eCauchyIntegral

Example Evaluate the following function

$\displaystyle \frac{1}{z^2 + 1}, C:$ $\displaystyle \mbox{circle of the radius $r > 1$\ with the center at the origin}$

Answer The contour $C$ is a circle of the radius $r > 1$ with the center at the origin. Then $f(z) = \frac{1}{z^2 + 1}$ is not analytic inside the $C$ ．Now we use partial fraction expasion of $f(z) = \frac{1}{z^2 + 1}$ .

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

By the Cauchy's Integral, we have

$\displaystyle \int_{\vert z-a\vert=r}\frac{1}{(z-a)^{n}} = \left\{\begin{array}{ll} 2\pi i & n = 1\\ 0 & n \neq 1 \end{array}\right.$

Therefore,

$\displaystyle \int_{\vert z\vert = r}\frac{1}{z^2 + 1}\ dz$	$\displaystyle =$	$\displaystyle \frac{1}{2i}[\int_{\vert z\vert=r}\frac{1}{z-1}\ dz - \int_{\vert z\vert=r}\frac{1}{z-1}\ dz$
	$\displaystyle =$	$\displaystyle \frac{1}{2i}[2\pi i - 2\pi i] = 0$

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