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Calculate integral $\int_{|z|=1} \frac{z}{(z-2)^2}dz$

I think it is equal to $0$ from Cauchy theorem... since the circle $|z|=1$ is contained in some region where $f(z)$ is holomorphic, is it?

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    $\begingroup$ I think you mean the right thing, but the relevant bit is not "the circle $\lvert z\rvert = 1$", it is the disk $\lvert z\rvert \leqslant 1$. $\endgroup$ Nov 30, 2014 at 14:35
  • $\begingroup$ Oh thats true, thank you $\endgroup$
    – lksz23
    Nov 30, 2014 at 14:52

1 Answer 1

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Yes you are correct. The function $f(z)=\frac{z}{(z-2)^2}$ has singularity at $2$ and it's outside the disc $|z|\leq1$.

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