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Please can somebody give me hint for this?

For $n\ne -1$ $$\frac{1}{2\pi i}\int_{C} z^ndz=0$$

Where C is a simple closed curve with the usual positive orientation and its inside.

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    Where is $f$ mentioned in the problem?2012-04-19
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    @AlexBecker: It is just a typo, I fix it. Thanks.2012-04-19
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    Is $n$ an integer? For $n=1/2$, this is false...2012-04-19

2 Answers 2

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let my $C(t)= re^{it}$, $0\le t\le 2\pi$ and n is an integer, It follows that $$I=r^{n+1} \int_{0}^{2\pi}ie^{(n+1)it}dt= r^{(n+1)}\frac{e^{it(n+1)}}{n+1}|_{0}^{2\pi} \text {when }n\ne -1$$ and $2\pi i$ when $n=-1$, that is $\int f(z)dz= 0$ when $n\ne -1$

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For $n\neq -1$, $f(z)=z^n$ has a primitive $F(z)=\frac{z^{n+1}}{n+1}$. Parameterize $\gamma$ with $t$ from 0 to 1.

$\displaystyle \int_\gamma f(z) \ dz = \displaystyle \int_\gamma F'(z)\ dz = \displaystyle \int_0^1 F'(\gamma(t)) \gamma'(t)\ dt = \displaystyle \int_0^1 (F(\gamma(t)))'\ dt =F(\gamma(1))-F(\gamma(0))=0$ because the path is closed.

This works even if the curve is not simple.

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    Can you give the value of the parameterize $\gamma$?2012-04-19
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    Yes. You can apply linear transformations to change the parameterization at will. Alternatively, just note that because the curve is closed, the values at the starting point and end point must be the same.2012-04-19