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May i ask you for a little help about calculating a curve integral?

The exercise is:

Calculate the integral $\int_{\partial D_{r}(z_{0})}(\xi -z_{0})^nd\xi $ for all possible $n\in \mathbb{Z}$ and $r\in \mathbb{R}_{>0}$.

I am not sure how should i start calculting in this generalized form, clearly it's Cauchy integral... I would be very happy if someone could help me or give me an idea.

Thank you very much in advance.

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    $\begingroup$ $\partial D_r(z_0)$ is a closed curve (when parametrised). So if $(\xi-z_0)^n$ has a primitive, the integral is $0$. That reduces the work much. $\endgroup$ – Daniel Fischer Sep 3 '13 at 20:56
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Using the hint by njguliyev $$ \int_{\partial D_r(z_0)}(\xi -z_0)^n\,d\xi=\int_0^{2\pi}\,r^{n+1}\,e^{i \,(n +1)\, t}\,i\,\,dt=0 \; \mbox{ if } n\neq 1 $$ If $n=-1$, then you can use Cauchy's integral formula $$ \int_{\partial D_r(z_0)}\frac{1}{\xi -z_0}\,d\xi= 2 \pi i $$

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  • $\begingroup$ Actually it's totally easy...I am sorry for the stupid question and thank you for helping me. $\endgroup$ – Lullaby Sep 3 '13 at 21:17
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Hint: $$\xi = z_0 + r e^{it}, \quad t \in [0,2\pi].$$

Then $$\int_{\partial D_{r}(z_{0})}(\xi -z_{0})^nd\xi = \int_0^{2\pi} ir^{n+1}e^{i(n+1)t} dt = \begin{cases} ?, & n \ne -1 \\ ?, & n = -1. \end{cases}$$

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