I have been asked to evaluate the following integral;
$$\int_C \frac{2dz}{z^2 -1}$$
Where $C$ is the circle of radius $1/2$, centre $1$, positively oriented.
I've determined that the function $\frac{2}{z^2 -1}$ is analytic everywhere except where $z$ is equal to $1$ or $-1$, and that the point $z = 1$ lies within the given contour.
Now, I know I can use partial fraction decomposition to acquire the following;
$$\frac{2}{z^2 – 1} = \frac{1}{z-1} – \frac{1}{z+1}$$
Which makes the integral become;
$$\int_C \frac{2dz}{z^2 -1} = \int_c \frac{1}{z-1} – \int_c \frac{1}{z+1}$$
From this point, I think it's correct for me to say that, by the Cauchy-Goursat Theorem, that the first integral will give $2\pi i$, however, I'm unsure of what to do with the second integral. Does it simply become zero??
Best Answer
As I stated within my question, the first integral will reduce to $2 \pi i$, by the Cauchy-Goursat Theorem, while the second will reduce to give zero, as it is analytic within the closed contour $C$. Thus, the entire integral reduces to $2 \pi i$.