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

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? – Alex Becker Apr 19 '12 at 2:42
@AlexBecker: It is just a typo, I fix it. Thanks. – Hassan Muhammad Apr 19 '12 at 3:19
Is $n$ an integer? For $n=1/2$, this is false... – J. M. Apr 19 '12 at 3:25

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 $\gamma$? – Hassan Muhammad Apr 19 '12 at 7:21
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. – Potato Apr 19 '12 at 18:28

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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