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) …