I am tring to find $$\int_{\Gamma}\frac{\cos(z)}{z^2(z^2+1)} \ dz,$$ where $\Gamma$ is the circle $|z|=2$.
My attempt:
Let $f(z)=\frac{\cos(z)}{z^2(z^2+1)}.$ Now, $f$ has singularities at $z=0,\pm i$.
$z=0$ is a pole of order $2$ and the residue is:
$$\lim_{z\to 0}\frac{\partial}{\partial z}\left(\frac{\cos(z)}{(z^2+1)}\right)=\frac{-z^2\sin(z)-\sin(z)-2z\cos(z)}{(z+1)^2}=0.$$
$z=i$ is a simple pole and has residue:
$$\lim_{z\to i} \frac{\cos(z)}{z^2(z+i)}=\frac{\cos(i)}{-2i}=\frac{\cosh(1)}{-2i}.$$
$z=-i$ is a simple pole and similar to above, has residue $\frac{\cosh(1)}{2i}.$
Now by the Residue theorem (all singularities lie inside $\Gamma$), $$\int_\Gamma f(z) \ dz=2\pi i\left(0-\frac{\cosh(1)}{2i}+\frac{\cosh(1)}{2i}\right)=0.$$
Is this a correct use of the Residue theorem?
Best Answer
Yes this is correct. Just to note, if your contour is more complicated you must factor in the winding number for each pole as well. Here they are all trivially 1.