Confusing integral

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April 20th 2010, 09:21 AM
piglet

Confusing integral

Show $\int_{C}\frac{1}{z-2}dz = 2\pi.i$ where $C$ is a circle of radius $1$ centred at $2$
using a parametrization.

I can show this using Cauchy's Integral formula i.e.
$\int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0)$ but i dont't think this is using a parametrization?

Any ideas?
April 20th 2010, 09:34 AM
chisigma

Setting $s=z-2$ the integral becomes...

$\int_{\gamma} \frac{ds}{s}$ (1)

... where $\gamma$ is now the unit circle centered at $s=0$ . Now You set $s=e^{i\cdot \omega}$ and solve the integral in $\omega$ ...

Kind regards

$\chi$ $\sigma$

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