Question

The contour $C$ given below is on the complex plane $z=x+iy$, where $j=\sqrt{-1}$.



The value of the integral $\frac{1}{\pi j}{\oint }_C\frac{dz}{z^2-1}$ is

• A. 2

Answer 1

The correct option is A 2

Let $f\left(z\right)=\frac{1}{z^2-1}$ then poles are $z=\pm 1$
But here simple poles and lies inside the given contour
$R_1=\underset{(z=1)}{Resf\left(z\right)}=\underset{z\to 1}{lim}(z-1)f\left(z\right)$$=li{m}_{z\to 1}\left(\frac{1}{z+1}\right)=\frac{1}{2}$
$R_2=\underset{(z=-1)}{Resf\left(z\right)}=\underset{z\to -1}{lim}(z+1)f\left(z\right)$
$=li{m}_{z\to -1}\left(\frac{1}{z-1}\right)=-\frac{1}{2}$
But we will take $R_2=+\frac{1}{2}$ because second contour is in clockwise direction.
So by Cauchy's residue theorem
$\frac{1}{\pi j}{\oint }_c\frac{dz}{z^2-1}=\frac{1}{\pi j}[2\pi j$ (sum of residues)]
$=2[\frac{1}{2}+\frac{1}{2}]=2$