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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?

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1… – anon Jan 4 '13 at 7:13
Do you know how to integrate by finding antiderivatives? – Jonas Meyer Jan 4 '13 at 7:17
@JonasMeyer Na. – La Belle Noiseuse Jan 4 '13 at 7:23

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 $0$.

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Would it be $z_0\in\Gamma$? – La Belle Noiseuse Jan 4 '13 at 7:28
@Kuttus Yes. The closed curve is $\Gamma$ and not $\Omega$. I have changed it. – user17762 Jan 4 '13 at 7:32
@Kuttus: $\Gamma$ denotes the closed curve, not the region it bounds. Note that $z_0$ is in the interior of this region, so $z_0 \not\in \Gamma$. – Michael Albanese Jan 4 '13 at 7:34
will there be a $n!\over 2\pi i$ in the formula u have written? and I think something wrong with your answer – La Belle Noiseuse Jun 3 '13 at 17:28
@TaxiDriver Yes. There should be $n!/(2 \pi i)$. But that won't change the fact that the answer is $0$. – user17762 Jun 3 '13 at 17:35

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 $\mathbb C\setminus\{1\}$.

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