written 6.2 years ago by | • modified 2.3 years ago |
Where C is the circle,
(i) |z|=1
(ii) |z-2|=1
(iii) |z+2|=1
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 6.2 years ago by | • modified 2.3 years ago |
Where C is the circle,
(i) |z|=1
(ii) |z-2|=1
(iii) |z+2|=1
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 5.9 years ago by | • modified 5.9 years ago |
(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 $