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Math Help - circle and integral

  1. #1
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    circle and integral

    True or False:  \int_{C} \frac{1}{z^2+1} \ dz = 0 for every circle given by  |z| = r, \ r > 1 .

    So that singular points are at  z^2 = -1 . Thus  z = \pm i. But these are both inside the circle  |z| = r, \ r > 1 . Thus the statement is false?
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     \int_{C} \frac{1}{z^2+1} \ dz = 2\pi i(\text{Res}_{z=i} \frac{1}{z^2+1} + \text{Res}_{z=-i} \frac{1}{z^2+1})
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    Quote Originally Posted by chiph588@ View Post
     \int_{C} \frac{1}{z^2+1} \ dz = 2\pi i(\text{Res}_{z=i} \frac{1}{z^2+1} + \text{Res}_{z=-i} \frac{1}{z^2+1})
    So the statement is false?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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     \text{Res}_{z=i} \frac{1}{z^2+1} = \frac{1}{2i}

    and

     \text{Res}_{z=-i} \frac{1}{z^2+1} = -\frac{1}{2i}
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  5. #5
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    Quote Originally Posted by manjohn12 View Post
    True or False:  \int_{C} \frac{1}{z^2+1} \ dz = 0 for every circle given by  |z| = r, \ r > 1 .

    So that singular points are at  z^2 = -1 . Thus  z = \pm i. But these are both inside the circle  |z| = r, \ r > 1 . Thus the statement is false?
    This is a special case of a more general result. If f,g are polynomials functions with g non-constant and \gcd(f,g)=1 with \deg (g) \geq \deg (f) + 2 then the sum of residues of \tfrac{f}{g} is always zero. Try proving it!
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    Quote Originally Posted by ThePerfectHacker View Post
    This is a special case of a more general result. If f,g are polynomials functions with g non-constant and \gcd(f,g)=1 with \deg (g) \geq \deg (f) + 2 then the sum of residues of \tfrac{f}{g} is always zero. Try proving it!
    Suppose that the sum of residues of \tfrac{f}{g} was not  0 . Then somehow arrive at a contradiction. Is this something you discovered?
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    Quote Originally Posted by manjohn12 View Post
    Suppose that the sum of residues of \tfrac{f}{g} was not  0 . Then somehow arrive at a contradiction. Is this something you discovered?
    No, I had this problem on a complex analysis exam. I just remembered the result.
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