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?
On
Use residue theorem to find integrals of this type. And observe that the only singular point for this analytic function in the disk above is at $z=1$.
On
The easiest way to proceed is to use the residue theorem. Since so many respondents have already tackled the problem along that line, we will proceed here directly and evaluate the contour integral using brute force.
We will use the parameterization $z=1+\frac12 e^{it}$, $0\le t< 2\pi$. Then $dz = ie^{it}dt$. To that end, we have
$$\begin{align} \oint_C \frac{2}{z^2-1}dz&=\oint_C \left(\frac{1}{z-1}-\frac{1}{z+1}\right) dz\tag1\\\\ &=\int_0^{2\pi} \left(\frac{1}{(1+\frac12 e^{it})-1}-\frac{1}{(1+\frac12 e^{it})+1}\right)\frac12ie^{it}dt\tag 2\\\\ &=2\pi i-\int_0^{2\pi} \frac{ie^{it}}{4+ e^{it}}dt\tag3\\\\ &=2\pi i-\log(4+e^{it})|_0^{2\pi}\tag4\\\\ &=2\pi i\tag5 \end{align}$$
For (1), we used partial fraction expansion of the integrand.
For (2), we used the aforementioned parameterization.
For (3), we evaluated the first integral as $\int_0^{2\pi} i dt=2\pi i.
For (4), we observed that the anti-derivative of the integrand was $\log (4+e^{it})$.
For (5), we evaluated that anti-derivative between the limits, and noted that $e^{i2\pi}=e^{i0}=1$ and so the integral is zero.
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: $$\oint_{|z-1|=\frac{1}{2}}f(z)\,dz = 2\pi i\cdot\text{Res}\left(f(z),z=1\right)=2\pi i\cdot\lim_{z\to 1}\frac{2}{z+1}=\color{red}{2\pi i}. $$