eCauchyIntegralForm

Example Evaluate the following．

$\displaystyle \oint_{\vert z\vert = 2}\frac{\cos{z}}{z^2 + 1}$

Answer A function $f(z)$ is analytic in the region $\Omega$ ． $C$ is a simply connected cureve in $\Omega$ and the region inside of $C$ is contained in $\Omega$ . Then for any point $a$ in $C$ , we have Cauchy's Integral Formula

$\displaystyle f(a) = \frac{1}{2\pi i}\int_{C}\frac{f(z)}{z-a};dz$

$\displaystyle f^{(n)}(a) = \frac{n!}{2\pi i}\int_{C}\frac{f(z)}{(z-a)^{n+1}}\;dz$

We solve the problem. A function $f(z) = \cos{z}$ is analytic inside of $\vert z\vert = 2$ ．Now to use Cauchy's Integral Formula, we use a partial fraction expansion.

$\displaystyle \frac{\cos{z}}{z^2 + 1} = \frac{1}{2i}\cos{z}(\frac{1}{z-i} - \frac{1}{z+i})$

$\displaystyle \oint_{\vert z\vert = 2}\frac{\cos{z}}{z^2 + 1} = \frac{1}{2i}(2\pi i \cos(i) - 2\pi i \cos(-i)) = \pi(\cos(i) - \cos(-i)) = 0$

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