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I am having difficulty evaluating the complex integral

$$\int_C \frac{dz}{(z-1)^2(z-i)} $$

over the contour $C: |z-1| = 1$.

The singularities are $z_0 = 1$ and $z_1=i$, with $z_0$ lying in the contour, so Cauchy Goursat theorem does not apply correct? Is there a way to apply Cauchy Integral Formula to this?(maybe by splitting it?)

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    $\begingroup$ Do you know Residue theorem ? $\endgroup$ – Empty Oct 22 '15 at 17:24
  • $\begingroup$ We have not learned Residue theorem yet $\endgroup$ – Vivid Oct 22 '15 at 17:25
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    $\begingroup$ If you don't know the Residue theorem you can get it from what's sometimes called Cauchy's Integral Formula; that's that theorem that gives $f^{(n)}(z)$ in terms of a contour integral over the boundary. You'll need to think a bit about what function $f$ to use... $\endgroup$ – David C. Ullrich Oct 22 '15 at 17:29
  • $\begingroup$ Ok that is all I needed thank you! $\endgroup$ – Vivid Oct 22 '15 at 17:31
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Hint :

$z_1=i$ is NOT a singular point in the given region.

Take , $f(z)=\frac{1}{z-i}$. Then use Cauchy's integral formula.

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