Question

Evaluate the integral $\oint$ 1/(z²+1)(z²-4)dz where |z|=1.5

Answer 1

First, we suppose that the circle $|z|=1.5$ is oriented counterclockwise, as the orientation was not specified. Second, as the function $f(z)=\frac{1}{(z^2+1)(z^2-4)}$ is meromorphic inside the disc $|z|\leq 1.5$, we can represent this integral as the sum of residues inside the disc :

$\int_\mathcal{C} \frac{1}{(z^2+1)(z^2-4)}dz=2\pi i\sum_{|a|<1.5}Res_a \frac{1}{(z^2+1)(z^2-4)}dz$

The residues of a meromorphic function are its poles, and the poles inside the disc are $z_1=i, z_2=-i$ (the poles are the zeros of the denominator with $|z|<1.5$ ). As both poles are simple, we can calculate the respective residues using the formula $Res_af(z)dz=\lim_{z\to a} (z-a)f(z)$ :

$Res_i f(z)dz=[\frac{1}{(z+i)(z^2-4)}]_{z=i}=\frac{i}{10}$

$Res_{-i}f(z)dz=[\frac{1}{(z-i)(z^2-4)}]_{z=-i}=\frac{-i}{10}$

Therefore, the intergal is $\int_\mathcal{C} \frac{1}{(z^2+1)(z^2-4)}dz=0$ and is, in fact, independent of the orientation.