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Why the path independence theorem does not work?

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data
    where C is the contour given with direction marked by increasing y, and where -2≤y≤2 , compute itgeral(z^2-2z+1)dz. With the condition x=5;

    Firstly I solved the auestion with the classical way ;
    taking z= 5 + it where -2≤t≤2;
    we take the i*integral((5+it)^2-2(5+it) +1)dt from t=-2 to t=2 and I found the result as
    176i/3.

    Then I tried to use the path independence theorem. The antiderivative of z^2-2z+1 is z^3/3 - z^2 + z then taking the integral from 2i to -2i yields -4i/3. So the two answers are different. Why the path independence theorem did not work here?[/QUOTE]
     
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2

    Hurkyl

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    That makes no sense; you have to integrate with respect to some variable. It looks like you just erased the dz from the original integral before getting to this point!
     
  4. Nov 7, 2007 #3
    sorry , I made a mistyping now I corrected it. There I used the formula
    integral(f(z)dz) over a contour c = integral(f(z(t))* z'(t)) dt .
     
  5. Nov 7, 2007 #4

    Hurkyl

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    Your paths don't have the same endpoints, do they?

    Incidentally,
    It's not the antiderivative; it's an antiderivative. (But that's not relevant here)
     
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