Is Cauchy's integral theorem affected by integral direction? 
Hello everyone,I have an exam question about the integration ~
as shown below
I know $\frac{1}{z^2-1}=\frac{1}{2}(\frac{1}{z-1}-\frac{1}{z+1})$. The integration around $1$ should be $2\pi i$, but i am confused if the integration around $-1$ is $2\pi i$ or $-2\pi i$? and why?
can anyone help me?
 A: It should be $+2\pi i$. Why? Imagine if you went the other way around the $-1$ pole. Then the integral would be the same as if your curve was just going around one big counter-clockwise circle $R_a$ of radius $a > 1$ (this is where the Cauchy integral theorem comes in).
Now, the triangle inequality gives us
$$
\left|\oint_{R_a}\frac{dz}{z^2 - 1}\right| \leq \oint_{R_a}\left|\frac{1}{z^2 - 1}\right|dz
$$
but as $a \to \infty$, the second integral goes to $0$ (because the integrand decreases much more rapidly than the domain of integration, i.e. length of $R_a$, increases), while the first remains unchanged (again, due to Cauchy), so we must have $\oint_{R_a}\frac{dz}{z^2 - 1} = 0$. This means that each of the poles cancel out. Since we know that the $+1$ pole contribute $2\pi i$, the $-1$ pole must contribute $-2\pi i$.
But in our integral we're evaluating around the $-1$ pole the other way, so its contribution to the integral has the opposite sign. Therefore it contributes $2\pi i$ and the whole integral evaluates to $4\pi i$.
