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Thread: Contour Integrals

  1. #1
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    Contour Integrals

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

    $\displaystyle z \in \mathbb{C} $

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

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

    Not sure how to start this.

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

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

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

    I have tried finding where $\displaystyle f= \frac{1}{(z-2)^{3}}dz$ 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.

    $\displaystyle z \in \mathbb{C} $

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

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

    Not sure how to start this.

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

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

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

    I have tried finding where $\displaystyle f= \frac{1}{(z-2)^{3}}dz$ 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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