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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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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 didn't know that sorry, lol.2012-03-13