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.
Best Answer
For the second pole: $${\rm Res}_{z = -i} f(z) = \lim_{z \to -i} \frac{(z+i) z}{(z+i)(z-i)} = \lim_{z \to -i} \frac{z}{z-i} = \frac{-i}{-2i} = \frac{1}{2}$$ so in fact two poles contribute the same to the final result which is $2\pi i \cdot \left( \frac{1}{2} + \frac{1}{2} \right) = 2\pi i$.