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This question already has an answer here:

Let $a,b$ be complex number and $|a| < r < |b|$, compute.

$\int_{\rho} \frac{dz}{(z-a)(z-b)}$

where $\rho$ is the circle with radius $r$ and the usual orientation.

I've tried the common path $\int_{\rho} \frac{dz}{(z-a)(z-b)} = \int_0^{2\pi}\frac{rie^{i \theta} d\theta}{(re^{i \theta} -a)(re^{i \theta}-b)}$ but in this point I don't know how use the $r$

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marked as duplicate by Arnaud D., Lord Shark the Unknown, Mr Pie, Cesareo, Joel Reyes Noche Apr 8 at 8:39

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By residue theorem the integral is $$ 2\pi i\text {Res}_{z=a}\frac1 {(z-a)(z-b)}=\frac {2\pi i}{a-b}, $$ since only the pole $z=a $ is inside the integration contour.

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