Contour Integrals

**nmatthies1** · December 5th 2009, 06:37 PM

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.

**mr fantastic** · December 6th 2009, 02:45 AM

Originally Posted by nmatthies1

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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