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Question: Evaluate $\int_{0}^{2 \pi} \frac{d \theta}{3 + 2cos \theta}$
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 77 views
 modified 15 days ago  • written 3 months ago by
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Put $z = e^{i \theta} \\ cos\theta = \frac{z^2 + 1}{2-z} \\ d\theta = \frac{dz}{iz}$

The equation can be written as,

$\int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\ = \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\ = \int_c \frac{dz}{i(z^2+3z+1)} \\ = \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)}$

Let, $f(z) = \frac{1}{z^2+3z+1}$

For poles $z^2+3z+1 = 0$

$z = \frac{-3 \pm \sqrt{9-4}}{2} \\ = \frac{-3 \pm \sqrt{5}}{2} \\ \therefore z = \frac{-3 + \sqrt{5}}{2} \hspace{0.20cm} \& \hspace{0.20cm} \frac{-3 - \sqrt{5}}{2}$

Let, $\alpha = \frac{-3 + \sqrt{5}}{2} \\ \beta = \frac{-3 - \sqrt{5}}{2}$

$z = \alpha$ is the only pole which lies inside the circle

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

$\lim_{z \to \alpha} (z-\alpha)[\frac{1}{(z-\alpha)(z-\beta)}] \\ = \frac{1}{\alpha - \beta} \\ \alpha - \beta = (\frac{-3 + \sqrt{5}}{2}) + (\frac{3 + \sqrt{5}}{2}) = \sqrt{5}$

$\therefore \int_c \frac{1}{z^2+3z+1} dz = 2 \pi i [\frac{1}{\sqrt{5}}] = \frac{2 \pi i}{\sqrt{5}} \\ \therefore \int_c \frac{1}{i}\frac{1}{z^2+3z+1} dz = \frac{1}{i}[\frac{2 \pi i}{\sqrt{5}}] = \frac{2 \pi}{\sqrt{5}}$

 modified 15 days ago  • written 15 days ago by
0

Put $z = e^{i \theta} \\ cos\theta = \frac{z^2 + 1}{2-z} \\ d\theta = \frac{dz}{iz}$

The equation can be written as,

$\int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\ = \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\ = \int_c \frac{dz}{i(z^2+3z+1)} \\ = \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)}$

Let, $f(z) = \frac{1}{z^2+3z+1}$

For poles $z^2+3z+1 = 0$

$z = \frac{-3 \pm \sqrt{9-4}}{2} \\ = \frac{-3 \pm \sqrt{5}}{2} \\ \therefore z = \frac{-3 + \sqrt{5}}{2} \hspace{0.20cm} \& \hspace{0.20cm} \frac{-3 - \sqrt{5}}{2}$

Let, $\alpha = \frac{-3 + \sqrt{5}}{2} \\ \beta = \frac{-3 - \sqrt{5}}{2}$

$z = \alpha$ is the only pole which lies inside the circle

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

$\lim_{z \to \alpha} (z-\alpha)[\frac{1}{(z-\alpha)(z-/beta)}] \\ = \frac{1}{\alpha - \beta} \\ \alpha - \beta = (\frac{-3 + \sqrt{5}}{2}) + (\frac{3 + \sqrt{5}}{2}) = \sqrt{5}$

$\therefore \int_c \frac{1}{z^2+3z+1} dz = 2 \pi i [\frac{1}{\sqrt{5}}] = \frac{2 \pi i}{\sqrt{5}} \\ \therefore \int_c \frac{1}{i}\frac{1}{z^2+3z+1} dz = \frac{1}{i}[\frac{2 \pi i}{\sqrt{5}}] = \frac{2 \pi}{\sqrt{5}}$