Contour Integrals for positively circular contour

Question

Find the contour integral of $\frac{1}{(z^2+1)^2}$ for the positively oriented circular contour $|z-Ri|=R$, for every positive real number $R>\frac{1}{2}$.

I don't know how to set up the integral.

Answer 1

Consider

$$\oint_C \frac{dz}{(z^2+1)^2}$$

I will illustrate via the residue theorem. Because $R>1/2$, the pole $z=i$ is always included within $C$. Therefore, the integral is $i 2 \pi$ times the residue at $z=i$. Because it is a double pole, the integral is

$$i 2 \pi \text{Res}_{z=i} \frac{1}{(z^2+1)^2} = i 2 \pi \left[\frac{d}{dz} \frac{1}{(z+i)^2}\right ]_{z=i}$$

Taking the derivative, we get

$$i 2 \pi \left( \frac{-2}{(i + i)^3} \right ) = i 2 \pi \left ( \frac{-i}{4}\right ) = \frac{\pi}{2}$$

as the value of the integral.