/2 * i

    Ditto for (z3-i)=0? amidoinitrite?
     
  21. Jan 16, 2012 #20

    Dick

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    Now can't you think of a simpler way to do it than that? You are trying to find the six sixth roots of -1. Try using polar form r*exp(i*theta).
     
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