How to solve this contour integral

Question

I am trying to solve this contour integral.I know that it could be evaluated using Cauchy's Theorem or by its residues and find that it is $0$, but I would like to solve it the old-fashioned way. $$\int_C\frac{dz}{z^2-1}$$ where $C$ is the circle $|z|=2$.

I know that I need to parametrize, but I genuinely do not know how to parametrize something like this, as I am used to regular lines or easy curves. What would be the best way to parametrize this?

Answer 1

Let $\gamma: [0,2\pi] \to \Bbb{C}$ be given by $\gamma(t):=2e^{it}$. Note that $\gamma$ parametrizes $|z|=2$. Then, $$ \int_{|z|=2}\frac{dz}{z^2-1} = \int_{0}^{2\pi} \frac{2ie^{it}}{4e^{2it}-1}dt = \frac{1}{2}\left(\log(1 - 2 e^{i t}) - \log(1 + 2 e^{i t})\right)\mid_{t=0}^{2\pi} = 0 $$ This is the same answer you get if you use the Residue theorem: $$ \int_{|z|=2}\frac{dz}{z^2-1} = 2\pi i ( \textrm{Res}(f, 1) + \textrm{Res}(f,- 1) )= 2\pi i\left(\frac{1}{2}-\frac{1}{2}\right)=0 $$ where $f(z):=\frac{1}{z^2-1}$.

Answer 2

I would go with $z=2e^{i\theta},$ for $0\le \theta\le 2\pi.$ Then $$dz=2ie^{i\theta}\,d\theta, $$ and you write $$\int_C\frac{dz}{z^2-1}=\int_0^{2\pi}\frac{2ie^{i\theta}}{4e^{2i\theta}-1}\,d\theta. $$ Can you finish from here?