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

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 148 views
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modified 4 months ago  • written 7 months ago by gravatar for Manan Bothra Manan Bothra0
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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}} $

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modified 4 months ago  • written 4 months ago by gravatar for Manan Bothra Manan Bothra0
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}} $

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written 4 months ago by gravatar for Manan Bothra Manan Bothra0
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