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How would you integrate to show $$\int_{\partial B(1,2)} \frac{1}{(z-2)^3}dz = 0$$ where $B(1,2)$ is a ball centre 1, radius 2 in the complex plane. Apparently you're meant to use the FTC but I can't see how to use the FTC in the complex plane?

Thanks!

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    Is FTC the [Fundamental theorem of calculus](http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)?2012-04-10
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    Yeah. This question is on a past paper and the answers are really brief and literally say 'FTC' above the equals sign. Oh and this integral is $0$ by the way, I'll edit that in.2012-04-10
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    Follows from Cauchys Integral theorem or some of the more primitive versions of it (which might have been meant by "FTC"?). Wikipedia also compares it to the fundamental theorem of calculus: http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem#Discussion2012-04-10
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    Is there any possibility that you intended $(z-2)^3$ instead of $(z-2^3)$?2012-04-10

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