TITLE: Finding a method to calculate the line integral of a certain function with a certain domain
QUESTION [0 upvotes]: For example, let us calculate the line integral $\int_{\partial} \frac{1}{(z-1)} $ where the domain is $D(0,|2|)$.
I first generalize the whole reasoning into: for integral $\int\frac{1}{(z-a)} $, if the domain contains a, then its value is 2$\pi i$, on the other hand, if

the $\int\frac{1}{(z-a)^n} $ with n equal or great than 2
for integral $\int\frac{1}{(z-a)} $ the point a does not contain in the domain

The integral goes directly to 0.
Is it valid?

REPLY [1 votes]: As a consequence of the analyticity of holomorphic functions you get the formula:
\begin{equation}
\frac{f^{n)}(z_{0})}{n!} = \frac{1}{2 \pi i} \int_{\partial}\frac{f(w)}{(w-z_{0})^{n+1}}dw
\end{equation}
$\partial$ is oriented counterclockwise, $z_{0}$ is inside $\partial$ and $f(w)$ is analytic on and inside $\partial$.
You can see a proof here (only for circles): https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions

You can also see this proof (not only for circles but the proof is not complete):
http://mathonline.wikidot.com/cauchy-s-integral-formula-for-derivatives
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