Evaluating $\int_\Gamma \frac{2z^2-z+1}{(z-1)^2(z+1)}dz$ along the contour that is shaped by the figure-8 centered at $z=-1$ and $z=1$.

Question

In my answer key, it says this equals $0$, but I get $4 \pi i$.

Here's why: $$ \int_\Gamma \frac{2z^2-z+1}{(z-1)^2(z+1)}dz = \int_\Gamma\biggl[\frac{1}{(z-1)^2}+\frac{1}{z-1}+\frac{1}{z+1}\biggr]dz $$ The last two terms are both equal to $2 \pi i$, while the first is $0$... Summing those, you get $4 \pi i$. But my textbook says the answer is $0$.

Thanks!

Answer 1

When taking contour integrals, we consider the positive direction to be counterclockwise. If we went clockwise, we would take the negative of the answer. With a figure eight, the contour will be counterclockwise in one instance and clockwise in the other. What does this tell you about the signs of the values of the integrands for their respected directions?