How to evaluate $ \int_c (\sin z)/ z^6$? Evaluate:
$$ \int_c {\sin z\over z^6} \, dz$$
Where, $c$ is a circle of radius $2, |z| = 2$. 
Don't understand why ${\pi i \over 5!} $ (Answer from book).
$$ {\sin z \over z^6} = {1 \over z^6}\left [ z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+\frac{z^9}{362880}-\frac{z^{11}}{39916800}+O(z)^{12} \right ]$$
What happens to $ \int_c {1 \over z^5}\,dz$ and $- {1\over 3!} \int_c {1 \over z^3} \, dz $?
 A: Your series expansion is the same as
$$
\frac{1}{z^5} - \frac{1}{6z^3} + \frac{1}{120z} - \frac{z}{5040} + \frac{z^3}{362880} -\cdots
$$
If you integrate term-by-term, you get $0$ in every case except the third term.
(I'd have mentioned the identity that DonAntonio posted but for the fact that he already posted it, so I'm giving you a different point of view here.  Learning that identity is virtually obligatory if you want to say you know this topic.  And at least one way of proving it is very nice too: see the "Remarks" section of this article.)
Later edit: Your initial concern may have been why things like
$$
\int_C \frac{dz}{z^n}
$$
vanish when $n\in\{-2,-3,-4,\ldots\}$.  If one has established that it doesn't matter what $C$ is as long as it winds once counterclockwise around the point where the function blows up, then we may as well take it for convenience to be the unit circle centered at $0$.  Thus we have
$$
\int_C \frac{dz}{z^n} = \int_0^{2\pi} \frac{i\theta e^{i\theta}\,d\theta}{e^{ni\theta}} = i\theta\int_0^{2\pi} \frac{d\theta}{e^{(n-1)i\theta}} = i\theta\int_0^{2\pi} e^{(1-n)i\theta}\,d\theta.
$$
You can now apply the fundamental theorem of calculus, bearing in mind that $n-1>0$.  If you have qualms about whether the F.T.C. works for complex-valued functions of a real variable, you can separate real and imaginary parts.
A: Hint: for a function $\,f(z)\,$ holomorphic within the domain inclosed by a closed smooth path $\,\gamma\,$, we have
$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz$$
