Cauchy Integral with two poles on the contour

Question

We should find the Cauchy principal value integral of the form $$ I=\oint \frac{dz}{(z-z_1)(z-z_2)}~, $$ where both roots $z_1$ and $z_2$ lie on the contour path. My answer is: $$ I=a \left(-\oint \frac{dz}{z-z_1}+\oint \frac{dz}{z-z_2}\right)=a(-i\pi+i\pi)=0~, $$ where $a=1/(z_2-z_1)$. However, in a book, they do not find it to be zero, though they do not explain why. Is there any case for which I should find this integral to be "not zero"? In the specific example of the book, the poles are $z_1=e^{-ik}$, $z_2=e^{ik}$, and the contour is the unit circle.

Answer 1

First: putting $\;f(z):=\frac1{(z-z_1)(z-z_2)}\;$ , we obtain (since both poles are simple assuming $\;z_1\neq z_2\;$ )

$$\text{Res}_{z=z_i}(f)=\lim_{z\to z_i}\frac{z-z_i}{(z-z_1)(z-z_2)}=\begin{cases}&\;\;\;\;\;\;\;\;\;\frac1{z_1-z_2}&,\;\;i=1\\{}\\&-\frac1{z_1-z_2}=\frac1{z_2-z_1}&,\;\;i=2\end{cases}$$

Now using the lemma, and its corollary, in the most upvoted answer here, we get that the integral indeed equals zero.