# Computation of integral $\int_{\rho} \frac{dz}{(z-a)(z-b)}$ [duplicate]

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 $$r$$

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

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.