integrating $\frac{(2z-1)}{(z^2 - 1)}$ around the circle of radius $1$ centred at $1$, anticlockwise.

Question

integrating $\frac{(2z-1)}{(z^2 - 1)}$ around the circle of radius $1$ centred at $1$, anticlockwise.

so I used the residue theorem, and I got the answer a $i(\pi)$, so I was just wondering whether that was right.

Answer 1

Yes, that is right, since\begin{align}\operatorname{res}_{z=1}\frac{2z-1}{z^2-1}&=\lim_{z\to1}(z-1)\frac{2z-1}{z^2-1}\\&=\lim_{z\to1}\frac{2z-1}{z+1}\\&=\frac12.\end{align}Therefore,$$\int_{\lvert z-1\rvert=1}\frac{2z-1}{z^2-1}\,\mathrm dz=2\pi i\times\frac12=\pi i.$$