The integral along a circle of the inverse linear function is zero

Question

Assume ${\rm C}$ is a circle and $a,b$ are distinct points in the interior of ${\rm C}$.

How can we see that the complex integral

$$ \frac{1}{b - a} \int_{\rm C}\left(\frac{1}{z - a} - \frac{1}{z - b}\right)\,{\rm d}z = 0 $$

Edit: I want to see this from Cauchy's formula, or from first principle, since the above result is used in proving the existence of consecutive derivatives, which is needed for the residue theorem.

Answer 1

Let $f(z) = \frac{1}{z-a}-\frac{1}{z-b}$. Using Residue Theorem, one have that $$ \oint_c f(z)\, dz = 2\pi i ({\rm Res}( f, a ) + {\rm Res}( f, b )) = 2\pi i(1-1) = 0. $$