I want to evaluate the following integral,
$$\oint_C\frac{z^2-1}{z^2+2}\,dz,$$ where $C$ is a circle of radius $2$, centered at $z=0$.
For the first one I can use the deformation principle. We have roots at $\pm \sqrt{2}i$. So define two new contours $C_1$, which encloses $\sqrt{2}i$, and $C_2$ which encloses $-\sqrt{2}i$. The deformation principle says that the integral is equivalent to, $$\oint_{C_1}\frac{z^2-1}{z^2+2}\,dz+\oint_{C_2}\frac{z^2-1}{z^2+2}\,dz.$$ We can simply use Cauchy's integral formula to evaluate these two integrals. The first evaluates to, $$\oint_{C_1}\frac{z^2-1}{z^2+2}\,dz=2\pi i\,\frac{z^2-1}{z+\sqrt{2}i}\bigg|_{z=\sqrt{2}i}=-\frac{3\pi}{\sqrt{2}}.$$ The second similarly evaluates to, $$\frac{3\pi}{\sqrt{2}},$$ such that the entire integral evaluates to $0$.
Am I thinking about this problem correctly. Is there a reason why the root at $2i$ cancels the root at $-2i$ to give $0$?
The function that you are integrating is $\dfrac{z^2-1}{z^2+2}$, which has two simples poles, at $\pm\sqrt2\,i$. So, the residue ate each pole $z_0$ is $\dfrac{{z_0}^2-1}{2z_0}$. But this is an odd functions and the poles are symmetric. Therefore, the sum of the residues is $0$.