Why does the integral over the closed contour $\gamma$ be zero?

Question

I am going through the property (ii) of winding number from L.V.Ahlfors' book where I have failed to understand the reasoning "and if $\gamma$ does not meet the segment we must have

$$\int_{\gamma} \left ( \frac {1} {z-a} - \frac {1} {z-b} \right ) dz = 0;$$ hence..."

Would anybody please make it clear to me?

Thank you in advance.

Answer 1

It's because, as Ahlfors explained, the function that we are integrating has a primitive, which is $\log\left(\frac{z-a}{z-b}\right)$. Therefore, and since $\gamma$ is a loop, the integral is $0$: