Contour integration of: $\int_C \frac{2}{z^2-1}\,dz$ I want to calculate this (for a homework problem, so understanding is the goal)
$$\int_C \frac{2}{z^2-1}\,dz$$ where $C$ is the circle of radius $\frac12$ centre $1$, positively oriented.

My thoughts:
$$z=-1+\frac12e^{i\theta},\,0\leq\theta\leq2\pi$$
$$z'=\frac{i}{2}e^{i\theta},f(z) = \frac{}{}$$, from here it is going to get really messy, so I doubt that is the direction I should go. Where should I go?
Or is it meant to get messy from here on out?

Since the only singularity of the function $f(z)=\frac{2}{z^2-1}$ on the domain $|z-1|\leq\frac{1}{2}$ is the simple pole in $z=1$, by the Residue Theorem we have:
$$\oint_{|z-1|=\frac{1}{2}}f(z)\,dz = 2\pi i\cdot\text{Res}\left(f(z),z=1\right)=2\pi i\cdot\lim_{z\to 1}\frac{2}{z+1}=\color{red}{2\pi i}. $$
