Contour integral $\int_{|z|=1}\frac{2z^2+z}{z^2-1}\, dz$ using residues

Question

I am trying to evaluate the contour integral

$$\int_{|z|=1}\frac{2z^2+z}{z^2-1}\, dz.$$ In this case the two singular points lie on the boundary (on the contour). So do I count the residues at this points or do I ignore them ?

Answer 1

As @user_of_math stated, poles on the contours only contribute half; that is, $$ \oint_{\lvert z\rvert = 1}\frac{2z^2+z}{z^2-1}dz = 2\pi i\sum_{<\partial R}\text{Res} + \pi i\sum_{\partial R}\text{Res} $$ where $\partial R$ is the contour. So $<\partial R$ is inside the contour and $\partial R$ is on the contour. Since we have no poles in the contour, the first sum is zero. \begin{align} \pi i\sum_{\partial R}\text{Res} & = \pi i\biggl[\lim_{z\to -1}(z+1)\frac{2z^2+z}{z^2-1}+\lim_{z\to 1}(z-1)\frac{2z^2+z}{z^2-1}\biggr]\\ &=\pi i \end{align}