how to evaluate this complex integral I need to evaluate $$\int_{|z|=2}\frac{1}{(z-1)^3}dz.$$ At $z=1$, it has a pole of order $3$. I can not remember how to find the residue when there are poles with multiplicity, could any one tell me?
 A: Recall Cauchy integral formula, which goes as follows. Let $f(z)$ be analytic inside an open set $\Omega$. Let $\Gamma$ be a closed curve inside $\Omega$. Let $z_0$ lie inside the closed curve $\Gamma$, we then have that
$$f^{(n)}(z_0) = \int_{\Gamma} \dfrac{f(z)}{(z-z_0)^{n+1}} dz$$
In your case, $f(z) = 1$, $n=2$, $z_0 = 1$ and $\Gamma$ is the circle $\vert z \vert = 2$. Hence, $f^{(2)}(z) = 0$. Hence, the integral is $0$.
A: The residue is the coefficient $a_{-1}$ in the Laurent expansion $$\frac{1}{(z-1)^3}=\cdots+\frac{a_{-3}}{(z-1)^3}+\frac{a_{-2}}{(z-1)^2}+\frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)^2+\cdots.$$
Can you find a sequence $(a_n)$  making this equation hold?
Alternatively, the reason residues determine the integrals is that all terms of the form $a_n(z-a)^n$ with $n\neq -1$ have antiderivatives on $\mathbb C\setminus\{a\}$, hence integrate to $0$ on any closed curve not containing $a$.  In particular, the function you present has an antiderivative on $\mathbb C\setminus\{1\}$.
