GATE EE 2018 Set 1 — Question 23

The given integral is $I = \oint_C \frac{z+1}{z^2-4} dz$.
The integrand is $f(z) = \frac{z+1}{z^2-4} = \frac{z+1}{(z-2)(z+2)}$.
The singularities of the function are at $z=2$ and $z=-2$.The contour C is a circle of radius 1 with center at the point $z = -2$. The equation of the contour is $|z - (-2)| = 1$, which is $|z+2| = 1$.We need to determine which singularities lie inside the contour C. $|2+2| = |4| = 4$. Since $4 > 1$, the pole $z=2$ lies outside the contour C. $|-2+2| = |0| = 0$. Since $0 < 1$, the pole $z=-2$ lies inside the contour C.Since there is a simple pole inside the contour, we can use Cauchy's Integral Formula, which states:
$\oint_C \frac{g(z)}{z-z_0} dz = 2\pi i \cdot g(z_0)$, provided $g(z)$ is analytic inside and on the contour C.We can rewrite the integral as:
$I = \oint_C \frac{\frac{z+1}{z-2}}{z - (-2)} dz$
Here, $z_0 = -2$ and $g(z) = \frac{z+1}{z-2}$.
The function $g(z)$ is analytic everywhere except at $z=2$. Since $z=2$ is outside the contour C, $g(z)$ is analytic inside and on C.Now, we apply the formula:
$I = 2\pi i \cdot g(-2)$
$g(-2) = \frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4}$Therefore, the value of the integral is:
$I = 2\pi i \cdot \frac{1}{4} = \frac{\pi i}{2}$Alternatively, using the Residue Theorem:
$I = 2\pi i \times (\text{Sum of residues of poles inside C})$
The only pole inside C is at $z=-2$.
The residue at the simple pole $z=-2$ is:
$\text{Res}(f, -2) = \lim_{z \to -2} (z+2)f(z) = \lim_{z \to -2} (z+2) \frac{z+1}{(z-2)(z+2)}$
$= \lim_{z \to -2} \frac{z+1}{z-2} = \frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4}$
So, $I = 2\pi i \times \frac{1}{4} = \frac{\pi i}{2}$.Both methods yield the same result.
The correct option is (A).