# Compute an integral..

$\int_\gamma z^n dz$ 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

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

$$\oint\limits_{\gamma} z^n \; dz = i\int\limits_0^{2\pi} e^{i(n+1)\theta} \; d\theta = \frac{e^{i(n+1)\theta}}{n+1} \Bigg|_{0}^{2\pi} = \frac{e^{i2\pi(n+1)}-1}{n+1}.$$
The conclusion follows when one considers the cases $n=-1$ and $n\neq -1$.
• But when $n=-1$ the integral is classic and is equals to $2\pi i$, how can you explain this in your computations? – DiegoMath Oct 31 '14 at 3:36
• @DiegoMath What happens to the RHS when $n=-1$? – Gahawar Oct 31 '14 at 3:40
• 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? – DiegoMath Oct 31 '14 at 3:55