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$\int_\gamma z^n dz$ where $\gamma$ is the unit circle $|z|=1$ oriented counter clockise and $n$ is an integer. Hint: the answer will depend on $n$.

What I don't get about it is, if I just apply Cauchy's Integral Theorem, don't I just get $0$? How come I should get an answer that depends on $n$?

Thanks

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    $\begingroup$ I think what they mean is if $n$ is negative, like $n = -1$, for example. Then the integral is not just $\, 0$. $\endgroup$ – layman Oct 31 '14 at 3:16
  • $\begingroup$ What if $n$ is negative? $\endgroup$ – azarel Oct 31 '14 at 3:17
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When one parameterises the contour, one has that

$$ \oint\limits_{\gamma} z^n \; dz = i\int\limits_0^{2\pi} e^{i(n+1)\theta} \; d\theta = \frac{e^{i(n+1)\theta}}{n+1} \Bigg|_{0}^{2\pi} = \frac{e^{i2\pi(n+1)}-1}{n+1}. $$

The conclusion follows when one considers the cases $n=-1$ and $n\neq -1$.

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  • $\begingroup$ But when $n=-1$ the integral is classic and is equals to $2\pi i$, how can you explain this in your computations? $\endgroup$ – DiegoMath Oct 31 '14 at 3:36
  • $\begingroup$ @DiegoMath What happens to the RHS when $n=-1$? $\endgroup$ – Gahawar Oct 31 '14 at 3:40
  • $\begingroup$ Wow, I understanding now, you are considering the case $n\neq-1$ and assume the case $n=-1$ as I've mentioned, am I right? $\endgroup$ – DiegoMath Oct 31 '14 at 3:55

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