Question

The value of the integral ${\oint }_C\frac{z+1}{z^2-4}dz$ in counter clockwise direction around a circle $C$ of radius $1$ with center at the point $z=-2$

• A. $\frac{\pi i}{2}$
• B. $2\pi i$
• C. $-\frac{\pi i}{2}$
• D. $-2\pi i$

Answer 1

The correct option is A $\frac{\pi i}{2}$

$C:|z+2|=1$ and $f\left(z\right)=\frac{z+1}{z^2-4}$



Only $z=-2$ lies inside $C$.
$R=\underset{(Z=-2)}{Resf\left(z\right)}$$={lim}_{z\to -2}(z+1)f\left(z\right)$
$={lim}_{z\to -2}\left(\frac{z+1}{z-2}\right)=\frac{-2+1}{-2-2}=\frac{1}{4}$
So, $I={\oint }_{C}f\left(z\right)dx$
$=2\pi i\left(\frac{1}{4}\right)$
[By Cauchy Residue Thoerem]
$\frac{\pi i}{2}$