written 6.2 years ago by | • modified 2.3 years ago |
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 6.2 years ago by | • modified 2.3 years ago |
Subject: Applied Mathematics 4
Topic: Complex Integration
Difficulty: Medium
written 2.3 years ago by | • modified 2.3 years ago |
Let $I = \int_{0}^{2 \pi} \frac{d \theta}{5+4cos \theta}$
We put, $z = e^{iθ}$
Then, $cos\ θ = \frac{1}{2}(z+\frac{1}{z}), d\theta = \frac{dz}{iz} $
Therefore,
$ I = \int_{0}^{2 \pi} \frac{1}{5+4cos \theta} dz $
$ I = \int_{|z|=1} \frac{1}{5+\frac{4}{2}(z+\frac{1}{z})} . \frac{dz}{iz} $
$ I = \int_{|z|=1} \frac{1}{5+2(z+\frac{1}{z})} . \frac{dz}{iz} $
$ I = \frac{1}{2i}\int_{|z|=1} \frac{1}{5z+2z^2+ 2}.dz $
Now, Poles are determined by $5z+2z^2+ 2 = 0$
That implies, (2z+1) (z+2) = 0
Or, Z = -1/2, -2
So inside the contour C there is a simple pole at z = $\frac{-1}{2}.$
Residue at the simple pole $(z = \frac{-1}{2})$
i.e. $\implies lim_{-1/2}(z + \frac{-1}{2}) \frac{1}{(2z+1) (z+2)}.dz$
$\implies lim_{-1/2}(z + \frac{-1}{2}) \frac{1}{2 (z+2)} = \frac{1}{2*(3/2)} = \frac{1}{3}$
$\therefore\ Residue\ at\ \int_{0}^{2 \pi} \frac{d \theta}{5+4cos \theta} = \pi \frac{1}{3} = \frac{\pi}{3}.$