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I was wondering if anyone can check my integration. I evaluated this $\int_{|z|=3} \! \frac{dz}{z^{2}(z-4)^{2}} \,$ to be zero but then when I checked it by using another integration method, I got a different answer.

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up vote 2 down vote accepted

You can use Cauchy's integral formula with $f(z) = \frac{1}{(z - 4)^2}$ and then your integral is precisely given by the formula

$$f'(0) = \frac{1!}{2 \pi i} \int_{|z| = 3} \frac{f(z)}{(z - 0)^{1 + 1}} \mathrm{d}z = \frac{1}{2 \pi i} \int_{|z| = 3} \frac{f(z)}{z^2} \mathrm{d}z$$

Then since $$f'(z) = \frac{-2}{(z - 4)^3}$$ you get

$$\int_{|z|=3} \! \frac{dz}{z^{2}(z-4)^{2}} = 2 \pi i f'(0) = \frac{ \pi i}{16} $$

so you don't get $0$ for an answer.

Note in particular that the function $f(z)$ is holomorphic inside the closed disk $|z| = 3$ because it only has a double pole at $z = 4$, which is outside that disk.

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I used that formula but then I assumed that the poles were inside the disk $|z|=3$ which is a circle of the form $x^2 + y^2 = 9$. Thanks for additional confirmation. – CalculusStudent Mar 13 '12 at 2:20
@CalculusStudent I'm glad that it helped you. Now, remember that you can accept answers by clicking on the check mark that's below the upvote/downvote arrows next to an answer. I see that you have a few questions which have been answered but you haven't accepted any answers yet, and since you are a new user you may not even be aware that it is possible and actually encouraged to accept an answer once you feel that your question has been successfully answered. – Adrián Barquero Mar 13 '12 at 2:34
I didn't know that sorry, lol. – CalculusStudent Mar 13 '12 at 6:30

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