The integral in question is
$\int_{_C} \frac{z}{z^2+1}\,dz,$ where $C$ is the path $|z-1| = 3.$
The two pole of $f(x)$ where $f(x)=\frac{z}{z^2+1}$ is $-j$ and $j$
${\rm Res}_{z=z_0}f(x)=\lim_{z\rightarrow\infty}(z-z_0)f(z)$
For the first pole:
${\rm Res}_{z=j}f(z)= \lim_{z\rightarrow\\j}(z-j)\frac{z}{z^2+1} \\ = \lim_{z\rightarrow\\j}\frac{(z-j)z}{(z+j)(z-j)}\\ =\lim_{z\rightarrow\\j}\frac{z}{(z+j)} =\frac{j}{(j+j)}$
${\rm Res}_{z=j}f(z)= \frac{1}{2}$.
For the second pole:
${\rm Res}_{z=-j}f(z)= \lim_{z\rightarrow\\-j}(z+j)\frac{z}{z^2+1} \\ = \lim_{z\rightarrow\\-j}\frac{(z+j)z}{(z+j)(z-j)}\\ = \lim_{z\rightarrow\\-j}\frac{z}{(z-j)}\\ = \frac{j}{(-j-j)}$
${\rm Res}_{z=-j}f(z)= \frac{-1}{2}$.
Sum:
${\rm Res}_{z=j}f(z)+ {\rm Res}_{z=-j}f(z)= \frac{1}{2}-\frac{1}{2} = 0$
Now I have always been under the impression that when integrating inside a path, the only time when the result is 0 is when there are no pole in or on the path.
Am I mistaken? or have I made an error in the calculation? Or should I not be trying to use the Residue Theorem all together?
Any help would be much appreciated.