written 6.1 years ago by | • modified 2.2 years ago |
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 6.1 years ago by | • modified 2.2 years ago |
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 5.8 years ago by | • modified 5.8 years ago |
Put z = e$^{i \theta}$
dz = i e$^{i \theta}$ d$\theta$
$d \theta = \frac{dz}{iz} $
Therefore, $ sin\theta = \frac{z-1/z}{2i} = \frac{z^2-1}{2iz} $
The integration becomes,
$ =\int_c \frac{1}{5+3(z^2-1)/(2iz)} \,\, \frac{dz}{iz} \\ = \int_c \frac{2iz}{10iz + 3z^2 - 3} \frac{dz}{iz} \\ = \int_c \frac{2}{3z^2 + 10iz - 3} \,\, dz $
Consider f(z) = $ \frac{2}{3z^2 + 10iz - 3} $
For poles,
$ 3z^2 + 10iz - 3 = 0 \\ 3z^2 + 9iz + iz - 3 = 0 \\ 3z(z+3i) + i(z+3i) = 0 \\ \therefore 3z + i = 0; z+ 3i = 0 $
Therefore, $ z = \frac{-i}{3}; z = -3i $ are poles of which $ z = \frac{-i}{3} $ is the only pole which lies inside |z| = 1
Therefore, residue of f(z) at $z = \frac{-i}{3} $ is
$ lim_{z \to (\frac{-i}{3})} (z + \frac{i}{3})[\frac{2}{(3z+i)(z+3i)}] \\ = lim_{z \to (\frac{-i}{3})} (z + \frac{i}{3})[\frac{2}{3(z+i/3)(z+3i)}] \\ = \frac{2}{3[-i/3 + 3i]} = \frac{2}{3(8i/3)} = \frac{2}{8i} = \frac{1}{4i} $
By Cauchy's Residue theorem,
$ \int_c \frac{2}{3z^2 + 10iz - 3} \,\, dz = 2 \pi i(\frac{1}{4 i}) = \frac{\pi}{2} $
$\therefore \int_0^{2\pi} \frac{d\theta}{5 + 3 sin\theta} = \frac{\pi}{2} $