Cauchy-Goursat Theorem Problem

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??

• Indeed, that is the application of Cauchy-Goursat. Since $\frac{1}{z+1}$ is analytic inside the contour $C$, the integral is zero. – Alex Wertheim Apr 26 '14 at 7:03
• As far as I know, what is usually termed 'Cauchy-Goursat' will allow you to evaluate the second integral. The value of the first integral (which you've correctly stated) is just a standard result, which can be seen by the parameterisation $z(t) = 1 + \frac{1}{2}e^{it}$, $t \in [0,2\pi]$. – ah11950 Apr 26 '14 at 7:03

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 $2 \pi i$.