Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

Evaluate $\int _{c}\frac {Z+6}{z^2-4}dz$ , where c is (i) |z|=1 (ii) |z−2|=1 (ii) |z+2|=1

$Given,\ \int_c \frac{z+6}{z^2-4} \,\, dz$

(i) |z| = 1

$f(z) = \frac{z+6}{z^2 - 4} = \frac{z+6}{(z-2)(z+2)}$

For poles, (z-2)(z+2) = 0.

Therefore, z = 2,-2 are two poles both of which lie outside the circle |z| = 1.

Therefore, by Cauchy's theorem,

$\int_c \frac{z+6}{z^2-4} \,\, dz = 0 $

(ii) |z-2| = 1

For centre, z = 2 = (2,0); r = 1; |z-2| = 1

Therefore, z = 2 is the only pole which lies inside.

Therefore, residue of f(z) at z = 2 is

$ \lim_{z \to 2} (z-2) [\frac{z+6}{(z-2)(z+2)}] = \frac{2+6}{2+2} = 2 \\ \therefore \int_c \frac{z+6}{z^2-4} \,\, dz = 2 \pi i (2) = 4\pi i.$

(iii) |z+2| = 1

Centre is z = -2 and is the only pole which is inside.

Therefore, residue of f(z) at z = -2 is

$\lim_{z \to -2} (z+2) [\frac{z+6}{(z-2)(z+2)}] = \frac{-2+6}{-2-2} = -1 \\ \therefore \int_c \frac{z+6}{z^2-4} \,\, dz = 2 \pi i (-1) = -2\pi i$