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Question: Evaluate $\int_{0}^{2 \pi} \frac{d \theta}{(2+cos \theta)^2 }$
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 102 views
 modified 10 weeks ago  • written 5 months ago by
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$\int_{\pi}^{2\pi} \frac{d\theta}{(2 + cos\theta)^2} \\ = \int_c \frac{1}{(2 + (z^2+1)/2z)^2} \frac{dz}{iz} \\ = \int_c \frac{1}{((4z+z^2+1)/2z)^2} \frac{dz}{iz} \\ = \int_c \frac{4z^2}{z^2+4z+1} \frac{dz}{iz} \\ = \frac{4}{i} \int_c \frac{z}{(z^2+4z+1)^2} dz$

Consider, $f(z) \ \frac{z}{(z^2+4z+1)^2}$

For Poles,

$z^2 + 4z + 1 = 0 \\ z = \frac{-4 \pm \sqrt{16-4}}{2} \\ = \frac{-4 \pm \sqrt{12}}{2} \\ = \frac{-4 \pm 2\sqrt{3}}{2} \\ = -2 \pm \sqrt{3}$

$\therefore z = -2+\sqrt{3} \hspace{0.20cm} \& \hspace{0.20cm} z = -2-\sqrt{3}$ are the two poles of which $z = -2+\sqrt{3}$ is the only pole which lies inside.

Let $\alpha = -2+\sqrt{3} \\ \beta = -2 -\sqrt{3}$

$\therefore \alpha$ lies inside while $\beta$ lies outside.

$\therefore f(z) = \frac{z}{[(z-\alpha)(z-\beta]^2} = \frac{z}{(z-\alpha)^2(z-\beta)^2}$

$z = \alpha$ is a pole of order '2'

Residue of f(z) at z = $\alpha$ is

$\frac{1}{(2-1!)} \lim_{z \to \alpha} \frac{d}{dz}[(z-\alpha)^2 [\frac{z}{(z-\alpha)^2(z-\beta)^2}]] \\ = \lim_{z \to \alpha} \frac{d}{dz}[\frac{z}{(z-\beta)^2}] \\ = \lim_{z \to \alpha} \frac{(z-\beta)^2(1) - (z)(2)(z-\beta)}{(z-\beta)^4} \\ = \lim_{z \to \alpha} \frac{(z-\beta)[(z-\beta) - 2z]}{(z-\beta)^4} \\ = \lim_{z \to \alpha} \frac{-z-\beta}{(z-\beta)^3} = - \frac{z+\beta}{(z - \beta)^3} \\ = - \frac{\alpha + \beta}{(\alpha - \beta)^3} = - \frac{-4}{(2\sqrt{3})^3} = \frac{4}{8(3)(\sqrt{3})} \\ = \frac{1}{6 \sqrt{3}} \\ \therefore \int_c \frac{z}{(z^2 + 4z + 1)^2} dz = 2 \pi i [\frac{1}{6 \sqrt{3}}] = \pi i[\frac{1}{3 \sqrt{3}}] \\ \therefore \frac{4}{i} \int_c \frac{z}{(z^2 + 4z + 1)^2} dz = \frac{4}{i} [\frac{\pi i}{3 \sqrt{3}}] = \frac{4 \pi}{3 \sqrt{3}}$