Evaluate $\int_{\gamma} \left(\frac{\ 1}{z^2-1}\right) dz$ around the curve ${\gamma}(t)=2e^{it}$

Question

Let ${\gamma}(t)=2e^{it}$ for $-\pi\leq t\leq \pi$ and find $$\int_{\gamma} \left(\frac{\ 1}{z^2-1}\right) dz.$$

Answer 1

The contour $\gamma$ encircles the poles at $z=\pm 1$. Using the residue theorem, we have

$$\oint_{\gamma}\frac{1}{z^2-1}\,dz=2\pi i\left(\frac{1}{2}+\frac{1}{-2}\right)=0$$

Answer 2

Notice that $\gamma$ is just the circle of center $0$ and radius $2$. $z^2-1 = (z-1)(z+1)$. $-1$ and $+1$ lie in the interior of the circle. Simply apply Cauchy's Integral Formula.