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Math Help - Contour Integrals

  1. #1
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    Contour Integrals

    Each of the following integrals is zero. Give a brief reason for each example.

     z \in \mathbb{C}

    Not sure about B(1,2) But I think it is the circle centered at 1+0i=1 with radius 2.

    (i) \int_{\partial B(1,2)} \frac{sinz}{z}dz

    Not sure how to start this.

    (ii) \int_{\partial B(1,2)} \frac{1}{z+2}dz

    Here,Cauchy's theorem can be applied. Since f=\frac{1}{z+2} is differentiable in \mathbb{C}\{-2} and B(1,2) is a closed contour in \mathbb{C}\{-2} which does not wind round any points outside \mathbb{C}\{-2}, \int_{\partial B(1,2)} \frac{1}{z+2}dz=0 by Cauchy's Theorem.

    (iii) \int_{\partial B(1,2)} \frac{1}{(z-2)^{3}}dz

    I have tried finding where f= \frac{1}{(z-2)^{3}}dz is differentiable in order to use Cauchy's theorem as in (ii) but haven't really managed to do that.
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  2. #2
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    Quote Originally Posted by nmatthies1 View Post
    Each of the following integrals is zero. Give a brief reason for each example.

     z \in \mathbb{C}

    Not sure about B(1,2) But I think it is the circle centered at 1+0i=1 with radius 2.

    (i) \int_{\partial B(1,2)} \frac{sinz}{z}dz

    Not sure how to start this.

    (ii) \int_{\partial B(1,2)} \frac{1}{z+2}dz

    Here,Cauchy's theorem can be applied. Since f=\frac{1}{z+2} is differentiable in \mathbb{C}\{-2} and B(1,2) is a closed contour in \mathbb{C}\{-2} which does not wind round any points outside \mathbb{C}\{-2}, \int_{\partial B(1,2)} \frac{1}{z+2}dz=0 by Cauchy's Theorem.

    (iii) \int_{\partial B(1,2)} \frac{1}{(z-2)^{3}}dz

    I have tried finding where f= \frac{1}{(z-2)^{3}}dz is differentiable in order to use Cauchy's theorem as in (ii) but haven't really managed to do that.
    (i) z = 0 is a removable singularity. Therefore ....

    (ii) Looks OK.

    (iii) There is a singularity at z = 2 (a pole of order 3). And the residue is zero (why?). Therefore ....
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