Complex integral over closed path with points outside path (Cauchy Integral)

Question

I am just looking for solution verification of the following integral.

$$\int_{|z-1|=1}\frac{dz}{z^2-1}$$

The denominator can be factored into $(z+1)(z-1)$, but $z_0=-1$ is not inside the path $|z-1|=1$, so we can rewrite as

$$\int_{|z-1|=1}\frac{(z+1)^{-1}dz}{z-1}$$

By Cauchy's Integral Formula, then we know that

$$\int_{|z-1|=1}\frac{(z+1)^{-1}dz}{z-1}=\frac{2\pi i}{0!}f^{(0)}(1)=\frac{2\pi i}{1+1}=\pi i$$

Is this correct?

Answer 1

Yes, your solution is perfectly fine. Another way to approach a problem such as this is to write

$$\begin{align} \oint_{|z-1|=1}\frac{1}{z^2-1}\,dz&=\oint_{|z-1|=1}\left(\frac{1/2}{z-1}-\frac{1/2}{z+1}\right)\,dz\\\\ &=\frac12\oint_{|z-1|=1}\frac{1}{z-1}\,dz-\frac12\oint_{|z-1|=1}\frac{1}{z+1}\,dz\\\\ &=\pi i +0\\\\ &=\pi i \end{align}$$